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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Find the area of the region enclosed by $y^2=x$ and $y=x-2$.
Solution:
The points of intersection of the parabola $y^2=x$ and the line $y=x-2$ are $(1,-1)$ and $(4,2)$

To compute the region [shown in figure] by integrating with respect to $\mathrm{x}$, we would have to split the region into two parts, because the equation of the lower boundary changes at $x=1$. However if we integrate with respect toy no splitting is necessary.
$
\begin{aligned}
\text { Required 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}-\frac{1}{2}\right)+(4+2)-\left(\frac{8}{3}+\frac{1}{3}\right) \\
& =\frac{3}{2}+6-\frac{9}{3}=\frac{9}{2} \text { sq.units. }
\end{aligned}
$

Question 2.
Find the area bounded by the curve $y=x^3$ and the line $y=x$.

Solution:
The line $y=x$ lies above the curve $y=x^3$ in the first quadrant and $y=x^3$ lies above the line $y=x$ in the third quadrant. To get the points of intersection, solve the curves $y=x^3, y=x \Rightarrow x^3=x$. We get $x=\{0, \pm$ $1\}$

$\begin{aligned}
\text { The required area } & =\mathrm{A}_1+\mathrm{A}_2=\int_{-1}^0[g(x)-f(x)] d x+\int_0^1[f(x)-g(x)] d x \\
& =\int_{-1}^0\left(x^3-x\right) d x+\int_0^1\left(x-x^3\right) d x \\
& =\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1 \\
& =\left(0-\frac{1}{4}\right)-\left(0-\frac{1}{2}\right)+\left(\frac{1}{2}-0\right)-\left(\frac{1}{4}-0\right) \\
& =-\frac{1}{4}+\frac{1}{2}+\frac{1}{2}-\frac{1}{4}=\frac{1}{2} \text { sq. units. }
\end{aligned}$

Question 3.
Find the area of the loop of the curve $3 a y^2=x(x-a)^2$.
Solution:
Put $\mathrm{y}=0$; we get $\mathrm{x}=0$, a
It meets the $\mathrm{x}$ - axis at $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{a}$
$\therefore$ Here a loop is formed between the points $(0,0)$ and $(\mathrm{a}, 0)$ about $\mathrm{x}$-axis. Since the curve is symmetrical about $\mathrm{x}$-axis, the area of the loop is twice the area of the portion above the $\mathrm{x}$-axis.

$
\begin{aligned}
\text { Required area } & =2 \int_0^a y d x \\
& =-2 \int_0^a \frac{\sqrt{x}(x-a)}{\sqrt{3 a}} d x=-\frac{2}{\sqrt{3 a}} \int_0^a\left[x^{3 / 2}-a \sqrt{x}\right] d x \\
& =-\frac{2}{\sqrt{3 a}}\left[\frac{2}{5} x^{5 / 2}-\frac{2 a}{3} x^{3 / 2}\right]_0^a=\frac{8 a^2}{15 \sqrt{3}}=\frac{8 \sqrt{3} a^2}{45} \\
\text { Required area } & =\frac{8 \sqrt{3} a^2}{45} \text { sq. units. }
\end{aligned}
$

Question 4.
Find the area between the line $y=x+1$ and the curve $y=x^2-1$
Solution:
To get the points of intersection of the curves we should solve the equations $y=x+1$ and $y=x^2-1$.
we get, $\mathrm{x}^2-1=\mathrm{x}+1$

$
x^2-x-2=0 \Rightarrow(x-2)(x+1)=0
$
$
\mathrm{x}=-1 \text { or } \mathrm{x}=2
$
$\therefore$ The line intersects the curve at $\mathrm{x}=-1$ and $\mathrm{x}=2$.
$
\begin{aligned}
\text { Required area } & =\int_a^b\left[\begin{array}{c}
f(x)-g(x) \\
\text { above below }
\end{array}\right] d x \\
& =\int_{-1}^2\left[(x+1)-\left(x^2-1\right)\right] d x \\
& =\int_{-1}^2\left[2+x-x^2\right] d x=\left[2 x+\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^2 \\
& =\left[4+2-\frac{8}{3}\right]-\left[-2+\frac{1}{2}+\frac{1}{3}\right]=\frac{9}{2} \text { sq. units. }
\end{aligned}
$