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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

#CBSE Class 12th
Komal MiglaniUpdated on 30 Mar 2025, 05:20 PM IST

Have you ever wondered how engineers measure the amount of work done in a machine or how to determine the total distance travelled by an object? The idea of Area under the Curve holds the solution! In NCERT, Area under the Curve is a crucial chapter that teaches students how to calculate numbers like work, distance, and even total revenue. To solve practical issues in physics, economics, and engineering, the area under the curve concept integrates mathematical functions.

This Story also Contains

  1. NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals
  2. The Sub-Topics That Are Covered in Class 12 Maths NCERT Exemplar Solutions Chapter 8 Are:
  3. NCERT Exemplar Class 12 Maths Solutions Chapter
  4. Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 8
  5. NCERT Solutions for Class 12 Maths: Chapter Wise
  6. NCERT solutions of class 12 - Subject-wise
  7. NCERT Notes of class 12 - Subject Wise
  8. NCERT Books and NCERT Syllabus

Students are advised to solve NCERT Solutions for Class 12 Maths regularly, as it helps the students to have a good grasp of the concepts. The syllabus of NCERT class 12 is provided here: NCERT Syllabus Class 12

NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals

Class 12 Maths Chapter 8 Exemplar Solutions Exercise: 8.3
Page number: 176-178
Total Questions: 34

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,
$y^{2} = 9x$ 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.


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, to find the coordinates of y.
Put x = 1 and x = 0 in the equation of y = 3x.
It is found that y coordinates 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 subtract 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 the parabola,
$y^{2 = }9x \\$
$\int y = 3 \sqrt x\\$
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. the 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 the equation mentioned above (1)
area between parabola and straight line $= 2 - \frac{3}{2} = 1/2\ unit^{2 }\\$
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 $y^{2 }= 2px$, which shows no negative values for x. Therefore, the graph will be on the right of the Y-axis and will pass through (0, 0)
Similarly, for equations $x^{2} = 2py$, there are no negative values for y, therefore, the graph will be above the X axis passing through (0, 0).
To find a 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
The area between the two parabolas = area under the parabola
$y^{2 }= 2px - \text{area under the parabola } x^{2 }= 2py \dots (1) \\$
For finding the area under$y^{2 }= 2px\: \: parabola \\$

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

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 $x^{2 }= 2py \text{ parabola} \\$

$\int y = x^{2}/ 2p \\$

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 $y=x^3$ along with the lines $\mathrm{y}=\mathrm{x}+6$ and $\mathrm{x}=0$ When $\mathrm{X}=0$, it means Y axis

The questions say to find the area between the curve and the line and Y axis
First solve the $y=x+6$ and $y=x^3$, in order to find the interaction point

$
\begin{aligned}
& \text { Put } y=x^3 \text { in } y=x+6 \\
& \Rightarrow x^3=x+6 \\
& \Rightarrow x^3-x-6=0
\end{aligned}
$
For checking $0,1,2$ satisfies this cubic, showing 2 is one factor, therefore $x-2$ is a factor.
Solving the equation,

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



The area bounded will be
Area bounded $=$ area by $y=x^3$ on Y axis - area by $\mathrm{y}=\mathrm{x}+6$ on Y axis $\ldots$

For finding the area under $y=x^3$

$
\Rightarrow x=\sqrt[3]{y}
$
Integrate the equation from 0 to 8

$
\begin{aligned}
& \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 x d y=\frac{3}{4}\left(2^3\right)^{\frac{4}{3}} \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\left[\frac{\mathrm{y}^{\frac{4}{3}}}{\frac{4}{3}}\right]_0^8 \Rightarrow \int_0^8 x d y=\frac{3}{4} 2^4 \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\frac{3}{4}\left[\mathrm{y}^{\frac{4}{3}}\right]_0^8 \Rightarrow \int_0^8 x d y=\frac{3}{4} 16 \\
& \Rightarrow \int_0^8 \mathrm{xdy}=\frac{3}{4}[8 \overline{4 / 3}-0]
\end{aligned}
$


Now, finding the area under $y=x+6$
For finding the area from 6 to 8 because the line passes through the $Y$ axis at 6 and extends up to 8, the point where curve and line intersect

$
\text { item } X=y-6
$


Integrating from 6 to 8

$
\Rightarrow \int_6^8 x d y=\int_6^8(y-6) d y
$

$
\begin{aligned}
& \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
\end{aligned}
$


Using the equation (1)
Area bound was found to be 12-2 = $10 \mathrm{unit}^2$
Therefore, the area was found to be 10 unit $^2$

Question:4

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

Answer:

The equation $y^{2 }= 4x$ 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 $x^{2} = 4y$, which is a parabola, not-defined negative values of y lie above the X-axis and pass through (0, 0)

To find a 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 $y^{2 }= 4x -\text{area under parabola } x^{2} = 4y \ldots . (1) \\$

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

$= y = 2 \sqrt{x}$
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 $x^{2}= 4y\\$

$= x^{2}= 4 y\\$

$\\\Rightarrow \int_{0}^{4} \mathrm{ydx}=\frac{4}{3}(2)^{3}\\$

$\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 $\frac{16}{3}$ unit $^2$

Question:5

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

Answer:

The equation$y^{2} = 9x$ has no negative values of x. Therefore it lies on the right of the Y axis passing through (0, 0).

And y = x depicts a straight line through the origin.
To find the area, see the picture shown below.





To find 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, the point of interaction is (9, 9)

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

Let us find the area under the 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, the area was found to be $13.5 unit^{2}. \\$

Question:6

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

Answer:

The equation $x^{2} = y$ depicts a parabola and has no negative values for y, therefore, it lies above the X-axis and passes through another origin (0, 0).

And y = x + 2 is a straight line.

The below figure shows the area to be found,

Let’s find the point of interaction of the two equations.

$\\\text{Put } y=x+2 inx^{2}=y\\ \Rightarrow x^{2}=x+2\\$

$ \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)

The area between the line and parabola = area under line – area under a parabola

Let us find the 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 the 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$\frac{9}{2}$ unit $^2$

Question:7

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

Answer:

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

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

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

On integrating the above equation from 0 to 2, it would give area enclosed under quadrant 1 only. Therefore, to find 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 the area under the parabola
$\\ \Rightarrow y^{2}=8 x\\ \Rightarrow y=2 \sqrt{2} \sqrt{x}\\ \text{Integrate from 0 to 2}\\ \Rightarrow \int_{0}^{2} y d x=\int_{0}^{2} 2 \sqrt{2} x^{\frac{1}{2}} d x\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{ \frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{2}\\ \Rightarrow \int_{0}^{2} \mathrm{ydx}=2 \sqrt{2}\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{2} \\$

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

The shaded areaOCBD $=2 \times \frac{16}{3}$

Hence area bounded $=\frac{32}{3}$ unit $^2$

Question:8

Sketch the region $\left\{(\mathrm{x}, 0) ; \mathrm{y}=\sqrt{4-\mathrm{x}^2}\right\}$and x-axis. Find the area of the region using integration.

Answer:

$
y=\sqrt{4-x^2}
$
Square both sides

$
\begin{aligned}
& \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 an origin and radius of 2
Equation of X axis is $\mathrm{y}=0$
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:

$y=2 \sqrt{x}$ will be the parabolic curve of $y^2=4 x$ only in $1^{\text {st }}$ quadrant $\mathrm{x}=0$ is the equation of the $Y$-axis and $\mathrm{x}=1$ is a line parallel to Y -axis passing through $(1,0)$. Plot equations $y=2 \sqrt{x}$ and $\mathrm{X}=1$

So we have to integrate $y=2 \sqrt{x}$ from 0 to 1 let us find area under parabola

$
\Rightarrow y=2 \sqrt{x}
$
Integrate from 0 to 1

$
\begin{aligned}
& \Rightarrow \int_0^1 \mathrm{ydx}=\int_0^1 2 \mathrm{x}^{\frac{1}{2}} \mathrm{dx} \\
& \Rightarrow \int_0^1 \mathrm{ydx}=2\left[\frac{\mathrm{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_0^1 \\
& \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}
\end{aligned}
$
Therefore, the area found to be

$
\frac{4}{3} \text { unit }^2
$

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 the 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 the Y axis.
On plotting all the above lines.

$\\\text{We have to find area under} \ the \ \mathrm{y}=5 \mathrm{x}+7$ 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, the 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 the area under the 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:

$
y=\sqrt{a^2-x^2}
$
Squaring both sides

$
\begin{aligned}
& \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 $1^{\text {st }}$ 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 the area under the 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 the 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 $\mathrm{v}=1$
Hence

$
\begin{aligned}
& \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}} \\
& \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 \\
& \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}
\end{aligned}
$

$
\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 \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}
$


We know that $\int_0^a \frac{1}{\sqrt{a^2-x^2}} d x=\sin ^{-1} \frac{x}{a}$

$
\begin{aligned}
& \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) \\
& \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}
\end{aligned}
$
Hence area bounded $=\frac{\pi a^2}{4}$ unit ${ }^2$

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, with no negative values of x, therefore it lies on the right of the Y axis, passing through the 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 the origin
We have to find area between $y=\sqrt{x}$ and y=x shown below

To find 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 the 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 line x + y + 2 = 0.

Answer:

$\\Y = -x^{2 }\\ X^{2 }= -y \\$
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 are (-1, -1) and (2, -4)
The below figure shows the area to be calculated


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


To find the area under the 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 a parabola, integrate it from -1 to 2
$Y = -x^{2}\\$
$\begin{aligned} &\text { Integrate from }-1 \text { to } 2\\ &\Rightarrow \int_{-1}^{2} y d x=\int_{-1}^{2}-x^{2} d x\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left[\frac{\mathrm{x}^{3}}{3}\right]_{-1}^{2}\\ &\Rightarrow \int_{-1}^{2} \mathrm{ydx}=-\left(\frac{2^{3}}{3}-\frac{(-1)^{3}}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-\left(\frac{8}{3}+\frac{1}{3}\right)\\ &\Rightarrow \int_{-1}^{2} y d x=-3 \end{aligned} \\$
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 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 the 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.
To find 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 points of interaction are (9, 3)
Below is the figure showing the area to be calculated
The point of the interaction of a 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 a 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:

$y^2=2 x$ depicts a parabola having no negative values for x and lying to the right of the Y axis while passing through the origin. The other equation of $x^2+y^2=4 x$ depicts a circle.

The general equation of a circle is given by $x^2+y^2+2 g x+2 f y+c=0$
Centre of circle is $(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}$

$
\text { In } x^2+y^2-4 x=0,2 g=-4 \Rightarrow g=-2 \text { 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.

$
\begin{aligned}
& \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
\end{aligned}
$


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



On integrating we will get an area for 1st quadrant only, but since it is symmetrical, we can multiply it by.
Area of shaded region $=$ area under the circle - area under the parabola $\ldots$ (1)To findg the area under circle

$
\begin{aligned}
& 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}
\end{aligned}
$
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 the parabola,

$
\begin{aligned}
& \Rightarrow y^2=2 x \\
& \Rightarrow y=\sqrt{2} \sqrt{x}
\end{aligned}
$

Integrate the above equation from 0 to 2

$
\begin{aligned}
& \Rightarrow \int_0^2 y d x=\sqrt{2} \int_0^2 x^{\frac{1}{2}} d x \\
& \Rightarrow \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 \\
& \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{ydx}=\frac{2^{\frac{3}{2}+\frac{3}{2}}}{3} \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{2^3}{3} \\
& \Rightarrow \int_0^2 \mathrm{ydx}=\frac{8}{3}
\end{aligned}
$


Using (i)
$\Rightarrow$ area of shaded in $1^{\text {st }}$ quadrant $=\pi-\frac{8}{3}$ unit $^2$
After multiplying it by 2
The area required of shaded region $=2\left(\pi-\frac{8}{3}\right)$ unit $^2$

Question:17

Find the area bounded by the curve y = six 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 the area under six from 0 to $\pi$ and multiply 2
2)Split the limit 0 to 2 $\pi$ into 0 to $\pi$ and $\pi$ to 2 $\pi$
Here we will solve this by splitting the limits
$Y = six \\$
$\begin{aligned} &\text { Integrating from } 0 \text { to } 2 \pi\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{2 \pi} \sin \mathrm{xdx}\\ &\Rightarrow \int_{0}^{2 \pi} \mathrm{ydx}=\int_{0}^{\pi} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi}^{2 \pi} \sin \mathrm{xdx}\\ &\text { Because } \int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{\mathrm{a}}^{\mathrm{c}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}+\int_{\mathrm{c}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x} \text { where } c \in(\mathrm{a}, \mathrm{b}) \end{aligned} \\$
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 six from 0 to 2 $\pi$ $\text{ is }4 unit $^{2}$ \\$

Question:18

Find the area of the 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.

$\Rightarrow$ Equation of PQ is $y-1=\frac{5-1}{0+1}(x+1)$thereforey $=4 x+5 \Rightarrow$ Equation of $Q R$ is $y-5=\frac{2-5}{3-0}(x-0) \therefore y=-x+5$
$\Rightarrow$ Equation of 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 $\mathrm{x}=\mathrm{a}$ and $\mathrm{x}=\mathrm{b}$, where $\mathrm{f}(\mathrm{x})$ 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

$
\begin{aligned}
& =\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 }
\end{aligned}
$

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 the 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 \quad \text { and } \quad x^2+y^2 \leq 16 a^2
$


Through substituting for $\mathrm{y}^2$

$
\begin{aligned}
& \Rightarrow x^2+6 a x=16 a^2 \\
& \Rightarrow(x-2 a)(x+8 a)=0 \\
& \therefore x=2 a
\end{aligned}
$

[ 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 $A=\int_a^b f(x) d x$ or $\int_a^b y d x$.
[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]
$
Required area $=2\left[\int_0^{2 a} \sqrt{6 a x} d x+\int_{2 a}^{4 a} \sqrt{(4 a)^2-x^2} d x\right]$

$
\begin{aligned}
& =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 }
\end{aligned}
$

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,

$
\begin{aligned}
& 2-2 y=y-1 \\
& 2+1=y+2 y \\
& 3=3 y \\
& y=1
\end{aligned}
$

Put the value of $y$ in (2) to get

$
\begin{aligned}
& x=1-1 \\
& =0
\end{aligned}
$


So, the intersection point is $(0,1)$
On solving the equations, the point of interaction is found to be $(0,1),(2,3)$ and $(4,-1)$The 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

$
\begin{aligned}
& =\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 }
\end{aligned}
$

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 is the figure,

By solving these equations,

$
\begin{aligned}
& y=4 x+5 \ldots \ldots \\
& y=5-x \ldots \ldots(2) \\
& 4 y=x+5 \ldots
\end{aligned}
$


From (1) and (2)

$
\begin{aligned}
& 4 x+5=5-x \Rightarrow x=0 \\
& \therefore y=5-x=5
\end{aligned}
$


From (2) and (3)

$
\begin{aligned}
& 4(5-x)=x+5 \\
& \Rightarrow x=3 ; \therefore y=5-x=2 \\
& \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 $[\mathrm{a}, \mathrm{b}]$, is given by $A=\int_a^b f(x) d x$ or $\int_a^b y d x$

Required area

$
\begin{aligned}
& =\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 }
\end{aligned}
$

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 xx-axis from x = 0 to x = 2 $\pi$
Below is the figure,

The 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 $y = 1 + \vert x +1 \vert , x = -3, x = 3, y = 0$
$|x+1|=\left\{\begin{array}{l}-x-1, x<-1 \\ x+1, x \geq-1\end{array}\right. \therefore y=1+|x+1| =\left\{\begin{array}{c}-x, x<-1 \\ x+2, x \geq-1\end{array}\right.$
AreThe 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:

$y-axis, y = cosx$ 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,

By substituting for $4 \mathrm{y} ; \Rightarrow x=x^2-2$

$
\begin{aligned}
& \Rightarrow x^2-x-2=0 \\
& \Rightarrow(x+1)(x-2)=0 \\
& \therefore x=-1,2
\end{aligned}
$
Required area

$
\begin{aligned}
& =\int_{-1}^2\left(\frac{x+2}{4}\right) d x-\int_{-1}^2 \frac{x^2}{4}=\frac{1}{4}\left[\frac{x^2}{2}+2 x\right]_{-1}^2-\frac{1}{4}\left[\frac{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 }
\end{aligned}
$

Question:26

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

Answer:
$\text { The curve } y=\sqrt{16-x^{2}} \text { and } x \text { -axis; } y=0$

$
\begin{aligned}
& \Rightarrow \sqrt{16-x^2}=0 \\
& \therefore x= \pm 4
\end{aligned}
$
Required area

$
\begin{aligned}
& =\int_{-4}^4\left(\sqrt{16-x^2}\right) \mathrm{dx} \\
& {\left[\int \sqrt{\mathrm{a}^2-\mathrm{x}^2} \mathrm{dx}=\frac{\mathrm{x} \sqrt{\mathrm{a}^2-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 \text { sq.units }
\end{aligned}
$

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 $x^2 + y^2 = 32$.

By substitution; $\Rightarrow x^2+x^2=32$

$
\therefore x=4
$


Radius of the circle $=\sqrt{32}=4 \sqrt{2}$
Required area

$
\begin{aligned}
& =\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{dx}=\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]} \\
& =\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

The 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 question mentions curve y = cos x and x = 0 and x = π

Required area

$
\begin{aligned}
& =\int_0^\pi|\cos x| d x \\
& =2 \int_0^{\frac{\pi}{2}} \cos x d x \\
& =2[\sin x]_0^{\frac{\pi}{2}} \\
& =2(1-0) \\
& =2 \text { sq. units }
\end{aligned}
$

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

$
\begin{aligned}
& \Rightarrow y^2=2 y \\
& \Rightarrow y(y-2)=0 \\
& \therefore y=0,2 \\
& \therefore x=2 \\
& y=0,4
\end{aligned}
$
Required area

$
\begin{aligned}
& =\int_0^4\left[\sqrt{\mathrm{x}}-\frac{\mathrm{x}}{2}\right] \mathrm{dx} \\
& =\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} \text { sq. units }
\end{aligned}
$

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 mentioned = sin x between the ordinates x = 0, x = π/2 and x 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 the 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
$\\ =\int_{-1}^{1}(2 y+3) d x \\ =\left[y^{2}+3 y\right]_{-1}^{1} \\ =(1+3-1+3) \\ =6 \text { sq.units } \\$

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

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 Chapter

NCERT Exemplar Class 12 Mathematics Chapter 1: Relations and Functions

NCERT Exemplar Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions

NCERT Exemplar Class 12 Mathematics Chapter 3: Matrices

NCERT Exemplar Class 12 Mathematics Chapter 4: determinants

NCERT Exemplar Class 12 Mathematics Chapter 5: Continuity and Differentiability

NCERT Exemplar Class 12 Mathematics Chapter 6: Applications of derivatives

NCERT Exemplar for Class 12 Mathematics Chapter 7: Integrals

NCERT Exemplar Class 12 Mathematics Chapter 8: Applications of Integrals

NCERT Exemplar Class 12 Mathematics Chapter 9: Differential equations

NCERT Exemplar Class 12 Mathematics Chapter 10: Vector Algebra

NCERT Exemplar Class 12 Mathematics Chapter 11: Three-Dimensional Geometry

NCERT Exemplar Class 12 Mathematics Chapter 12: Linear Programming

NCERT Exemplar Class 12 Mathematics Chapter 13: Probability

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Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 8

Class 12 Maths NCERT exemplar solutions Chapter 8 Application of Integrals touches upon an exhaustive explanation of how the integrals can be used to find the area under curves. Several topics 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's 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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NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths: Chapter Wise

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions

NCERT solutions for class 12 maths chapter 3 Matrices

NCERT solutions for class 12 maths chapter 4 Determinants

NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability

NCERT solutions for class 12 maths chapter 6 Application of Derivatives

NCERT solutions for class 12 maths chapter 7 Integrals

NCERT solutions for class 12 maths chapter 8 Application of Integrals

NCERT solutions for class 12 maths chapter 9 Differential Equations

NCERT solutions for class 12 maths chapter 10 Vector Algebra

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry

NCERT solutions for class 12 Maths chapter 12 Linear Programming

NCERT solutions for class 12 maths chapter 13 Probability

NCERT solutions of class 12 - Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Notes of class 12 - Subject Wise

Given below are the subject-wise NCERT Notes of class 12 :

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

Frequently Asked Questions (FAQs)

Q: What are the important topics in the NCERT Exemplar Class 12 Maths Chapter 8?
A:

The use of Integrals is all about integration to determine the area under curves and two functions. It is concerned with definite integrals to determine exact areas with emphasis on graphical representation to visualize regions that are bounded. Students are instructed to use these concepts in daily applications such as physics, economics, and engineering. The ability to draw curves and determine areas facilitates solving problems effectively. Solving NCERT Exemplar problems makes one effective in problem-solving, hence easy to solve integration-based area problems in examinations and practical applications.

Q: What are the real-life applications of the application of integrals in Class 12?
A:

Physics and Engineering are basically applied sciences to find and compute different significant values like area and volume, the amount of work done within a system, and finding the center of mass in mechanics and electric circuits.

Economics and Business – Helps in finding cost, revenue, and profit functions over time.

Biology and Medicine – Used in population growth models, blood flow analysis, and medical imaging, like CT scans.

Architecture and Design – Helps in designing curved structures, bridges, and roads by calculating surface areas.

Astronomy - Applied in the calculation of planetary motion and the space covered by celestial objects.

Q: Why is Chapter 8 Application of Integrals, important in Class 11 Maths?
A:

Class 12 Maths consists of Chapter 8, Application of Integrals, which is important as it is employed in determining the area under curves and between two curves, which is of practical use in real applications like physics, engineering, economics, and architecture. It is a continuation of the concept of integration and uses definite integrals to solve problems of daily life. It is a very important chapter for competitive examinations such as JEE and NEET, in which questions involving integrals in relation to the area are generally asked. These principles are used to solve intricate problems in curved surfaces, motion, and the study of real-life data and thus is an important topic in higher mathematics.

Q: What are some common mistakes students make in Class 12 Application of Integrals?
A:

Class 12 - Application of Integrals has improper limits of integration, where students get the upper and lower limits wrong, and do not use absolute values, since the area is always positive. Most also confuse curves with limits, which results in an improper setup of integration. Overlooking graphical representation results in misinterpretation of problems, whereas errors in calculation during integration, particularly with trigonometric and logarithmic functions, result in incorrect answers. To prevent these errors, students must practice regularly, thoroughly examine the provided curves, and recheck their calculations to avoid errors.

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Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD
Bristol Baptist College, Bristol
The Promenade, Clifton Down, Bristol BS8 3NJ

Questions related to CBSE Class 12th

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Have a question related to CBSE Class 12th ?
Can i improve my 12 result this year in PCM

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

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R
Rahul Mandal
8 Sep'25
how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

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G
gopi
4 Sep'25
Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

  • B.Tech in Computer Science Engineering (CSE) – most popular choice.

  • BCA (Bachelor of Computer Applications) – good for software and IT jobs.

  • B.Sc. Computer Science / IT – good for higher studies and research.

  • B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

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P
Prachi Kumari
4 Sep'25
I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

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G
gopi
3 Sep'25
12 arts previous years board paper 2025

To find Class 12 Arts board papers, go to the official website of your education board, then click on the Sample Papers, Previous Years Question Papers(PYQ) or Model Papers section, and select the Arts stream. You will find papers for the various academic year. You can then select the year of which you want to solve and do your practice. There are many other educational websites that post pyqs on their website you can also visit that.

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A
Anshika Gupta
10 Sep'25
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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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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 Complete Answer

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