Exercise 9.8 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.8$
Question 1.

Find the area of the region bounded by $3 x-2 y+6=0, x=-3, x=1$ and $x$-axis.

Solution:
$
\begin{aligned}
3 x-2 y+6 & =0 \\
y & =\frac{3 x+6}{2}
\end{aligned}
$

To find further limit put $y=0$
We get, $x=-2$
$
\begin{aligned}
\text { Area } & =\int_{-3}^{-2}-y d x+\int_{-2}^1 y d x \\
& =\frac{-1}{2} \int_{-3}^{-2}(3 x+6) d x+\frac{1}{2} \int_{-2}^1(3 x+6) d x \\
& =\frac{-1}{2}\left[\frac{3 x^2}{2}+6 x\right]_{-3}^{-2}+\frac{1}{2}\left[\frac{3 x^2}{2}+6 x\right]_{-2}^1 \\
& =\frac{-1}{2}\left[-6+\frac{9}{2}\right]+\frac{1}{2}\left[\frac{15}{2}+6\right] \\
& =\frac{1}{2}\left[6-\frac{9}{2}+\frac{15}{2}+6\right] \\
& =\frac{1}{2}[12+3]=\frac{15}{2}=7.5 \text { Square units. }
\end{aligned}
$

Question 2.
Find the area of the region bounded by $2 x-y+1=0, y=-1, y=3$ and $y-a x i s$.
Solution:

$
\begin{gathered}
2 \mathrm{x}-\mathrm{y}+1=0 \\
x=\frac{y-1}{2}
\end{gathered}
$

To find further limit put $x=0$, we get $y=1$
$
\begin{aligned}
\text { Area } & =\int_{-1}^1-x d y+\int_{-1}^3 x d y \\
& =\frac{-1}{2} \int_{-1}^1(y-1) d y+\frac{1}{2} \int_1^3(y-1) d y \\
& =\frac{-1}{2}\left[\frac{y^2}{2}-y\right]_{-1}^1+\frac{1}{2}\left[\frac{y^2}{2}-y\right]_1^3 \\
& =\frac{-1}{2}\left[\left(\frac{1}{2}-1\right)-\left(\frac{1}{2}+1\right)\right]+\frac{1}{2}\left[\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right] \\
& =\frac{-1}{2}-\left[\frac{1}{2}-\frac{3}{2}\right]+\frac{1}{2}\left[\frac{3}{2}+\frac{1}{2}\right]=\frac{1}{2}\left[\frac{1}{2}+\frac{3}{2}+\frac{3}{2}+\frac{1}{2}\right] \\
& =\frac{1}{2}[4]=2 \text { square units. }
\end{aligned}
$

Question 3.
Find the area of the region bounded by the curve $2+x-x^2+y=0, x-$ axis, $x=-3$ and $x=3$.

Solution:
To find further limits put $y=0$, we get $x=-1$ and $x=2$

$
\begin{aligned}
\text { Required Area } & =\int_{-3}^{-1} y d x+\int_{-1}^2-y d x+\int_2^3 y d x \\
& =\int_{-3}^{-1}\left(x^2-x-2\right) d x+\int_{-1}^2\left(2+x-x^2\right) d x+\int_2^3\left(x^2-x-2\right) d x \\
& =\left(\frac{x^3}{3}-\frac{x^2}{2}-2 x\right)_{-3}^{-1}+\left(2 x+\frac{x^2}{2}-\frac{x^3}{3}\right)_{-1}^2+\left(\frac{x^3}{3}-\frac{x^2}{2}-2 x\right)_2^3 \\
& =\frac{52}{6}+\frac{27}{6}+\frac{11}{6}=\frac{90}{6} \\
& =15 \text { Square units. }
\end{aligned}
$

Question 4.
Find the area of the region bounded by the line $y=2 x+5$ and the parabola $y=x^2-2 x$.

Solution:
To find point of intersection of the curves $y=2 x+5$ and $y=x^2-2 x$ we get $(-1,3)$ and $(5,15)$

$
\begin{aligned}
& \text { Area }=\int_{-1}^5[f(x)-g(x)] d x \\
& =\int_{-1}^5\left[(2 x+5)-\left(x^2-2 x\right)\right] d x \\
& =\int_{-1}^5\left(4 x+5-x^2\right) d x \\
& =\left[\frac{4 x^2}{2}+5 x-\frac{x^3}{3}\right]_{-1}^5 \\
& =\left(50+25-\frac{125}{3}\right)-\left(2-5+\frac{1}{3}\right) \\
& =75-\frac{125}{3}+3-\frac{1}{3} \\
& =78-42=36 \text { Sq. units. } \\
&
\end{aligned}
$

Question 5.
Find the area of the region bounded between the curves $y=\sin x$ and $y=\cos x$ and the lines $x=0$ and $x=$ $\pi$.
Solution:
To find the points of intersection,
$
\begin{aligned}
& \sin x=\cos x=\frac{1}{\sqrt{2}} \Rightarrow x=\frac{\pi}{4} \\
& \sin x=\cos x=\frac{1}{\sqrt{2}} \Rightarrow x=\frac{5 \pi}{4}
\end{aligned}
$
From the diagram, we see that $\cos x>\sin x$ for $0 \leq x \leq \frac{\pi}{4}$ and $\sin x>\cos x$ for $\frac{\pi}{4} $
\begin{aligned}
\text { Area } & =\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^\pi(\sin x-\cos x) d x \\
& =(\sin x+\cos x)_0^{\frac{\pi}{4}}+(-\cos x-\sin x)_{\frac{\pi}{4}}^\pi \\
& =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)+(1-0)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right) \\
& =2 \sqrt{2} \text { sq. units. }
\end{aligned}
$

Question 6.
Find the area of the region bounded by $y=\tan x, y=\cot x$ and the line $x=0, x=\frac{\pi}{2}, 0$
Solution:
To find the points of intersection of these two curves between 0 to $\frac{\pi}{2}$ is $\frac{\pi}{4}$

$
\begin{aligned}
& =\int_0^{\frac{\pi}{4}} \tan x d x+\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \cot x d x \\
& =(\log \sec x)_0^{\frac{\pi^4}{4}}+(\log \sin x)_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\
& =\log \sec \left(\frac{\pi}{4}\right)-\log \sec (0)+\log \sin \left(\frac{\pi}{2}\right)-\log \sin \left(\frac{\pi}{4}\right) \\
& =\log \sqrt{2}-0+0-\log \frac{1}{\sqrt{2}} \\
& =\log \sqrt{2}+\log \sqrt{2} \\
& =\log (\sqrt{2} \times \sqrt{2})=\log 2
\end{aligned}
$

Question 7.
Find the area of the region bounded by parabola $y^2=x$ and the line $y=x-2$

Solution:

To find the points of intersection solve the two equations $y^2=x$ and $y=x-2$

$
\begin{aligned}
\text { Area } & =\int_{-1}^2[f(y)-g(y)] d y \\
& =\int_{-1}^2\left[y+2-y^2\right] d y=\left[\frac{y^2}{2}+2 y-\frac{y^3}{3}\right]_{-1}^2 \\
& =\left(\frac{4}{2}+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)=\frac{9}{2} \text { Sq. units. }
\end{aligned}
$

Question 8.
Father of a family wishes to divide his square field bounded by $x=0, x=4, y=4$ and $y=0$ along the curve $y^2=4 x$ and $x^2=4 y$ into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
Solution:
To find the points of intersection of the two curves, $y^2=4 x$ and $x^2=4 y$ are $(0,0)$ and $(4,4)$.
Area of the square field $=4 \times 4=16$ sq. units

$\therefore$ Each of three portions must be $=\frac{16}{3}$ sq.units.
Now, Area of the middle portion
(Area between two curves)
$
\begin{aligned}
& =\frac{4}{3}(4)^{3 / 2}-\frac{(4)^3}{12} \\
& =\frac{32}{3}-\frac{64}{12} \\
& =\frac{32}{3}-\frac{16}{3}=\frac{16}{3} \text { sq.units. }
\end{aligned}
$
So, the remaining each of the two parts must be $\frac{16}{3}$ sq.units.
$\therefore$ Yes, It is possible to divide the square field into three equal parts.

Question 9.
The curves $=(x-2)^2+1$ has a minimum point at $P$. A point $Q$ on the curve is such that the slope of $P Q$ is 2. Find the area bounded by the curve and the chord $P Q$.
Solution:
$
\begin{aligned}
& y=(x-2)^2+1 \\
& \frac{d y}{d x}=2(x-2)
\end{aligned}
$

$
\begin{aligned}
\frac{d^2 y}{d x^2} & =2 \\
\text { Put } \frac{d y}{d x} & =0, \text { we get } x \\
\text { At } x & =2 \frac{d^2 y}{d x^2}>0
\end{aligned}
$
$\therefore \mathrm{x}=2$ is a minimum point
$\therefore$ The point $\mathrm{P}$ is $(2,1)$
But slope of $\mathrm{PQ}$ is 2
$\therefore$ Equation of the chord $\mathrm{PQ}$
$
\begin{aligned}
& y-y_1=m\left(x-x_1\right) \\
& y-1=2(x-2) \\
& y-1=2 x-4 \\
& y=2 x-3
\end{aligned}
$

On solving the curve and line we get the point $\mathrm{Q}(4,5)$
$
\begin{aligned}
\text { Required Area } & =\int_2^4(f(x)-g(x)) d x \\
& =\int_2^4\left(2 x-3-\left[(x-2)^2+1\right]\right) d x \\
& =\int_2^4\left(6 x-x^2-8\right) d x \\
& =\left[\frac{6 x^2}{2}-\frac{x^3}{3}-8 x\right]_2^4 \\
& =\left(48-\frac{64}{3}-32\right)-\left(12-\frac{8}{3}-16\right) \\
& =20-\frac{56}{3}=\frac{4}{3} \text { sq. units. }
\end{aligned}
$

Question 10.
Find the area of the region common to the circle $x^2+y^2=16$ and the parabola $y^2=6 x$.

Solution:
To find points of intersection of $x^2+y^2=16$ and $y^2=6 x$ are $\left.(2,2 \sqrt{3})\right)$ and $\left.(2,-2 \sqrt{3})\right)$ Due to symmetrical property,

$
\begin{aligned}
& \text { Area }=2 \int_0^2 \sqrt{6 x} d x+2 \int_2^4 \sqrt{16-x^2} d x \\
& \quad=2 \sqrt{6}\left[\frac{x^{3 / 2}}{3 / 2}\right]_0^2+2\left[\frac{x}{2} \sqrt{16-x^2}+\frac{16}{2} \sin ^{-1}\left(\frac{x}{4}\right)\right]_2^4 \\
& =\frac{8 \sqrt{12}}{3}-2 \sqrt{12}+8 \pi-\frac{8 \pi}{3} \\
& =\frac{4}{3}(4 \pi+\sqrt{3}) \text { sq.units. }
\end{aligned}
$