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NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals

NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:08 PM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT exemplar Class 12 Maths solutions chapter 8 Application Of Integrals is one of vital chapters of the NCERT Class 12 Maths Solutions. For those students who are planning to score well in their Class 12 Mathematics exam, they should not ignore integrals. Integrals and integration hold a crucial part of the question paper that one will receive in their exam. Not only integrals are important for CBSE board exams, but it is highly important for competitive examinations and higher education. Integrals are one of the base topics in calculus that is helpful in both Maths and physics. Chapter 7 of NCERT Class 12 Maths book is integrals, where students are made aware of what it is and what are its operations. In the NCERT exemplar Class 12 Maths solutions chapter 8, the topic is discussed in detail.

This Story also Contains
  1. The Sub-Topics That Are Covered in Class 12 Maths NCERT Exemplar Solutions Chapter 8 Are:
  2. What can you learn in NCERT Exemplar Solutions for Class 12 Maths Chapter 8?
  3. NCERT Exemplar Class 12 Maths Solutions
  4. Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 8

Question:1

Find the area of the region bounded by the curves y^2 = 9x, y = 3x.

Answer:

It is mentioned in the question that,
y\textsuperscript{2} = 9x belongs to a parabola and y = 3x belong to a straight line which passes through origin.
Starting with a rough figure showing those equations below,
From the parabola equations it is seen that x cannot be negative, so the graph would be on the right of the X-axis. The parabola would thus be opening to the right.

In order to find the points of the two equations mentioned, you have to solve the two equations simultaneously.
\begin{array}{l} \text { Put } y=3 x \text { in } y^{2}=9 x \\ \Rightarrow(3 x)^{2}=9 x \\ \Rightarrow 9 x^{2}=9 x \\ \Rightarrow x^{2}-x=0 \\ \Rightarrow x(x-1)=0 \\ \Rightarrow x=0 \text { and } x=1 \end{array}
We got the coordinates of x, for finding the coordinates of y.
Put x = 1 and x = 0 in the equations of y = 3x.
It is found that y coordinated are y = 3 and y = 0 respectively.
Therefore, the point of interaction of parabola and straight line is (1, 3) and (0, 0)
Now, calculate the area enclosed between the parabola and the straight line.
For calculating the area we have to minus the area under the straight line which extends from x = 0 to x = 1 from the area under the parabola.
It can be written as,
Area between parabola and straight line = area under parabola – area under straight line …. (1)

On calculating the area under parabola,
y\textsuperscript{2 = }9x \\
\int y = 3 \sqrt x\\
On integrating the above equation from 0 to 1
\\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} \sqrt{x} d x \\ \Rightarrow \int_{0}^{1} y d x=3 \int_{0}^{1} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} y d x=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} \\
\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2\left[1^{\frac{3}{2}}-0\right] \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2
Now, for calculating the area under line y = 3x i.e. area of triangle OAB
\int y = 3x \\
On integrating the above equation from 0 to 1
\\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\int_{0}^{1} 3 \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=3\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{3}{2}

Using equation mentioned above (1)
area between parabola and straight line = 2 - \frac{3}{2} = 1/2\ unit\textsuperscript{2 }\\
Therefore,
It is found that the area was 1/2 unit2 between the curves y^2 = 9x\: \: \: and\: \: \: y = 3x.

Question:2

Find the area of the region bounded by the parabola y^2 = 2px, x^2 = 2py

Answer:

The equation of the parabola is y\textsuperscript{2 }= 2px which shows no negative values for x. Therefore, the graph will be on right of Y-axis will be passing through (0, 0)
Similarly, for equations x\textsuperscript{2} = 2py no negative values for y, therefore, the graph will be above the X axis passing through (0, 0).
For finding point of interaction, let us solve the equations simultaneously.
\\ \text { Put } y=\frac{x^{2}}{2 p} \text { in } y^{2}=2 p x \\ \Rightarrow\left(\frac{x^{2}}{2 p}\right)^{2}=2 p x \\ \Rightarrow \frac{x^{4}}{4 p^{2}}=2 p x \\ \Rightarrow x^{4}=8 p^{3} x \\ \Rightarrow x^{3}=8 p^{3} \\
\\ \Rightarrow x=2 p \\ \text { Put } x=2 p \text { in } y^{2}=2 p x \\ \Rightarrow y^{2}=2 p(2 p) \\ \Rightarrow y=2 p
The interaction point was found to be (2p, 2p)

Now, we have to find out the areas between the two parabolas.
The equation can be written as
Area between the two parabola = area under parabola
y\textsuperscript{2 }= 2px - \text{area under the parabola } x\textsuperscript{2 }= 2py $ \ldots $ (1) \\
For finding the area undery\textsuperscript{2 }= 2px\: \: parabola \\

\int y = \sqrt 2p \sqrt x \\

Integrating the equation from 0 to 2p
\\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} \sqrt{x} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \int_{0}^{2 p} x^{\frac{1}{2}} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\sqrt{2 p} \frac{2}{3}\left[(2 p)^{\frac{3}{2}}-0\right]
\\\Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}}(2 p)^{\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{\frac{1}{2}+\frac{3}{2}} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{2}{3}(2 p)^{2} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{8 p^{2}}{3}
For finding the area under x\textsuperscript{2 }= 2py \text{ parabola} \\

\int y = x\textsuperscript{2}/ 2p \\

Integrating from 0 to 2p
\\\Rightarrow \int_{0}^{2 p} y d x=\int_{0}^{2 p} \frac{x^{2}}{2 p} d x \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{x^{3}}{3}\right]_{0}^{2 p} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{1}{2 p}\left[\frac{(2 p)^{3}}{3}\right] \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{(2 p)^{2}}{3} \\ \Rightarrow \int_{0}^{2 p} y d x=\frac{4 p^{2}}{3}
Using
(i)
\\\Rightarrow$ area bounded by two parabolas given $=\frac{8 \mathrm{p}^{2}}{3}-\frac{4 \mathrm{p}^{2}}{3}$ $\\\Rightarrow$ area bounded by two parabolas given $=\frac{4 \mathrm{p}^{2}}{3}$\\ Hence area is $\frac{4 \mathrm{p}^{2}}{3}$ unit ${ }^{2}$

Question:3

Find the area of the region bounded by the curve y = x^3 and y = x + 6 and x = 0.

Answer:

Let’s start with a rough plot of the curve y = x\textsuperscript{3} along with the lines y = x + 6 and x = 0
When x = 0, it means Y axis
Questions says to find the area between curve and the line and Y axis
First solve the y = x + 6 and y = x\textsuperscript{3}, in order to find the interaction point
\\\text { Put } y=x^{3} \text { in } y=x+6 \\ \Rightarrow x^{3}=x+6 \\ \Rightarrow x^{3}-x-6=0 \\ \\

For checking is 0, 1, 2 satisfy this cubic, shows 2 is one factor, therefore x – 2 is a factor.

Solving the equation,

\\ \Rightarrow(x-2)\left(x^{2}+2 x+3\right)=0 \\ \text{Observe that } x^{2}+2 x+3 \text{ don't have real roots} \\
Therefore, x = 2
Substituting this x = 2 in y = x + 6, y = 8
Therefore, curves intersect at (2, 8)


The area bounded will be

item Area bounded = area by y = x\textsuperscript{3} on Y axis – area by y = x + 6 on Y axis \ldots (1) \\





For finding area under y = x\textsuperscript{3} \\


\Rightarrow x=\sqrt[3]{y} \\
Integrate the equation from 0 to 8
\\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\int_{0}^{8} \mathrm{y} \frac{1}{3} \mathrm{dy} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{4}{3}}}{\frac{4}{3}}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\frac{3}{4}\left[\mathrm{y}^{\frac{4}{3}}\right]_{0}^{8} \\ \Rightarrow \int_{0}^{8} \mathrm{xdy}=\frac{3}{4}[8 \overline{4/3}-0] \\ \\ \\\\\Rightarrow \int_{0}^{8} x d y=\frac{3}{4}\left(2^{3}\right)^{\frac{4}{3}} \\ \Rightarrow \int_{0}^{8} x d y=\frac{3}{4} 2^{4} \\ \Rightarrow \int_{0}^{8} x d y=\frac{3}{4} 16 \\ \Rightarrow \int_{0}^{8} x d y=12 \\ \\

Now, finding area under y = x + 6

For finding the area from 6 to 8 because line passes through Y axis at 6 and extends upto 8, the point where curve and line intersects.

\int X = y - 6 \\
Integrating from 6 to 8
\\ \Rightarrow \int_{6}^{8} x d y=\int_{6}^{8}(y-6) d y \\ \Rightarrow \int_{6}^{8} x d y=\left[\frac{y^{2}}{2}-6 y\right]_{6}^{8} \\ \Rightarrow \int_{6}^{8} x d y=\left[\left(\frac{8^{2}}{2}-6(8)\right)-\left(\frac{6^{2}}{2}-6(6)\right)\right] \\ \Rightarrow \int_{6}^{8} x d y=[32-48-18+36] \\ \Rightarrow \int_{6}^{8} x d y=2 \\ \\
Using the equation (1)
Area bound was found to be 12 – 2 = 10 unit\textsuperscript{2 }\\

Therefore, the area was found to be 10 unit\textsuperscript{2 }\\

Question:4

Find the area of the region bounded by the curve y^2 = 4x, x^2 = 4y.

Answer:

The equations y\textsuperscript{2 }= 4x is a parabola and no negative values of x are seen, therefore, this parabola lies to the right of the Y axis passing through (0, 0)

Similarly, for x\textsuperscript{2} = 4y which is a parabola, not defined negative values of y lies above X axis and passing through (0, 0)

For finding point of interaction, solve simultaneously.
\begin{aligned} &\text { Put } y=\frac{x^{2}}{4} \text { in } y^{2}=4 x\\ &\Rightarrow\left(\frac{x^{2}}{4}\right)^{2}=4 x\\ &\Rightarrow \frac{x^{4}}{16}=4 x\\ &\Rightarrow x^{4}=64 x\\ &\Rightarrow x^{3}=64\\ &\Rightarrow x=4\\ &\text { Put } x=4 \text { in } y^{2}=4 x\\ &\Rightarrow y^{2}=4(4)\\ &\Rightarrow y=4 \end{aligned} \\

The point of interaction is therefore (4, 4)
For finding the area between two parabolas
Area between two parabolas = area under parabola y\textsuperscript{2 }= 4x -\text{area under parabola } x\textsuperscript{2} = 4y \ldots . (1) \\


Let us find area under parabola y\textsuperscript{2}= 4x\\

= y = 2 \sqrt{x}

Integrate from 0 to 4



Let us find area under parabola \mathrm{y}^{2}=4 \mathrm{x}
\Rightarrow y=2 \sqrt{x}
Integrate from 0 to 4
\Rightarrow \int_{0}^{4} \mathrm{ydx}=2 \int_{0}^{4} \sqrt{\mathrm{x}} \mathrm{dx} \\ \begin{aligned} &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2 \int_{0}^{4} \mathrm{x} \frac{1}{2} \mathrm{dx}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=2\left[\frac{\mathrm{x} \frac{3}{2}}{\frac{3}{2}}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{4}\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left[4^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}\left(2^{2}\right)^{\frac{3}{2}} \end{aligned} \\


\\ \Rightarrow \int_{0}^{4} y d x=\frac{4}{3}(2)^{3} \\ \Rightarrow \int_{0}^{4} y d x=\frac{32}{3} \\ \\
To find area under parabola x\textsuperscript{2}= 4y\\

= x\textsuperscript{2}= 4 y\\

\\\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}(2)^{3}\\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{32}{3}\\ \text{Now let us find area under parabola} x^{2}=4 y\\ \Rightarrow x^{2}=4 y\\ \Rightarrow \mathrm{y}=\frac{\mathrm{x}^{2}}{4} \\ \text{Integrate from 0 to 4}\\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\int_{0}^{4} \frac{\mathrm{x}^{2}}{4} \mathrm{~d} \mathrm{x} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{0}^{4} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{1}{4}\left[\frac{4^{3}}{3}\right] \\ \Rightarrow \int_{0}^{4} y d x=\frac{4^{2}}{3} \\ \Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{16}{3} \\ \text{Using (i)} \\ \Rightarrow \text{area bounded by two parabolas given} =\frac{32}{3}-\frac{16}{3} \\ \Rightarrow \text{area bounded by two parabolas given} =\frac{16}{3} \\
Therefore, the area is found to be

Question:5

Find the area of the region included between y^2 = 9x and y = x

Answer:

The equations y\textsuperscript{2} = 9x has no negative values of x. Therefore it lies on right of Y axis passing through (0, 0).

And y = x depicts a straight line through origin.
For finding the area, picture shown below.





For finding the point of interaction solve the two equations simultaneously.
\\Put \: \: y=x \text{ in } y^{2}=9 x \Rightarrow x^{2}=9 x \Rightarrow x=9\\Put\: \: x=9 in y=x \text{ we get } y=9 \\

Therefore, point of interaction is (9, 9)

The area between the parabola and line = area under parabola – area under line \ldots (1) \\

Let us find area under parabola
\\\Rightarrow y^{2}=9 x \\ \Rightarrow y=3 \sqrt{x} \\ \text{Integrate from 0 to 9}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} 3 \mathrm{x}^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=3\left[\frac{\mathrm{x} \frac{1}{2}+1}{\frac{1}{2}+1}\right]_{0}^{9}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ \\
\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=3 \frac{2}{3}\left[9 \frac{3}{2}-0\right]\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{2}\right)^{\frac{3}{2}}\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=2\left(3^{3}\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=54\\
Now let us find area under straight line \mathrm{y}=\mathrm{x}
y=x
Integrate from 0 to 9
\\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\int_{0}^{9} \mathrm{xdx} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{9} \\\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left(\frac{9^{2}}{2}-0\right)\\ \Rightarrow \int_{0}^{9} \mathrm{ydx}=\frac{81}{2}\\ \Rightarrow \int_{0}^{9} y \mathrm{dx}=40.5
Using (i)
\Rightarrow area between parabola and line =54-40.5=13.5 unit ^{2}
Therefore, area was found to be 13.5 unit\textsuperscript{2}. \\

Question:6

Find the area of the region enclosed by the parabola x^2 = y and the line y = x + 2

Answer:

The equation x\textsuperscript{2} = y depicts a parabola and has no negative values for y, therefore, it lies above X axis and passing through origin (0, 0).

And y = x + 2 is a straight line.

Below figure shows the area to be found out,

Let’s find the point of interaction of the two equations.
\\\text{Put } y=x+2 in x^{2}=y\\ \Rightarrow x^{2}=x+2\\ \Rightarrow x^{2}-x-2=0\\ \Rightarrow x^{2}-2 x+x-2=0\\ \Rightarrow x(x-2)+1(x-2)=0\\ \Rightarrow(x+1)(x-2)=0\\ \Rightarrow x=-1 \text{ and } x=2\\ \text{Put } x=-1 \text{ and } x=2 in x^{2}=y \text{ we get } y=1 \text{ and } y=4 respectively \\
The point of interaction was found to be (-1, 1) and (2, 4)

Area between the line and parabola = area under line – area under parabola


Let us find area under line y=x+2
\\ \Rightarrow y=x+2\\ \text{Integrate from -1 to 2} \Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}(x+2) d x\\ \Rightarrow \int_{-1}^{2} y \mathrm{~d} x=\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}\\ \Rightarrow \int_{-1}^{2} y d x=\left[\left(\frac{2^{2}}{2}+2(2)\right)-\left(\frac{(-1)^{2}}{2}+2(-1)\right)\right]\\ \Rightarrow \int_{-1}^{2} y d x=6-\left(\frac{1}{2}-2\right) \\ \Rightarrow \int_{-1}^{2} \mathrm{ydx}=6-\left(\frac{-3}{2}\right)\\ \Rightarrow \int_{-1}^{2} y \mathrm{dx}=6+\frac{3}{2}\\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=\frac{15}{2}\\
Now let us find area under the parabola
\\ x^{2}=y \Rightarrow y=x^{2}\\ \text{Integrate from -1 to 2 }\\ \Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2} x^{2} d x \\
\\\begin{aligned} &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=3\\ &\text { Using (i) }\\ &\text { area enclosed by line and parabola }=\frac{15}{2}-3=\frac{9}{2} \text { unit }^{2} \end{aligned} \\
Therefore, the area was found to be

Question:7

Find the area of region bounded by the line x = 2 and the parabola y^2 = 8x

Answer:

The equation y\textsuperscript{2} = 8x is a parabola not defining negative values of x, therefore it lies to the right of the Y axis passing through origin.
And x = 2 is a straight line parallel to Y axis.
Below figure shows the area to be calculated.

We have to integrate y\textsuperscript{2} = 8x \\

\\Y = 2 \sqrt 2 \sqrt x \text{ from 0 to 2} \\

On integrating the above equation from 0 to 2 it would give area enclosed under quadrant 1 only. Therefore, for finding the area in quadrant 2 we have to multiply it by 2 since it is symmetrical.
Therefore the area ODBC = 2 x area OBC \ldots .(1) \\
let us find area under parabola
\\ \Rightarrow y^{2}=8 x\\ \Rightarrow y=2 \sqrt{2} \sqrt{x}\\ \text{Integrate from 0 to 2}\\ \Rightarrow \int_{0}^{2} y d x=\int_{0}^{2} 2 \sqrt{2} x^{\frac{1}{2}} d x\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{ \frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2}\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\

\\\begin{aligned} &\Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2} \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{2}\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{\frac{1}{2}}\right)\left(2^{\frac{3}{2}}-0\right)\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{\frac{1}{2}+\frac{3}{2}}\right)\\ &\Rightarrow \int_{0}^{2} y d x=\frac{4}{3}\left(2^{2}\right)\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{16}{3}\\ &\Rightarrow \operatorname{areaOBC}=\frac{16}{3}\\ &\text { Using (i) } \end{aligned} \\

Question:8

Sketch the region and x-axis. Find the area of the region using integration.

Answer:

\begin{aligned} &y=\sqrt{4-x^{2}}\\ &\text { Square both sides }\\ &\Rightarrow y^{2}=4-x^{2}\\ &\Rightarrow x^{2}+y^{2}=4\\ &\Rightarrow x^{2}+y^{2}=2^{2} \end{aligned} \\
The above equation depicts a circle with origin and radius 2
Now in \mathrm{y} =\sqrt{4-\mathrm{x}^{2}} \ \ -2 \leq \mathrm{x} \leq 2 \ and \ \mathrm{y} \geq 0 \\ \text{ which means }\mathrm{x} \: \ and \: \ \mathrm{y} \text{ both positive or } \mathrm{x} \text{ negative and } \mathrm{y} \text{ positive} \\ \text{ hence the curve } \mathrm{y} = \sqrt{4-\mathrm{x}^{2}} \text{has to be above } \mathrm{X} -axis \ in \ 1^{\mathrm{st} } and \: \: 2^{\text {nd }} quadrant\text{Hence the graph of y}=\sqrt{4-x^2} \text{will be graph of circle } \mathrm{x}^{2}+\mathrm{y}^{2}=2^{2} \text{lying only above X -axis }
Equation of X axis is y = 0
For point of interaction,
Y = 0
\\ \int X \textsuperscript{2} = 4 \\ \int X = \pm 2 \\
Point of interaction are (-2, 0) and (2, 0)
The below figure shows the area

Now let us find the area
y=\sqrt{4-x^{2}}
Integrate from -2 to 2
\Rightarrow \int_{-2}^{2} \mathrm{ydx}=\int_{-2}^{2} \sqrt{4-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x} \\
\\\begin{aligned} &\text { Using uv rule of integration where u and } v \text { are functions of } x\\ &\int_{a}^{b} u v d x=\left[u \int \operatorname{vdx}\right]_{a}^{b}-\int_{a}^{b}\left(u^{\prime} \int v d x\right) d x\\ &\text { Here } u=\sqrt{4-x^{2}} \text { and } v=1\\ &\text { Hence } u^{\prime}=\frac{1}{2}\left(4-x^{2}\right)^{\frac{1}{2}-1}(-2 x)=\frac{-x}{\sqrt{4-x^{2}}}\\ &\Rightarrow \int_{-2}^{2} y d x=\left[\sqrt{4-x^{2}} \int 1 d x\right]_{-2}^{2}-\int_{-2}^{2}\left(\frac{-x}{\sqrt{4-x^{2}}}\right)(1 d x) d x\\ &\Rightarrow \int_{-2}^{2} y d x=\left[\sqrt{4-x^{2}}(x)\right]_{-2}^{2}-\int_{-2}^{2}\left(\frac{-x^{2}}{\sqrt{4-x^{2}}}\right) d x\\ &\Rightarrow \int_{-2}^{2} \mathrm{ydx}=\left(\sqrt{4-2^{2}}(2)\right)-\left(\sqrt{4-(-2)^{2}}(-2)\right)-\int_{-2}^{2}\left(\frac{4-\mathrm{x}^{2}-4}{\sqrt{4-\mathrm{x}^{2}}}\right) \mathrm{d} \mathrm{x} \end{aligned}
\\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2}\left(\frac{4-x^{2}}{\sqrt{4-x^{2}}}-\frac{4}{\sqrt{4-x^{2}}}\right) d x \\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2} \sqrt{4-x^{2}} d x+\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \text { But } y=\sqrt{4-x^{2}} \\ \Rightarrow \int_{-2}^{2} y d x=-\int_{-2}^{2} y d x+\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \Rightarrow \int_{-2}^{2} y d x+\int_{-2}^{2} y d x=\int_{-2}^{2} \frac{4}{\sqrt{4-x^{2}}} d x \\ \Rightarrow \int_{-2}^{2} y d x=2 \int_{-2}^{2} \frac{1}{\sqrt{2^{2}-x^{2}}} d x
\\\begin{aligned} &\text { We know that } \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1} \frac{x}{a}\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left[\sin ^{-1} \frac{\mathrm{x}}{2}\right]_{-2}^{2}\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\sin ^{-1} \frac{2}{2}-\sin ^{-1} \frac{-2}{2}\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\\ &\Rightarrow \int_{-2}^{2} \mathrm{y} \mathrm{dx}=2 \pi\\ &\text { Hence area is } 2 \pi \text { unit }^{2} \end{aligned}

Question:9

Calculate the area under the curve y=2\sqrt x included between the lines x = 0 and x = 1.

Answer:

\\$y=2 \sqrt{x}$\\ \text{squaring both sides we get}\\ $\Rightarrow y^{2}=4 x$\\ $\mathrm{y}^{2}=4 \mathrm{x}$ \text{ is a equation of parabola }\\\text{The equation show no negative values for x, therefore it lies to the right of Y axis passing through origin. }\\ \text{Now for} y=2 \sqrt x \text{ x and y both has to be greater than 0 that is both positive hence both lie in}\\ $1^{\text {st }}$ quadrant
\\ \text{Hence} \ y=2 \sqrt{x}$ \text{ will be parabolic curve of } $y^{2}=4 x \ $ only in $ 1^{\text {st }}$ \text{quadrant}\\ $\mathrm{x}=0$ \text{is equation of} $\mathrm{Y}$ -axis and \\$\mathrm{x}=1$\text{ is a line parallel to} $\mathrm{Y}$ \text{-axis passing through (1,0)}\\ \text{Plot equations} $y=2 \sqrt{x}$ and $x=1$

\\\text{So we have to integrate} y=2 \sqrt{x}$ from 0 to 1\\ \text{let us find area under parabola}
\\$\Rightarrow y=2 \sqrt{x}$
\text{Integrate from 0 to 1}
\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} 2 \mathrm{x}^{\frac{1}{2}} \mathrm{dx}
\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}
\\\begin{aligned} &\Rightarrow \int_{0}^{1} \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=2 \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\frac{4}{3}\left(1^{\frac{3}{2}}-0\right)\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{4}{3}\\ &\text { Therefore, the area found to be }\\ &\frac{4}{3} \text { unit }^{2} \end{aligned}

Question:10

Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8.

Answer:

The equation of the line is 2y = 5x + 7
Now, we need point of the line, so we substitute x = 0 and then y = 0
\\Put \: \: x=0 \Rightarrow 2 y=5(0)+7\\ \Rightarrow \mathrm{y}=\frac{7}{2}\\ Put\: \: y=0\\ \\\Rightarrow 2(0)=5 x+7 \\\Rightarrow 5 x=-7 \\\Rightarrow x=-\frac{7}{5}
\\Hence \left(0, \frac{7}{2}\right) and \left(-\frac{7}{5}, 0\right)$ are the required two points to draw the line $2 \mathrm{y}=5 \mathrm{x}+7$
The other two x = 2 and x = 8 are straight lines parallel to Y axis.
On plotting all the above lines.

\\\text{We have to find area under} 2 \mathrm{y}=5 \mathrm{x}+7$\\ that is $\mathrm{y}=1 / 2(5 \mathrm{x}+7)$ from 2 to 8\\ $\Rightarrow y=1 / 2(5 x+7)$\\ Integrate from 2 to 8\\ $\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2} \int_{2}^{8}(5 \mathrm{x}+7) \mathrm{dx}$ $\\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}\left[\frac{5 \mathrm{x}^{2}}{2}+7 \mathrm{x}\right]_{2}^{8}$ $\\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}\left[\left(\frac{5(8)^{2}}{2}+7(8)\right)-\left(\frac{5(2)^{2}}{2}+7(2)\right)\right]$ $\\\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}[(160+56)-(10+14)]
\\\begin{aligned} &\Rightarrow \int_{2}^{8} \mathrm{ydx}=\frac{1}{2}(192)\\ &\Rightarrow \int_{2}^{8} \mathrm{ydx}=96\\ &\text { Hence the area bounded by given lines is } 96 \text { unit }^{2} \end{aligned}

Question:11

Draw a rough sketch of the curve y=\sqrt{x-1} in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.

Answer:

\\ \mathrm{y}=\sqrt{\mathrm{x}-1} \\
Squaring both sides
\Rightarrow y^{2}=x-1 \\ \mathrm{y}^{2}=\mathrm{x}-1\text{ is equation of a parabola} \\

The above equation has no values for x less than 1, therefore parabola will be right of x = 1

Now observe that in \mathrm{y}=\sqrt{\mathrm{x}-1} \mathrm{x} \geq 1 \: \: and\: \: \mathrm{y} has to positive because of square root hence \mathrm{x} \ and\ \mathrm{y} both positive hence the parabola will be drawn only in 1^{\text {st }} quadrant

We have to plot the curve in [1,5] so just draw the parabolic curve from x=1 to x=5 in 1^{\text {st }} quadrant
x=1 and x=5 are lines parallel to Y-axis



So we have to integrate \mathrm{y}=\sqrt{\mathrm{x}-1} from 1 to 5
let us find area under parabolic curve
\Rightarrow y=\sqrt{x-1}
Integrate from 1 to 5
\Rightarrow \int_{1}^{5} y \mathrm{dx}=\int_{1}^{5}(\mathrm{x}-1)^{\frac{1}{2}} \mathrm{dx}\\ \Rightarrow \int_{1}^{5} y d x=\left[\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{1}^{5}\\ \Rightarrow \int_{1}^{5} \mathrm{ydx}=\left[\frac{(\mathrm{x}-1)^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{5} \\

\\\begin{array}{l} \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left[(x-1)^{\frac{3}{2}}\right]_{1}^{5} \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left((5-1)^{\frac{3}{2}}-0\right) \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3}\left(2^{2}\right)^{\frac{3}{2}} \\ \Rightarrow \int_{1}^{5} y d x=\frac{2}{3} \times 2^{3} \\ \Rightarrow \quad \int_{1}^{5} y d x=\frac{16}{3} \\ \text { Hence area bounded }=\frac{16}{3} \text { unit }^{2} \end{array} \\

Question:12

Determine the area under the curve y=\sqrt{a^2-x^2} included between the lines x = 0 and x = a

Answer:

\begin{aligned} &y=\sqrt{a^{2}-x^{2}}\\ &\text { Squaring both sides }\\ &\Rightarrow y^{2}=a^{2}-x^{2}\\ &\Rightarrow x^{2}+y^{2}=a^{2} \end{aligned} \\
The above equation is of a circle having centre as (0, 0) and radius a
Now in \mathrm{y}=\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} -a \leq \mathrm{x} \leq \mathrm{a} and \mathrm{y} \geq 0 which means x and y both positive or x negative and y positive hence the curve y=\sqrt{a^{2}-x^{2}} has to be above X -axis in 1^{s t} and 2^{\text {nd }} quadrant
x=0 is equation of Y -axis and x=a is a line parallel to Y-axis passing through (a, 0)




So we have to integrate y=\sqrt{a^2-x^2} from 0 to a


So we have to integrate y=\sqrt{a^2-x^2} from 0 to a
let us find area under curve
\\ y=\sqrt{a^{2}-x^{2}} \\
Integrate from 0 to a
\Rightarrow \int_{0}^{a} \mathrm{ydx}=\int_{0}^{a} \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{dx}
Using uv rule of integration where u and v are functions of
\int_{a}^{b} u v d x=\left[u \int v d x\right]_{a}^{b}-\int_{a}^{b}\left(u^{\prime} \int v d x\right) d x
Here u=\sqrt{a^{2}-x^{2}} and v=1
Hence
\mathrm{u}^{\prime}=\frac{1}{2}\left(\mathrm{a}^{2}-\mathrm{x}^{2}\right)^{\frac{1}{2}-1}(-2 \mathrm{x})=\frac{-\mathrm{x}}{\sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}
\\ \begin{array}{l} \Rightarrow \int_{0}^{a} y d x=\left[\sqrt{a^{2}-x^{2}} \int_{0}^{a} d x\right]_{0}^{a}-\int_{0}^{a}\left(\frac{-x}{\sqrt{a^{2}-x^{2}}} \int_{0}^{a} d x\right) d x \\ \Rightarrow \int_{0}^{a} y d x=\left[\sqrt{a^{2}-x^{2}}(x)\right]_{0}^{a}-\int_{0}^{a}\left(\frac{-x^{2}}{\sqrt{a^{2}-x^{2}}}\right) \mathrm{d} x \\ \Rightarrow \int_{0}^{a} y d x=\left(\sqrt{a^{2}-a^{2}}(a)\right)-\left(\sqrt{a^{2}-0^{2}}(0)\right)-\int_{0}^{a}\left(\frac{a^{2}-x^{2}-a^{2}}{\sqrt{a^{2}-x^{2}}}\right) \mathrm{d} x \\ \Rightarrow \int_{0}^{a} y d x=-\int_{0}^{a}\left(\frac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}-\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}\right) d x \\ \end{array}
\\ \begin{array}{l} \Rightarrow \int_{0}^{a} y d x=-\int_{0}^{a} \sqrt{a^{2}-x^{2}} d x+\int_{0}^{a} \frac{a^{2}}{\sqrt{a^{2}-x^{2}}} d x \\ \text { But } y=\sqrt{a^{2}-x^{2}} \\ \Rightarrow \int_{0}^{a} y d x=-\int_{0}^{a} y d x+\int_{0}^{a} \frac{a^{2}}{\sqrt{a^{2}-x^{2}}} d x \end{array}
\begin{aligned} &\Rightarrow \int_{0}^{a} y d x+\int_{0}^{a} y d x=\int_{0}^{a} \frac{a^{2}}{\sqrt{a^{2}-x^{2}}} d x\\ &\Rightarrow 2 \int_{0}^{a} y d x=a^{2} \int_{0}^{a} \frac{1}{\sqrt{a^{2}-x^{2}}} d x\\ &\Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2} \int_{0}^{a} \frac{1}{\sqrt{a^{2}-x^{2}}} d x\\ \end{aligned} \\
\text{We know that } \int_{0}^{a} \frac{1}{\sqrt{a^{2}-x^{2}}} d x = \sin ^{-1} \frac{x}{a}
\\ \Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2}\left[\sin ^{-1} \frac{x}{a}\right]_{0}^{a}\\ \Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2}\left(\sin ^{-1} \frac{a}{a}-\sin ^{-1} \frac{0}{2}\right) \\
\begin{array}{l} \Rightarrow \int_{0}^{a} y d x=\frac{a^{2}}{2}\left(\frac{\pi}{2}-0\right) \\ \Rightarrow \int_{0}^{a} y d x=\frac{\pi a^{2}}{4} \\ \text { Hence area bounded }=\frac{\pi a^{2}}{4} \text { unit }^{2} \end{array} \\ \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip} \vspace{\baselineskip}

Question:13

Find the area of the region y=\sqrt x bounded by and y = x.

Answer:

\begin{aligned} &y=\sqrt{x}\\ &\text { squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\
This is a parabola, no negative values of x, therefore it lies on the right of Y axis passing through origin.
Now y=\sqrt{x} means y and x both has to be positive hence both lie in 1^{\text {st }} quadrant hence y=\sqrt{x} will be part of \mathrm{y}^{2}=\mathrm{x} which is lying only in 1^{\text {st }} quadrant
And y=x is a straight line passing through origin
We have to find area between y=\sqrt{x} and y=x shown below

For finding the point of interaction, solve two equations simultaneously.
\\Put y=x in y^{2}=x \\ \Rightarrow x^{2}=x \\ \Rightarrow x=1 \\
Put x=1 in y=x we get y=1
The point of interaction is (1, 1)
Area between the parabolic curve and line = area under parabolic curve – area under line \ldots (1) \\


For the area under parabolic curve

\int Y = \sqrt x \\
Integrating from 0 to 1
\begin{aligned} &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{x}^{\frac{1}{2}} \mathrm{~d} \mathrm{x}\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}\\ &\Rightarrow \int_{0}^{1} \mathrm{y} \mathrm{dx}=\frac{2}{3}\left[1^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{1} \mathrm{ydx}=\frac{2}{3} \end{aligned} \\

For area under straight line y = x
On integrating from 0 to 1
\Rightarrow \int_{0}^{1} \mathrm{ydx}=\int_{0}^{1} \mathrm{xdx} \\ \Rightarrow \int_{0}^{1} y d x=\left[\frac{x^{2}}{2}\right]_{0}^{1} \\ \Rightarrow \int_{0}^{1} \mathrm{ydx}=\left(\frac{1^{2}}{2}-0\right) \\ \Rightarrow \int_{0}^{1} y d x=\frac{1}{2} \\
Using (i)
\Rightarrow \text{area between parabolic } = \frac{2}{3}-\frac{1}{2}= \frac{1}{6} unit ^{2} \\
Hence area bounded is \frac{1}{6} unit ^{2} \\ \\

Question:14

Find the area enclosed by the curve y = -x^2 and the straight lilne x + y + 2 = 0.

Answer:

\\Y = -x\textsuperscript{2 }\\ X\textsuperscript{2 }= -y \\
This equation depicts a parabola defining no positive values of y therefore it lies below X axis and passes through origin \\
X + y + 2 = 0 depicts a straight line
For the point of interaction, both of the equations are solved simultaneously.
\begin{array}{l} \text { Put } y=-(x+2) \text { in } x^{2}=-y \\ \Rightarrow x^{2}=-(-(x+2)) \\ \Rightarrow x^{2}=x+2 \\ \Rightarrow x^{2}-x-2=0 \\ \Rightarrow(x+1)(x-2)=0 \\ \Rightarrow x^{2}-2 x+x-2=0 \\ \Rightarrow x(x-2)+1(x-2)=0 \\ \Rightarrow x=-1 \text { and } x=2 \\\end{array} \\

\begin{array}{l} \Rightarrow y=-1 \\ \text { Put } x=2 \text { in } x^{2}=-y \\ \Rightarrow 2^{2}=-y \\ \Rightarrow y=-4 \end{array} \\
Therefore, the point of interaction is (-1, -1) and (2, -4)
Below figure shows the area to be calculated


Area between the line and parabola = Area under line – area under parabola \ldots (1) \\


For finding area under line, integrate it from -1 to 2
Y = -(x + 2)
\begin{array}{l} \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=\int_{-1}^{2}-(\mathrm{x}+2) \mathrm{d} \mathrm{x} \\ \Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2} \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{dx}=-\left[\left(\frac{2^{2}}{2}+2(2)\right)-\left(\frac{-1^{2}}{2}+2(-1)\right)\right] \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=-\left[6-\left(\frac{1}{2}-2\right)\right] \\ \Rightarrow \int_{-1}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=-\frac{15}{2} \end{array} \\
For the area under parabola, integrate it from -1 to 2
Y = -x\textsuperscript{2}\\
\begin{aligned} &\text { Integrate from }-1 \text { to } 2\\ &\Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}-x^{2} d x\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-3 \end{aligned} \\
Substituting both values in (1)
area enclosed by line and parabola =-\frac{15}{2}-(-3)=-\frac{9}{2} \text { unit }^{2}
The negative sign depicts the area is below the X axis
Hence the area enclosed is \frac{9}{2} \text { unit }^{2}

Question:15

Find the area bounded by the curve y=\sqrt x, x = 2y + 3 in the first quadrant and x-axis.

Answer:

\begin{aligned} &y=\sqrt{x}\\ &\text { Squaring both sides }\\ &\Rightarrow y^{2}=x \end{aligned} \\
The above equation depicts a parabola which does not define negative values for x and therefore it lies to the right of the Y axis and passes through origin.
For plotting y = \sqrt x, both the values of x and y have to be positive hence this graph lies in the 1st quadrant only.
The point cannot be negative because the square root symbol itself depicts the positive root.
The equation x = 2y + 3 is a straight line.
For finding the point of interaction, solve the two equations simultaneously.
\\ \text { Put } x=2 y+3 \text { in } y^{2}=x \\ \Rightarrow y^{2}=2 y+3 \\ \Rightarrow y^{2}-2 y-3=0 \\ \Rightarrow y^{2}-3 y+y-3=0 \\ \Rightarrow y(y-3)+1(y-3)=0 \\ \Rightarrow(y+1)(y-3)=0 \\ \Rightarrow y=-1 \text { and } y=3 \\ \text { Put } y=3 \text { in } y^{2}=x \\ \Rightarrow x=3^{2} \\ \Rightarrow x=9 \\
Y = -1 won’t make a difference since we are looking for the 1st quadrant only.
Therefore, the point of interaction are (9, 3)
Below is the figure shows the area to be calculated

Point of interaction of straight line on X axis can be calculated by substituting y = 0 in the equation of the line
\int X = 2(0) + 3 => x = 3

The area enclosed = area under y = \sqrt x – area under the straight line \ldots (1) \\


For the area under y = \sqrt x integrate the equation from 0 to 9
\begin{aligned} &\Rightarrow \int_{0}^{9} y d x=\int_{0}^{9} x^{\frac{1}{2}} d x\\ &\Rightarrow \int_{0}^{9} y d x=\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{9}\\ &\Rightarrow \int_{0}^{9} \mathrm{ydx}=\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{9}\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3}\left[9^{\frac{3}{2}}-0\right]\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3}\left[\left(3^{2}\right)^{\frac{3}{2}}\right]\\ &\Rightarrow \int_{0}^{9} y d x=\frac{2}{3} \times 3^{3} \end{aligned} \\
For the area under straight line
Y = (x-3) / 2

Integrating the equation from 3 to 9

\\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2} \int_{3}^{9}(x-3) d x \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left[\frac{x^{2}}{2}-3 x\right]_{3}^{9} \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left(\left(\frac{9^{2}}{2}-3(9)\right)-\left(\frac{3^{2}}{2}-3(3)\right)\right) \\ \Rightarrow \int_{3}^{9} y d x=\frac{1}{2}\left(\left(\frac{81}{2}-27\right)-\left(\frac{9}{2}-9\right)\right) \\ \Rightarrow \int_{0}^{9} y d x=\frac{1}{2} \times 18
\Rightarrow \int_{0}^{9} \mathrm{ydx}=9
Using (i)
\Rightarrow area bounded =18-9=9 unit ^{2}
Hence area bounded by the curve and a straight line is 9 unit ^{2}

Question:16

Find the area of the region bounded by the curve y^2 = 2x and x^2 + y^2 = 4x.

Answer:

y\textsuperscript{2} = 2x depicts a parabola having no negative values for x and lying to the right of the Y axis with passing through origin.\\

The other equation of x\textsuperscript{2} + y\textsuperscript{2} = 4x depicts a circle.
The general equation of circle is given by $x^{2}+y^{2}+2 g x+2 f y+c=0$
Centre of circle is $(-\mathrm{g},-\mathrm{f})$ \text{and radius is }$\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}
In $x^{2}+y^{2}-4 x=0,2 g=-4 \Rightarrow g=-2$ and $f=c=0$
Hence center is $(-(-2), 0)$ that is (2,0) and radius is $\sqrt{(-2)^{2}+0^{2}-0}$ which is 2
For the point of interaction, solve the two equations simultaneously.
\\ \text { Put } y^{2}=2 x \text { in } x^{2}+y^{2}=4 x \\ \Rightarrow x^{2}+2 x=4 x \\ \Rightarrow x^{2}-2 x=0 \\ \Rightarrow x(x-2)=0 \\ \Rightarrow x=0 \text { and } x=2 \\ \text { Put } x=2 \text { in } y^{2}=2 x \\ \Rightarrow y=\pm 2 \\

Therefore, the point of interaction have been found to be (0, 0), (2, 2), and (2, -2)
Below is the diagram of the area to be calculated.


On integrating we will get area for 1st quadrant only, but since it is symmetrical we can multiple it by 2.

Area of shaded region = area under the circle – area under the parabola \ldots (1) \\

For finding the area under circle
\\ x^{2}+y^{2}=4 x \\ \Rightarrow y^{2}=4 x-x^{2} \\ \Rightarrow y=\sqrt{4 x-x^{2}} \\ \Rightarrow y=\sqrt{-\left(-4 x+x^{2}+4-4\right)} \\ \Rightarrow y=\sqrt{-\left(\left(x^{2}-4 x+4\right)-2^{2}\right)} \\ \Rightarrow y=\sqrt{-\left((x-2)^{2}-2^{2}\right)} \\ \Rightarrow y=\sqrt{2^{2}-(x-2)^{2}} \\

Integrate the above equation from 0 to 2
\\ \Rightarrow \int_{0}^{2} y d x=\int_{0}^{2} \sqrt{2^{2}-(x-2)^{2}} d x \\ \text { Using } \int \sqrt{a^{2}-x^{2}}=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a} \\ \Rightarrow \int_{0}^{2} y d x=\left[\frac{x-2}{2} \sqrt{2^{2}-(x-2)^{2}}+\frac{2^{2}}{2} \sin ^{-1} \frac{x-2}{2}\right]_{0}^{2} \\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=\left[0-\left(\frac{0-2}{2} \sqrt{2^{2}-(0-2)^{2}}+\frac{2^{2}}{2} \sin ^{-1} \frac{0-2}{2}\right)\right]\\ \Rightarrow \int_{0}^{2} y d x=\left[-\left(0+2 \sin ^{-1}(-1)\right)\right] \\ \Rightarrow \int_{0}^{2} y d x=\pi \\

For the area under parabola,
\\ \Rightarrow y^{2}=2 x \\ \Rightarrow y=\sqrt{2} \sqrt{x} \\
Integrate the above equation from 0 to 2
\\ \Rightarrow \int_{0}^{2} y d x=\sqrt{2} \int_{0}^{2} x^{\frac{1}{2}} d x \\ \Rightarrow \quad \int_{0}^{2} y d x=\sqrt{2}\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2} \\ \Rightarrow \int_{0}^{2} y d x=2^{\frac{1}{2}}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\
\begin{aligned} &\Rightarrow \int_{0}^{2} \mathrm{ydx}=2^{\frac{1}{2}} \times \frac{2}{3}\left[\mathrm{x}^{\frac{3}{2}}\right]_{0}^{2}\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{2^{\frac{1}{2}+1}}{3}\left(2^{\frac{3}{2}}-0\right)\\ &\Rightarrow \int_{0}^{2} \mathrm{y} \mathrm{d} \mathrm{x}=\frac{2^{\frac{3}{2}+\frac{3}{2}}}{3}\\ &\Rightarrow \int_{0}^{2} \mathrm{y} \mathrm{dx}=\frac{2^{3}}{3}\\ &\Rightarrow \int_{0}^{2} \mathrm{ydx}=\frac{8}{3}\\ &\text { Using (i) } \end{aligned} \\
\Rightarrow$ area of shaded in $1^{\text {st }}$ quadrant $=\pi-\frac{8}{3}$ unit $^{2}
After multiplying it by 2
\text { The area required of shaded region }=2\left(\pi-\frac{8}{3}\right) \text { unit }^{2}

Question:17

Find the area bounded by the curve y = sinx between x = 0 and x = 2 \pi

Answer:

Below is the graph with the shaded region whose area has to be calculated

From the plot, it is seen that the area from 0 to \pi is positive whereas the area from \pi to 2\pi is negative.
But both the areas are equal in magnitude with opposite signs.
If we integrate them, the areas will cancel each other and the answer will be 0.

Therefore, we integrate the two areas separately and split the limits in two ways –

1) Find area under sinx from 0 to \pi and multiple by 2
2)Split the limit 0 to 2 \pi into 0 to \pi and \pi to 2 \pi
Here we will solve by splitting the limits
Y = sinx \\
\begin{aligned} &\text { Integrating from } 0 \text { to } 2 \pi\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\text { Because } \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x} \text { where } c \in(\mathrm{a}, \mathrm{b}) \end{aligned} \\
Now the limits become negative
\begin{aligned} &\begin{array}{l} \text { Hence for } \pi \text { to } 2 \pi \sin x \text { will become }-\sin x \end{array}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{y} \mathrm{dx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi}-\sin \mathrm{x} \mathrm{d} \mathrm{x}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{xdx}-\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=[-\cos \mathrm{x}]_{0}^{\pi}-[-\cos \mathrm{x}]_{\pi}^{2 \pi}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=(-(\cos \pi-\cos 0))-(-(\cos 2 \pi-\cos \pi))\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=(-((-1)-1))-(-(1-(-1))) \end{aligned} \\
\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=4 $$
Hence area bounded by sinx from 0 to 2 \pi \text{ is }4 unit $^{2}$ \\

Question:18

Find the area of region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration.

Answer:

The vertices of the triangle are given as (-1, 1), (0, 5) and (3, 2)
The three vertices are given alphabets as P, Q and R respectively.

\\\Rightarrow \text{Equation of } \mathrm{PQ } \: \: is \: \: y-1=\frac{5-1}{0+1}(x+1)\\
\\\therefore y=4 x+5$ $\Rightarrow$ Equation of $Q R$ is $y-5=\frac{2-5}{3-0}(x-0)$ $\therefore y=-x+5$ $\Rightarrow$ Equation of $\mathrm{RP}$ is $y-2=\frac{1-2}{-1-3}(x-3)$ $\therefore 4 y=x+5$ \\
Area of the region bounded by the curve $y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by $A=\int_{a}^{b} f(x) d x$ or $\int_{a}^{b} y d x$
Required area
=\int_{-1}^{0}(4 x+5) d x+\int_{0}^{3}(-x+5) d x-\int_{-1}^{3}\left(\frac{x}{4}+\frac{5}{4}\right) d x$
\\=\left[2 x^{2}+5 x\right]_{-1}^{0}+\left[-\frac{x^{2}}{2}+5 x\right]_{0}^{3}-\left[\frac{x^{2}}{8}+\frac{5 x}{4}\right]_{-1}^{3}\\=((0+0)-(2-5))+\left(\left(-\frac{9}{2}+15\right)-(0+0)\right)-\left(\left(\frac{9}{8}+\frac{15}{4}\right)-\left(\frac{1}{8}-\frac{5}{4}\right)\right)\\=3+\frac{21}{2}-\frac{39}{8}-\frac{9}{8}$ \\
=\frac{15}{2}\text{sq.units}

Question:19

Draw a rough sketch of the region (x, y) : y^2 \leq 6ax\: \: and \: \: x^2 + y^2 \leq 16a^2. Also find the area of the region sketched using method of integration.

Answer:

\text { The region }\left\{(x, y): y^{2} \leq 6 a x \text { and } x^{2}+y^{2} \leq 16 a^{2}\right\}



By solving the equations:
$y^{2} \leq 6 a x$ \: \: and\: \: $x^{2}+y^{2} \leq 16 a^{2}$
Through substituting for $\mathrm{y}^{2}$
\\ \Rightarrow x^{2}+6ax=16 a^{2} \\ \Rightarrow(x-2a)(x+8a)=0\\ \therefore x=2a \\ \text{as x=-8 a is not possible }
Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a,b], is given by $\mathrm{A}=\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$ or $\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{ydx} .$
[By the symmetry of the image w.r.t x axis]
\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]$ \\
\text { Required area }=2\left[\int_{0}^{2 \mathrm{a}} \sqrt{6 \mathrm{ax}} \mathrm{d} \mathrm{x}+\int_{2 \mathrm{a}}^{4 \mathrm{a}} \sqrt{(4 \mathrm{a})^{2}-\mathrm{x}^{2}} \mathrm{dx}\right]
\\=2\left[\sqrt{6 a}\left[\frac{2 x^{\frac{3}{2}}}{3}\right]_{0}^{2 a}+\left[\frac{x \sqrt{(4 a)^{2}-x^{2}}}{2}+\frac{(4 a)^{2}}{2} \sin ^{-1}\left(\frac{x}{4 a}\right)\right]_{2 a}^{4 a}\right] \\ =2\left(\frac{8}{3} \sqrt{3} a^{2}+4 \pi a^{2}-2 \sqrt{3} a^{2}-\frac{4 a^{2} \pi}{3}\right) \\ =\frac{4}{3} a^{2}[\sqrt{3}+4 \pi] \text { sq.units } \\

Question:20

Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.

Answer:

The lines given are x + 2y = 2
x = 2 - 2y \ldots .(1) \\
y – x = 1
x = y -1 \ldots .(2) \\
and 2x + y = 7 \ldots .(3) \\


Equate the values of x from 1 and 2 to get,
\\$2-2 y=y-1$\\ $2+1=y+2 y$\\ $3=3 y$\\ $y=1$\\
put the value of y in (2) to get
\\$x=1-1$\\ $=0$
So, intersection point is (0,1)
On solving the equations, the point of interaction is found to be (0, 1), (2, 3) and (4, -1)
Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by A=\int_{a}^{b} f(x) d x \: \: or\: \: \int_{a}^{b} y d x
Required area
$$ =\int_{0}^{2}\left(\mathrm{x}-1-\frac{2-\mathrm{x}}{2}\right) \mathrm{dx}+\int_{2}^{4}\left(7-2 \mathrm{x}-\frac{2-\mathrm{x}}{2}\right) \mathrm{dx} $$ \\
\\ =\int_{0}^{2}\left(\frac{3 x}{2}\right) d x+\int_{2}^{4}\left(6-\frac{3 x}{2}\right) d x \\ =\left[\frac{3 x^{2}}{4}\right]_{0}^{2}+\left[6 x-\frac{3 x^{2}}{4}\right]_{2}^{4} \\ =3+(24-12)-(12-3)=6 \text { sq.units }

Question:21

Find the area bounded by the lines y = 4x + 5, y = 5 - x and 4y = x + 5

Answer:
The lines mentioned are y = 4x + 5, y = 5 - x and 4y = x + 5
Below if the figure,

By solving these equations,
\\$y=4 x+5 \ldots \ldots(1)$\\ $y=5-x \ldots \ldots(2)$\\ $4 y=x+5 \ldots . .(3)$\\
From (1) and (2)
\\$4 x+5=5-x$ $\Rightarrow x=0 ; \\\therefore y=5-x=5$\\
From (2) and (3)
\\$4(5-x)=x+5$\\ $\Rightarrow x=3 ; \therefore y=5-x=2$ \\
\begin{aligned} &\text { From }(1) \text { and }(3)\\ &4(4 x+5)=x+5\\ &\Rightarrow x=-1 ; \therefore y=4 x+5=1 \end{aligned} \\
The point of interaction have found to be (0, 5), (3, 2) and (-1, 1)
Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by A=\int_{a}^{b} f(x) d x$ or $\int_{a}^{b} y d x$
Required area
\\=\int_{-1}^{0}(4 x+5) d x+\int_{0}^{3}(-x+5) d x-\int_{-1}^{3}\left(\frac{x}{4}+\frac{5}{4}\right) d x$\\ $=\left[2 x^{2}+5 x\right]_{-1}^{0}+\left[-\frac{x^{2}}{2}+5 x\right]_{0}^{3}-\left[\frac{x^{2}}{8}+\frac{5 x}{4}\right]_{-1}^{3}$ \\$=((0+0)-(2-5))+\left(\left(-\frac{9}{2}+15\right)-(0+0)\right)-\left(\left(\frac{9}{8}+\frac{15}{4}\right)-\left(\frac{1}{8}-\frac{5}{4}\right)\right)$ \\

\\=3+\frac{21}{2}-\frac{39}{8}-\frac{9}{8} \\ =\frac{15}{2} \text { sq. units } \\

Question:22

Find the area bounded by the curve y = 2cos x and the x-axis from x = 0 to x = 2 \pi

Answer:

The curve given is y = 2cosx and x axis from x = 0 to x = 2 \pi
Below is the figure,

Area of the region bounded by the curve y=f(x) , the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x
From the figure;
Required area =\int_{0}^{2 \pi}|2 \cos x| \mathrm{d} \mathrm{x}=4 \int_{0}^{\frac{\pi}{2}}(2 \cos \mathrm{x}) \mathrm{dx}
\\=8[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}\\ =8 sq.units \\

Question:23

Draw a rough sketch of the given curve y = 1 + \vert x +1 \vert , x = -3, x = 3, y = 0 and find the area of the region bounded by them, using integration.

Answer:

Given;
Curve y = 1 + \vert x +1 \vert , x = -3, x = 3, y = 0
|x+1|=\left\{\begin{array}{l}-x-1, x<-1 \\ x+1, x \geq-1\end{array}\right. \therefore y=1+|x+1| =\left\{\begin{array}{c}-x, x<-1 \\ x+2, x \geq-1\end{array}\right.
Area of the region bounded by the curve y=f(x) , the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x \\


Required area
=\int_{-3}^{-1}(-\mathrm{x}) \mathrm{dx}+\int_{-1}^{3}(\mathrm{x}+2) \mathrm{d} \mathrm{x}\\ =-\left[\frac{\mathrm{x}^{2}}{2}\right]_{-3}^{-1}+\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{3}\\ =-\left(\frac{1}{2}-\frac{9}{2}\right)+\left(\frac{9}{2}+6-\frac{1}{2}+2\right)\\ =4+12=16 sq.units \\

Question:24

The area of the region bounded by the y-axis, y = cos x and y = sin x, 0 \leq x \leq\frac{\pi}{2} is.
A. \sqrt{2} sq units
B. (\sqrt{2} + 1) sq units
C. (\sqrt{2} – 1) sq unit
D. (2\sqrt{2} – 1) sq unit

Answer:

The questions has given,

y-axis, y = cosx and y = sinx, 0 \leq x \leq \pi /2 \\




\begin{aligned} &\Rightarrow \sin x=\cos x\\ &\therefore x=\frac{\pi}{4}\\ &\text { Required area }\\ &=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x) d x\\ &=[\sin \mathrm{x}+\cos \mathrm{x}]_{0}^{\frac{\pi}{4}}\\ &=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right)\\ &=(\sqrt{2}-1) \text { sq.units } \end{aligned}

Question:25

The area of the region bounded by the curve x^2 = 4y and the straight line x = 4y – 2 is.
A. \frac{3}{8} sq units
B.\frac{5}{8} sq units
C. \frac{7}{8} sq units
D. \frac{9}{8} sq units

Answer:

The questions gives equation for curve x^2 = 4y and straight line x = 4y – 2
Below is the figure,

\\\text{By substituting for} $4 y ; $\Rightarrow x=x^{2}-2$\\ $\Rightarrow x^{2}-x-2=0$\\ $\Rightarrow(x+1)(x-2)=0$\\ $\therefore x=-1,2$\\ \text{Required area}\\ $=\int_{-1}^{2}\left(\frac{\mathrm{x}+2}{4}\right) \mathrm{d} \mathrm{x}-\int_{-1}^{2} \frac{\mathrm{x}^{2}}{4}$ $=\frac{1}{4}\left[\frac{\mathrm{x}^{2}}{2}+2 \mathrm{x}\right]_{-1}^{2}-\frac{1}{4}\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}$
\\ =\frac{1}{4}\left(\frac{4}{2}+4-\frac{1}{2}+2-\frac{8}{3}-\frac{1}{3}\right) \\ =\frac{9}{8} \text { sq. units }

Question:26

The area of the region bounded by the curve y=\sqrt{16-x^2} and x-axis is.
A. 8π sq units
B. 20π sq units
C. 16π sq units
D. 256π sq units

Answer:

A)
\text { The curve } y=\sqrt{16-x^{2}} \text { and } x \text { -axis; } y=0

\begin{array}{l} \Rightarrow \sqrt{16-\mathrm{x}^{2}}=0 \\ \therefore \mathrm{x}=\pm 4 \end{array}
\\\text{Required area}\\ =\int_{-4}^{4}\left(\sqrt{16-\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x}$ \\$\left[\int \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\frac{\mathrm{x} \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}{2}+\frac{\mathrm{a}^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]$ \\$=2 \int_{0}^{4}\left(\sqrt{4^{2}-x^{2}}\right) d x$ \\$=2\left[\frac{\mathrm{x} \sqrt{4^{2}-\mathrm{x}^{2}}}{2}+\frac{4^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{4}\right)\right]_{0}^{4}$ \\$=2\left(0+\frac{8 \pi}{2}-0-0\right)$ \\$=8 \pi$ sq.units

Question:27

Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x^2 + y^2 = 32 is
A. 16π sq units
B. 4π sq units
C. 32π sq units
D. 24 sq units

Answer:

B)
The x-axis, the line y = x and the circle x^2 + y^2 = 32 .

\\ \text{By substitution}; \Rightarrow x^{2}+x^{2}=32$\\ $\therefore x=4$\\ \text{Radius of the circle} $=\sqrt{32}=4 \sqrt{2}$\\ \text{Required area}\\ $=\int_{0}^{4} \mathrm{xdx}+\int_{4}^{4 \sqrt{2}}\left(\sqrt{32-\mathrm{x}^{2}}\right) \mathrm{dx}$\\ $\left[\int \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}} \mathrm{~d} \mathrm{x}=\frac{\mathrm{x} \sqrt{\mathrm{a}^{2}-\mathrm{x}^{2}}}{2}+\frac{\mathrm{a}^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]$
\begin{aligned} &=\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}}\left(\sqrt{(4 \sqrt{2})^{2}-x^{2}}\right) d x\\ &=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{4}+\left[\frac{\mathrm{x} \sqrt{(4 \sqrt{2})^{2}-\mathrm{x}^{2}}}{2}+\frac{(4 \sqrt{2})^{2}}{2} \sin ^{-1}\left(\frac{\mathrm{x}}{4 \sqrt{2}}\right)\right]_{4}^{4 \sqrt{2}}\\ &=\left(8-0+\left(\frac{32}{2} \times \frac{\pi}{2}\right)-\frac{16}{2}-\left(\frac{32}{2} \times \frac{\pi}{4}\right)\right)\\ &=(8+8 \pi-8-4 \pi)\\ &=4 \pi \text { sq.units } \end{aligned}

Question:28

Area of the region bounded by the curve y = cos x between x = 0 and x = π is
A. 2 sq units
B. 4 sq units
C. 3 sq units
D. 1 sq units

Answer:

A)
The questions mentions curve y = cos x and x = 0 and x = π

Required area
\\=\int_{0}^{\pi}|\cos x| d x$\\ $=2 \int_{0}^{\frac{\pi}{2}} \cos x d x$\\ $=2[\sin \mathrm{x}]_{0}^{\frac{\pi}{2}}$\\ $=2(1-0)$\\ $=2$ sq.units

Question:29

The area of the region bounded by parabola y^{2} = x and the straight line 2y = x is
A. \frac{4}{3} sq units
B. 1 sq units
C. \frac{2}{3} sq units
D. \frac{1}{3} sq units

Answer:

A)
The questions gives parabola y^2 = x and a straight line 2y = x


By substituting for x
\\ $\Rightarrow y^{2}=2 y$\\ $\Rightarrow y(y-2)=0$\\ $\therefore y=0,2$\\ $\therefore x=2 \\y=0,4$\\
Required area
\\ =\int_{0}^{4}\left[\sqrt{\mathrm{x}}-\frac{\mathrm{x}}{2}\right] \mathrm{d} \mathrm{x} \\ =\left[\frac{2 \mathrm{x} \frac{3}{2}}{3}-\frac{\mathrm{x}^{2}}{4}\right]_{0}^{4} \\ =\left(\frac{16}{3}-4-0+0\right) \\ =\frac{4}{3} \mathrm{sq} . \text { units }

Question:30

The area of the region bounded by the curve y = sin x between the ordinates x = 0, x = π/2 and the x-axis is
A. 2 sq units
B. 4 sq units
C. 3 sq units
D. 1 sq units

Answer:

D)
The questions mentions y = sin x between the ordinates x = 0, x = π/2 and x axis

\\ =\int_{0}^{\frac{\pi}{2}}|\cos x| d x \\ =\int_{0}^{\frac{\pi}{2}} \sin x d x \\ =[-\cos x]_{0}^{\frac{\pi}{2}} \\ =0-(-1) \\ =1 \text { sq.units }

Question:31

The area of the region bounded by the ellipse \frac{x^{2}}{25}+\frac{y^{2}}{16}=1 is
A. 20π sq units
B. 20π2 sq units
C. 16π2 sq units
D. 25π sq units

Answer:

A)
The ellipse \frac{x^{2}}{25}+\frac{y^{2}}{16}=1


Ellipse has a symmetry with x axis and y axis
The required area can be calculated as
\\ =4 \int_{0}^{5} \frac{4}{5}\left(\sqrt{25-x^{2}}\right) d x \\ {\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\ =\frac{16}{5} \int_{0}^{5}\left(\sqrt{5^{2}-x^{2}}\right) d x
\\ =\frac{16}{5}\left[\frac{x \sqrt{5^{2}-x^{2}}}{2}+\frac{5^{2}}{2} \sin ^{-1}\left(\frac{x}{5}\right)\right]_{0}^{5} \\ =\frac{16}{5}\left(0-\frac{25 \pi}{4}-0-0\right) \\ =20 \pi \text { sq.units }


Question:32

The area of the region bounded by the circle x^{}2 + y^{}2 = 1 is
A. 2π sq units
B. π sq units
C. 3π sq units
D. 4π sq units

Answer:

B)
The circle x^{}2 + y^{}2 = 1

The circle is symmetrical with the x axis and y axis
Required Area
\\ =4 \int_{0}^{1}\left(\sqrt{1-x^{2}}\right) d x \\ {\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\ =4 \int_{0}^{1}\left(\sqrt{1^{2}-x^{2}}\right) d x \\ =4\left[\frac{x \sqrt{1^{2}-x^{2}}}{2}+\frac{1^{2}}{2} \sin ^{-1}\left(\frac{x}{1}\right)\right]_{0}^{1} \\ =4\left(0-\frac{\pi}{4}-0-0\right) \\ =\pi \text { sq.units }

Question:33

The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is
A. \frac{7}{2} sq units
B. \frac{9}{2} sq units
C. \frac{11}{2} sq units
D. \frac{13}{2} sq units

Answer:

A)
The questions gives y = x + 1 and lines x = 2 and x = 3

Required area
\\ =\int_{2}^{3}(\mathrm{x}+1) \mathrm{d} \mathrm{x} \\ =\left[\frac{\mathrm{x}^{2}}{2}+\mathrm{x}\right]_{2}^{3} \\ =\left(\frac{9}{2}+3-2-2\right) \\ =\frac{7}{2} \mathrm{sq} . \text { units }

Question:34

The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is
A. 4 sq units
B. \frac{3}{2} sq units
C. 6 sq units
D. 8 sq units

Answer:

C)
The question mentions the curve x = 2y + 3 and the lines on y axis y = 1 and y = -1

Required area
\\ =\int_{-1}^{1}(2 y+3) d x \\ =\left[y^{2}+3 y\right]_{-1}^{1} \\ =(1+3-1+3) \\ =6 \text { sq.units } \\

Students can make use of NCERT exemplar Class 12 Maths solutions chapter 8 pdf download, to access it offline. We will help the students to understand the application of integrals by solving the questions given in NCERT.

NCERT exemplar Class 12 Maths solutions chapter 8 Application Of Integrals

  • Sub Topics Covered

The Sub-Topics That Are Covered in Class 12 Maths NCERT Exemplar Solutions Chapter 8 Are:

  • Introduction
  • The area under simple curves
  • Area of the region by a line and a curve
  • The area between two curves

What can you learn in NCERT Exemplar Solutions for Class 12 Maths Chapter 8?

  • Integrals are basically a way to add small parts up to infinite times. It can help find various measures like volume, area, work etc.
  • It is a way of adding slices of one thing to create the whole using sum-up. In the real world, there are several applications of integrals which ranges from acceleration, velocity, work, the centre of mass, area between curves, kinetic energy, probability, etc.
  • As the number of usages of integral is so varying, the need to understand integrals becomes even more significant. That is the sole reason why NCERT exemplar Class 12 Maths chapter 8 solutions gets such importance in exams and CBSE syllabus.
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NCERT Exemplar Class 12 Maths Solutions


Benefits of NCERT Exemplar Class 12 Maths Solutions Chapter 8

Class 12 Maths NCERT exemplar solutions chapter 8 Application of Integrals touch upon exhaustive explanation of how the integrals can be used and to find the area under curves. There are several topics that are covered, and there are plenty of many things that one will learn.

  • Some of the major topics that are covered under NCERT exemplar solutions for Class 12 Maths chapter 8 are properties of integrals (definite), applications of integrals (indefinite), limits of integration, etc.
  • Other than this, several topics encompass the area under curves and parabolas like the area between two curves and the area between one curve and one line.
  • There are also some major rules and formulas covered, like Walli's formula and Leibnitz rule. The NCERT exemplar Class 12 Maths solutions chapter 8 also covers the value of functions like gamma function and integral function.
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Frequently Asked Questions (FAQs)

1. How many questions are there in this chapter?

This chapter contains a total 4 questions in one exercise. The questions are based on problem solving.

2. Are these solutions helpful for competitive examinations?

Yes, these Class 12 Maths NCERT exemplar solutions chapter 8 are designed to help you prepare for competitive examinations like JEE, NEET.

3. How to read math chapters?

You need to refer to NCERT Books for solving math problems once you clarify the concept you can use formulae to solve numerous problems. Also prefer NCERT exemplar Class 12 Maths solutions chapter 8 pdf download using the webpage to pdf tools.

4. Who has prepared the solutions?

The NCERT exemplar solutions for Class 12 Maths chapter 8 are prepared by specialized mathematicians who refer various high-level books to find best possible answers.

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Questions related to CBSE Class 12th

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

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    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

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Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer
Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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