Exercise 3.1 - Chapter 3 Integral Calculus II 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Integral Calculus II
Ex $3.1$
Question 1.
Using Integration, find the area of the region bounded by the line $2 y+x=8$, the $\mathrm{x}$-axis, and the lines $\mathrm{x}=2, \mathrm{x}=4$
Solution:
The given lines are $2 y+x=8, x$-axis, $x=2, x=4$

$
\begin{aligned}
\text { Required area } & =\int_2^4 y d x \\
\text { Now } 2 y+x=8 & \Rightarrow y=\frac{8-x}{2} \\
\text { So area } & =\int_2^4\left(\frac{8-x}{2}\right) d x \\
& =\frac{1}{2}\left[8 x-\frac{x^2}{x}\right]_2^4 \\
& =\frac{1}{2}[32-8-16+2]=5 \text { sq. units }
\end{aligned}
$

Question 2.
Find the area bounded by the lines $y-2 x-4=0, y=1, y=3$, and the $y$-axis. Solution:
Given lines are $y-2 x-4=0, y=1, y=3, y$-axis $\mathrm{y}-2 \mathrm{x}=4=0 \Rightarrow \mathrm{x}=\frac{y-4}{2}$
We observe that the required area lies to the left of the y-axis

$
\begin{aligned}
\text { So Area } & =\int_1^3-x d y \\
& =-\int_1^3 \frac{y-4}{2} d y=-\frac{1}{2}\left[\frac{y^2}{2}-4 y\right]_1^3 \\
& =-\frac{1}{2}\left[\frac{9}{2}-12-\frac{1}{2}+4\right]=-\frac{1}{2}(-4)=2 \text { sq.units }
\end{aligned}
$
Question 3.
Calculate the area bounded by the parabola $y^2=4 a x$ and its latus rectum. Solution:
Given parabola is $y^2=4 a x$
Its focus is $(\mathrm{a}, 0)$
Equation of latus rectum is $\mathrm{x}=\mathrm{a}$
The parabola is symmetrical about the $\mathrm{x}$-axis
Required area $=2$ [Area in the first quadrant between the limits $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{a}$ ]

$
\begin{aligned}
& =2 \int_0^a y d x \\
& =2 \int_0^a \sqrt{4 a x} d x \\
& =2(2 \sqrt{a}) \int_0^a x^{\frac{1}{2}} d x \\
& =4 \sqrt{a}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^a \\
& =(4 \sqrt{a}) \frac{2}{3} a^{\frac{3}{2}}=\frac{8}{3} a^2 \text { sq.units }
\end{aligned}
$
Question 4.
Find the area bounded by the line $\mathrm{y}=\mathrm{x}$, the $\mathrm{x}$-axis and the ordinates $\mathrm{x}=1, \mathrm{x}=2$.
Solution:
Given lines are $y=x, x$-axis, $x=1, x=2$

$
\begin{aligned}
\text { Required area } & =\int_1^2 y d x=\int_1^2 x d x=\frac{x^2}{2} \int_1^2=2-\frac{1}{2} \\
& =\frac{3}{2} \text { sq.units }
\end{aligned}
$
Question 5.
Using integration, find the area of the region bounded by the line $y-1=x$, the $\mathrm{x}$ axis, and the ordinates $x=-2, x=3$
Solution:
Given lines are $\mathrm{y}-1=\mathrm{x}, \mathrm{x}$-axis, $\mathrm{x}=-2, \mathrm{x}=3$

$
\begin{aligned}
\text { Required area } & =\int_{-2}^{-1}-y d x+\int_{-1}^3 y d x \\
& =-\int_{-2}^{-1}(x+1) d x+\int_{-1}^3(x+1) d x \\
& =-\left[\frac{(x+1)^2}{2}\right]_2^{-1}+\left[\frac{(x+1)^2}{2}\right]_{-1}^3
\end{aligned}
$
$
\begin{aligned}
& =-\frac{1}{2}[-1]+\frac{1}{2}[16-0] \\
& =\frac{1}{2}+8=\frac{17}{2} \text { sq.units }
\end{aligned}
$
Question 6.
Find the area of the region lying in the first quadrant bounded by the region $y=4 x^2, x=0, y$ $=0$ and $y=4$.
Solution:
The given parabola is $\mathrm{y}=4 \mathrm{x}^2$
$
x^2=\frac{y}{4}
$
comparing with the standared form $\mathrm{x}^2=4$ ay
$
4 a=\frac{1}{4} \Rightarrow a=\frac{1}{16}
$
The parabola is symmetric about $\mathrm{y}$-axis
We require the area in the first quadrant.

$\begin{aligned}
\text { Area } & =\int_0^4 x d y=\int_0^4 \sqrt{\frac{y}{4}} d y=\frac{1}{2} \int_0^4 \sqrt{y} d y \\
& =\frac{1}{2}\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4=\frac{1}{3}(4)^{\frac{3}{2}}=\frac{8}{3} \text { sq.units }
\end{aligned}$

Question 7.
Find the area bounded by the curve $y=x^2$ and the line $y=4$
Solution:

Given the parabola is $y=x^2$ and line $y=4$
The parabola is symmetrical about the y-axis.
So required area $=2$ [Area in the first quadrant between limits $y=0$ and $y=4$ ]

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