Exercise 9.8-Additional Problems - Chapter 9 - Applications of Integration - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Exercise 9.8-Additional Problems - Chapter 9 - Applications of Integration - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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}
$