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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

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.

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: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

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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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Our experienced teachers have covered every question and have created the NCERT exemplar Class 12 Maths chapter 8 solutions for it. The solutions given by us are simple, methodical and step oriented. We want our students to understand every step so much so that solving questions will come super easy for them.

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 Question (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

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


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In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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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

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

 

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

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

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 .

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

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.

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.

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

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

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