NCERT Exemplar Class 12 Maths Solutions Chapter 8 Application Of Integrals

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.
$\begin{array}{l}\text{ Put }y=3x\text{ in }{y}^2=9x\\ \Rightarrow (3x{)}^2=9x\\ \Rightarrow 9x^2=9x\\ \Rightarrow x^2-x=0\\ \Rightarrow x(x-1)=0\\ \Rightarrow x=0\text{ and }x=1\end{array}$
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
$$\begin{gathered} \Rightarrow {\int }_0^{1}ydx=3{\int }_0^1\sqrt{x}dx \\ \Rightarrow {\int }_0^{1}ydx=3{\int }_0^1{x}^{\frac{1}{2}}dx \\ \Rightarrow {\int }_0^{1}ydx=3{\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^1 \\ \Rightarrow {\int }_0^{1}ydx=3{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1 \end{gathered}$$
$$\begin{gathered} \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}=3\frac{2}{3}{\left[{\mathrm{x}}^{\frac{3}{2}}\right]}_0^1 \\ \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}=2[1^{\frac{3}{2}}-0] \\ \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}=2 \end{gathered}$$
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
$$\begin{gathered} \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}={\int }_0^{1}3\mathrm{xdx} \\ \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}=3{\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^1 \\ \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}=3(\frac{1^2}{2}-0) \\ \Rightarrow {\int }_0^1\mathrm{y}\mathrm{dx}=\frac{3}{2} \end{gathered}$$

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

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.
$$\begin{gathered} \text{ Put }y=\frac{x^2}{2p}\text{ in }{y}^2=2px \\ \Rightarrow {\left(\frac{x^2}{2p}\right)}^2=2px \\ \Rightarrow \frac{x^4}{4p^2}=2px \\ \Rightarrow x^4=8p^{3}x \\ \Rightarrow x^3=8p^3 \end{gathered}$$
$$\begin{gathered} \Rightarrow x=2p \\ \text{ Put }x=2p\text{ in }{y}^2=2px \\ \Rightarrow y^2=2p(2p) \\ \Rightarrow y=2p \end{gathered}$$
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$y^2=2pxparabola$

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

Integrating the equation from 0 to 2p
$$\begin{gathered} \Rightarrow {\int }_0^{2p}ydx=\sqrt{2p}{\int }_0^{2p}\sqrt{x}dx \\ \Rightarrow {\int }_0^{2p}ydx=\sqrt{2p}{\int }_0^{2p}{x}^{\frac{1}{2}}dx \\ \Rightarrow {\int }_0^{2p}ydx=\sqrt{2p}{\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^{2p} \\ \Rightarrow {\int }_0^{2p}ydx=\sqrt{2p}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{2p} \\ \Rightarrow {\int }_0^{2p}ydx=\sqrt{2p}\frac{2}{3}[(2p{)}^{\frac{3}{2}}-0] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\int }_0^{2p}ydx=\frac{2}{3}(2p{)}^{\frac{1}{2}}(2p{)}^{\frac{3}{2}} \\ \Rightarrow {\int }_0^{2p}ydx=\frac{2}{3}(2p{)}^{\frac{1}{2}+\frac{3}{2}} \\ \Rightarrow {\int }_0^{2p}ydx=\frac{2}{3}(2p{)}^2 \\ \Rightarrow {\int }_0^{2p}ydx=\frac{8p^2}{3} \end{gathered}$$
For finding the area under $x^2=2py\text{ parabola}$

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

Integrating from 0 to 2p
$$\begin{gathered} \Rightarrow {\int }_0^{2p}ydx={\int }_0^{2p}\frac{x^2}{2p}dx \\ \Rightarrow {\int }_0^{2p}ydx=\frac{1}{2p}{\left[\frac{x^3}{3}\right]}_0^{2p} \\ \Rightarrow {\int }_0^{2p}ydx=\frac{1}{2p}\left[\frac{(2p{)}^3}{3}\right] \\ \Rightarrow {\int }_0^{2p}ydx=\frac{(2p{)}^2}{3} \\ \Rightarrow {\int }_0^{2p}ydx=\frac{4p^2}{3} \end{gathered}$$
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

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

$\Rightarrow (x-2)(x^2+2x+3)=0$
Observe that $x^2+2x+3$ 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 $\dots$

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

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

$\begin{array}{rl}& \Rightarrow {\int }_0^8\mathrm{xdy}={\int }_0^8\mathrm{y}\frac{1}{3}\mathrm{dy}\\ & \Rightarrow {\int }_0^8\mathrm{xdy}={\left[\frac{{\mathrm{y}}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]}_0^8\Rightarrow {\int }_0^{8}xdy=\frac{3}{4}{\left(2^3\right)}^{\frac{4}{3}}\\ & \Rightarrow {\int }_0^8\mathrm{xdy}={\left[\frac{{\mathrm{y}}^{\frac{4}{3}}}{\frac{4}{3}}\right]}_0^8\Rightarrow {\int }_0^{8}xdy=\frac{3}{4}{2}^4\\ & \Rightarrow {\int }_0^8\mathrm{xdy}=\frac{3}{4}{\left[{\mathrm{y}}^{\frac{4}{3}}\right]}_0^8\Rightarrow {\int }_0^{8}xdy=\frac{3}{4}16\\ & \Rightarrow {\int }_0^8\mathrm{xdy}=\frac{3}{4}[8\overline{4/3}-0]\end{array}$

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

$\Rightarrow {\int }_6^{8}xdy={\int }_6^8(y-6)dy$

$\begin{array}{rl}& \Rightarrow {\int }_6^{8}xdy=[(\frac{8^2}{2}-6(8))-(\frac{6^2}{2}-6(6))]\\ & \Rightarrow {\int }_6^{8}xdy=[32-48-18+36]\\ & \Rightarrow {\int }_6^{8}xdy=2\end{array}$

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.
$\begin{array}{rl}& \text{ Put }y=\frac{x^2}{4}\text{ in }{y}^2=4x\\ & \Rightarrow {\left(\frac{x^2}{4}\right)}^2=4x\\ & \Rightarrow \frac{x^4}{16}=4x\\ & \Rightarrow x^4=64x\\ & \Rightarrow x^3=64\\ & \Rightarrow x=4\\ & \text{ Put }x=4\text{ in }{y}^2=4x\\ & \Rightarrow y^2=4(4)\\ & \Rightarrow y=4\end{array}$

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

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

$=y=2\sqrt{x}$
Integrate from 0 to 4
$$\begin{gathered} \Rightarrow {\int }_0^4\mathrm{ydx}=2{\int }_0^4\sqrt{\mathrm{x}}\mathrm{dx} \\ \begin{array}{rl}& \Rightarrow {\int }_0^4\mathrm{ydx}=2{\int }_0^4\mathrm{x}\frac{1}{2}\mathrm{dx}\\ & \Rightarrow {\int }_0^4\mathrm{ydx}=2{\left[\frac{{\mathrm{x}}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^4\\ & \Rightarrow {\int }_0^4\mathrm{ydx}=2{\left[\frac{\mathrm{x}\frac{3}{2}}{\frac{3}{2}}\right]}_0^4\\ & \Rightarrow {\int }_0^4\mathrm{ydx}=\frac{4}{3}{\left[{\mathrm{x}}^{\frac{3}{2}}\right]}_0^4\\ & \Rightarrow {\int }_0^4\mathrm{ydx}=\frac{4}{3}[4^{\frac{3}{2}}-0]\\ & \Rightarrow {\int }_0^4\mathrm{ydx}=\frac{4}{3}{\left(2^2\right)}^{\frac{3}{2}}\end{array} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\int }_0^{4}ydx=\frac{4}{3}(2{)}^3 \\ \Rightarrow {\int }_0^{4}ydx=\frac{32}{3} \end{gathered}$$
To find area under parabola $x^2=4y$

$=x^2=4y$

$\Rightarrow {\int }_0^4\mathrm{ydx}=\frac{4}{3}(2{)}^3$

$\Rightarrow {\int }_0^4\mathrm{ydx}=\frac{32}{3}$

$\text{Now let us find area under parabola}$

$$\begin{gathered} x^2=4y \\ \Rightarrow x^2=4y \\ \Rightarrow \mathrm{y}=\frac{{\mathrm{x}}^2}{4} \end{gathered}$$ $\text{Integrate from 0 to 4}$

$\Rightarrow {\int }_0^4\mathrm{ydx}={\int }_0^4\frac{{\mathrm{x}}^2}{4}\mathrm{d}\mathrm{x}$

$\Rightarrow {\int }_0^4\mathrm{ydx}=\frac{1}{4}{\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^4$

$$\begin{gathered} \Rightarrow {\int }_0^4\mathrm{ydx}=\frac{1}{4}\left[\frac{4^3}{3}\right] \\ \Rightarrow {\int }_0^{4}ydx=\frac{4^2}{3} \\ \Rightarrow {\int }_0^4\mathrm{ydx}=\frac{16}{3} \\ \text{Using (i)} \end{gathered}$$

$\Rightarrow \text{area bounded by two parabolas given}=\frac{32}{3}-\frac{16}{3}$ $\Rightarrow \text{area bounded by two parabolas given}=\frac{16}{3}$
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.
$$\begin{gathered} Puty=x\text{ in }{y}^2=9x\Rightarrow x^2=9x\Rightarrow x=9 \\ Putx=9iny=x\text{ we get }y=9 \end{gathered}$$

Therefore, the point of interaction is (9, 9)

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

Let us find the area under the parabola
$$\begin{gathered} \Rightarrow y^2=9x \\ \Rightarrow y=3\sqrt{x} \\ \text{Integrate from 0 to 9} \\ \Rightarrow {\int }_0^9\mathrm{ydx}={\int }_0^{9}3{\mathrm{x}}^{\frac{1}{2}}\mathrm{dx} \\ \Rightarrow {\int }_0^9\mathrm{ydx}=3{\left[\frac{\mathrm{x}\frac{1}{2}+1}{\frac{1}{2}+1}\right]}_0^9 \\ \Rightarrow {\int }_0^{9}y\mathrm{dx}=3{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^9 \end{gathered}$$
$$\begin{gathered} \Rightarrow {\int }_0^9\mathrm{ydx}=3\frac{2}{3}[9\frac{3}{2}-0] \\ \Rightarrow {\int }_0^9\mathrm{ydx}=2{\left(3^2\right)}^{\frac{3}{2}} \\ \Rightarrow {\int }_0^9\mathrm{ydx}=2\left(3^3\right) \\ \Rightarrow {\int }_0^9\mathrm{ydx}=54 \end{gathered}$$
Now let us find area under straight line $\mathrm{y}=\mathrm{x}$
y=x
Integrate from 0 to 9
$$\begin{gathered} \Rightarrow {\int }_0^9\mathrm{ydx}={\int }_0^9\mathrm{xdx} \\ \Rightarrow {\int }_0^9\mathrm{ydx}={\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^9 \\ \Rightarrow {\int }_0^9\mathrm{ydx}=(\frac{9^2}{2}-0) \\ \Rightarrow {\int }_0^9\mathrm{ydx}=\frac{81}{2} \\ \Rightarrow {\int }_0^{9}y\mathrm{dx}=40.5 \end{gathered}$$
Using (i)
$\Rightarrow$ area between parabola and line $=54-40.5=13.5uni{t}^2$
Therefore, the area was found to be $13.5uni{t}^2.$

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.

$$\begin{gathered} \text{Put }y=x+2in{x}^2=y \\ \Rightarrow x^2=x+2 \end{gathered}$$

$\Rightarrow x^2-x-2=0$$$\begin{gathered} \Rightarrow x^2-2x+x-2=0 \\ \Rightarrow x(x-2)+1(x-2)=0 \end{gathered}$$

$\Rightarrow (x+1)(x-2)=0$

$\Rightarrow x=-1\text{ and }x=2$

$\text{Put }x=-1\text{ and }x=2in{x}^2=y\text{ we get }y=1\text{ and }y=4respectively$
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

$$\begin{gathered} \Rightarrow y=x+2 \\ \text{Integrate from -1 to 2}\Rightarrow {\int }_{-1}^{2}ydx={\int }_{-1}^2(x+2)dx \end{gathered}$$

$\Rightarrow {\int }_{-1}^{2}y\mathrm{d}x={[\frac{x^2}{2}+2x]}_{-1}^2$

$\Rightarrow {\int }_{-1}^{2}ydx=[(\frac{2^2}{2}+2(2))-(\frac{(-1{)}^2}{2}+2(-1))]$

$\Rightarrow {\int }_{-1}^{2}ydx=6-(\frac{1}{2}-2)$

$\Rightarrow {\int }_{-1}^2\mathrm{ydx}=6-\left(\frac{-3}{2}\right)$

$\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
$$\begin{gathered} {x}^2=y\Rightarrow y=x^2 \\ \text{Integrate from -1 to 2 } \\ \Rightarrow {\int }_{-1}^{2}ydx={\int }_{-1}^2{x}^{2}dx \end{gathered}$$
$\begin{array}{rl}& \Rightarrow {\int }_{-1}^2\mathrm{ydx}={\left[\frac{{\mathrm{x}}^3}{3}\right]}_{-1}^2\\ & \Rightarrow {\int }_{-1}^2\mathrm{ydx}=(\frac{2^3}{3}-\frac{(-1{)}^3}{3})\\ & \Rightarrow {\int }_{-1}^2\mathrm{ydx}=(\frac{8}{3}+\frac{1}{3})\\ & \Rightarrow {\int }_{-1}^2\mathrm{ydx}=3\\ & \text{ Using (i) }\\ & \text{ area enclosed by line and parabola }=\frac{15}{2}-3=\frac{9}{2}{\text{ unit }}^2\end{array}$
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 $\dots .(1)$
Let us find the area under the parabola
$$\begin{gathered} \Rightarrow y^2=8x \\ \Rightarrow y=2\sqrt{2}\sqrt{x} \\ \text{Integrate from 0 to 2} \\ \Rightarrow {\int }_0^{2}ydx={\int }_0^{2}2\sqrt{2}{x}^{\frac{1}{2}}dx \\ \Rightarrow {\int }_0^2\mathrm{ydx}=2\sqrt{2}{\left[\frac{{\mathrm{x}}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^2 \\ \Rightarrow {\int }_0^2\mathrm{ydx}=2\sqrt{2}{\left[\frac{{\mathrm{x}}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^2 \end{gathered}$$

$\begin{array}{rl}& \Rightarrow {\int }_0^2\mathrm{ydx}=2\sqrt{2}\frac{2}{3}{\left[{\mathrm{x}}^{\frac{3}{2}}\right]}_0^2\\ & \Rightarrow {\int }_0^{2}ydx=\frac{4}{3}\left(2^{\frac{1}{2}}\right)(2^{\frac{3}{2}}-0)\\ & \Rightarrow {\int }_0^{2}ydx=\frac{4}{3}\left(2^{\frac{1}{2}+\frac{3}{2}}\right)\\ & \Rightarrow {\int }_0^{2}ydx=\frac{4}{3}\left(2^2\right)\\ & \Rightarrow {\int }_0^2\mathrm{ydx}=\frac{16}{3}\\ & \Rightarrow \mathrm{areaOBC}=\frac{16}{3}\\ & \text{ Using (i) }\end{array}$

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

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

Question:8

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

Answer:

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

$\begin{array}{rl}& \Rightarrow y^2=4-x^2\\ & \Rightarrow x^2+y^2=4\\ & \Rightarrow x^2+y^2=2^2\end{array}$
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
$\Rightarrow {\int }_{-2}^2\mathrm{ydx}={\int }_{-2}^2\sqrt{4-{\mathrm{x}}^2}\mathrm{d}\mathrm{x}$
$\begin{array}{rl}& \text{ Using uv rule of integration where u and }v\text{ are functions of }x\\ & {\int }_a^{b}uvdx={[u\int \mathrm{vdx}]}_a^b-{\int }_a^b(u^{\prime }\int vdx)dx\\ & \text{ Here }u=\sqrt{4-x^2}\text{ and }v=1\\ & \text{ Hence }{u}^{\prime }=\frac{1}{2}{(4-x^2)}^{\frac{1}{2}-1}(-2x)=\frac{-x}{\sqrt{4-x^2}}\\ & \Rightarrow {\int }_{-2}^{2}ydx={[\sqrt{4-x^2}\int 1dx]}_{-2}^2-{\int }_{-2}^2\left(\frac{-x}{\sqrt{4-x^2}}\right)(1dx)dx\\ & \Rightarrow {\int }_{-2}^{2}ydx={[\sqrt{4-x^2}(x)]}_{-2}^2-{\int }_{-2}^2\left(\frac{-x^2}{\sqrt{4-x^2}}\right)dx\\ & \Rightarrow {\int }_{-2}^2\mathrm{ydx}=(\sqrt{4-2^2}(2))-(\sqrt{4-(-2{)}^2}(-2))-{\int }_{-2}^2\left(\frac{4-{\mathrm{x}}^2-4}{\sqrt{4-{\mathrm{x}}^2}}\right)\mathrm{d}\mathrm{x}\end{array}$

$\Rightarrow {\int }_{-2}^{2}ydx=-{\int }_{-2}^2(\frac{4-x^2}{\sqrt{4-x^2}}-\frac{4}{\sqrt{4-x^2}})dx$

$$\begin{gathered} \Rightarrow {\int }_{-2}^{2}ydx=-{\int }_{-2}^2\sqrt{4-x^2}dx+{\int }_{-2}^2\frac{4}{\sqrt{4-x^2}}dx \\ \text{ But }y=\sqrt{4-x^2} \end{gathered}$$

$\Rightarrow {\int }_{-2}^{2}ydx=-{\int }_{-2}^{2}ydx+{\int }_{-2}^2\frac{4}{\sqrt{4-x^2}}dx$

$\Rightarrow {\int }_{-2}^{2}ydx+{\int }_{-2}^{2}ydx={\int }_{-2}^2\frac{4}{\sqrt{4-x^2}}dx$

$\Rightarrow {\int }_{-2}^{2}ydx=2{\int }_{-2}^2\frac{1}{\sqrt{2^2-x^2}}dx$
$\begin{array}{rl}& \text{ We know that }\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\frac{x}{a}\\ & \Rightarrow {\int }_{-2}^2\mathrm{y}\mathrm{dx}=2{[{\sin }^{-1}\frac{\mathrm{x}}{2}]}_{-2}^2\\ & \Rightarrow {\int }_{-2}^2\mathrm{y}\mathrm{dx}=2({\sin }^{-1}\frac{2}{2}-{\sin }^{-1}\frac{-2}{2})\\ & \Rightarrow {\int }_{-2}^2\mathrm{y}\mathrm{dx}=2(\frac{\pi }{2}-(-\frac{\pi }{2}))\\ & \Rightarrow {\int }_{-2}^2\mathrm{y}\mathrm{dx}=2(\frac{\pi }{2}+\frac{\pi }{2})\\ & \Rightarrow {\int }_{-2}^2\mathrm{y}\mathrm{dx}=2\pi \\ & \text{ Hence area is }2\pi {\text{ unit }}^2\end{array}$

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 $y^2=4x$ 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

$\begin{array}{rl}& \Rightarrow {\int }_0^1\mathrm{ydx}={\int }_0^{1}2{\mathrm{x}}^{\frac{1}{2}}\mathrm{dx}\\ & \Rightarrow {\int }_0^1\mathrm{ydx}=2{\left[\frac{{\mathrm{x}}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_0^1\\ & \Rightarrow {\int }_0^1\mathrm{ydx}=2{\left[\frac{{\mathrm{x}}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1\\ & \Rightarrow {\int }_0^1\mathrm{ydx}=2\frac{2}{3}{\left[{\mathrm{x}}^{\frac{3}{2}}\right]}_0^1\\ & \Rightarrow {\int }_0^1\mathrm{ydx}=\frac{4}{3}(1^{\frac{3}{2}}-0)\\ & \Rightarrow {\int }_0^2\mathrm{ydx}=\frac{4}{3}\end{array}$
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
$$\begin{gathered} Putx=0\Rightarrow 2y=5(0)+7 \\ \Rightarrow \mathrm{y}=\frac{7}{2} \\ Puty=0 \\ \Rightarrow 2(0)=5x+7 \\ \Rightarrow 5x=-7 \\ \Rightarrow x=-\frac{7}{5} \end{gathered}$$
$Hence(0,\frac{7}{2})and(-\frac{7}{5},0)$ 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 $\Rightarrow y=1/2(5x+7)$ Integrate from 2 to 8

$\Rightarrow {\int }_2^8\mathrm{ydx}=\frac{1}{2}{\int }_2^8(5\mathrm{x}+7)\mathrm{dx}$

$\Rightarrow {\int }_2^8\mathrm{ydx}=\frac{1}{2}{[\frac{5{\mathrm{x}}^2}{2}+7\mathrm{x}]}_2^8$

$\Rightarrow {\int }_2^8\mathrm{ydx}=\frac{1}{2}[(\frac{5(8{)}^2}{2}+7(8))-(\frac{5(2{)}^2}{2}+7(2))]$

$\Rightarrow {\int }_2^8\mathrm{ydx}=\frac{1}{2}[(160+56)-(10+14)]$
$\begin{array}{rl}& \Rightarrow {\int }_2^8\mathrm{ydx}=\frac{1}{2}(192)\\ & \Rightarrow {\int }_2^8\mathrm{ydx}=96\\ & \text{ Hence the area bounded by given lines is }96{\text{ unit }}^2\end{array}$

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
$$\begin{gathered} \Rightarrow y^2=x-1 \\ {\mathrm{y}}^2=\mathrm{x}-1\text{ is equation of a parabola} \end{gathered}$$

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

Now observe that in $\mathrm{y}=\sqrt{\mathrm{x}-1}\mathrm{x}\ge 1and\mathrm{y}$ 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
$$\begin{gathered} \Rightarrow {\int }_1^{5}y\mathrm{dx}={\int }_1^5(\mathrm{x}-1{)}^{\frac{1}{2}}\mathrm{dx} \\ \Rightarrow {\int }_1^{5}ydx={\left[\frac{(x-1{)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_1^5 \\ \Rightarrow {\int }_1^5\mathrm{ydx}={\left[\frac{(\mathrm{x}-1{)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_1^5 \end{gathered}$$

$\begin{array}{l}\Rightarrow {\int }_1^{5}ydx=\frac{2}{3}{[(x-1{)}^{\frac{3}{2}}]}_1^5\\ \Rightarrow {\int }_1^{5}ydx=\frac{2}{3}((5-1{)}^{\frac{3}{2}}-0)\\ \Rightarrow {\int }_1^{5}ydx=\frac{2}{3}{\left(2^2\right)}^{\frac{3}{2}}\\ \Rightarrow {\int }_1^{5}ydx=\frac{2}{3}\times 2^3\\ \Rightarrow {\int }_1^{5}ydx=\frac{16}{3}\\ \text{ Hence area bounded }=\frac{16}{3}{\text{ unit }}^2\end{array}$

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

$\begin{array}{rl}& \Rightarrow y^2=a^2-x^2\\ & \Rightarrow x^2+y^2=a^2\end{array}$
The above equation is of a circle having centre as $(0,0)$ and radius a .Now in $\mathrm{y}=\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}-a\le \mathrm{x}\le \mathrm{a}$ and $\mathrm{y}\ge 0$ 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

${\int }_a^{b}uvdx={[u\int vdx]}_a^b-{\int }_a^b(u^{\prime }\int vdx)dx$

Here $u=\sqrt{a^2-x^2}$ and $\mathrm{v}=1$
Hence

$\begin{array}{rl}& {\mathrm{u}}^{\prime }=\frac{1}{2}{({\mathrm{a}}^2-{\mathrm{x}}^2)}^{\frac{1}{2}-1}(-2\mathrm{x})=\frac{-\mathrm{x}}{\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}}\\ & \Rightarrow {\int }_0^{a}ydx={\left[\sqrt{a^2-x^2}{\int }_0^{a}dx\right]}_0^a-{\int }_0^a\left(\frac{-x}{\sqrt{a^2-x^2}}{\int }_0^{a}dx\right)dx\\ & \Rightarrow {\int }_0^{a}ydx={[\sqrt{a^2-x^2}(x)]}_0^a-{\int }_0^a\left(\frac{-x^2}{\sqrt{a^2-x^2}}\right)\mathrm{d}x\\ & \Rightarrow {\int }_0^{a}ydx=(\sqrt{a^2-a^2}(a))-(\sqrt{a^2-0^2}(0))-{\int }_0^a\left(\frac{a^2-x^2-a^2}{\sqrt{a^2-x^2}}\right)\mathrm{d}x\\ & \Rightarrow {\int }_0^{a}ydx=-{\int }_0^a(\frac{a^2-x^2}{\sqrt{a^2-x^2}}-\frac{a^2}{\sqrt{a^2-x^2}})dx\\ & \Rightarrow {\int }_0^{a}ydx=-{\int }_0^a\sqrt{a^2-x^2}dx+{\int }_0^a\frac{a^2}{\sqrt{a^2-x^2}}dx\\ & \text{ But }y=\sqrt{a^2-x^2}\end{array}$