2 Marks Questions

Question 12 Marks

Find the area of the region bounded by the curve $y^2=9 x, x=2, x=4$ and the $x-$axis in the first quadrant.

Answer

Area of the region bounded by $x=2, x=4$ and the curve
$=\int_2^4 y d x$
$=\int_2^4 3 \sqrt{x} d x$
$=3 \times \frac{2}{3}\left(x^{\frac{3}{2}}\right)_2^4$
$=2\left(4^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right)$
$=2(8-2 \sqrt{2})=16-4 \sqrt{2}$ square units.

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Question 22 Marks

Find the area of the region bounded by the parabola $y^2=4 a x$ and its latus rectum.

Answer

Required Area $\text{OLL'O}$

$=2 \int_o^a y d x$
$=2 \int_o^a \sqrt{4 a x} d x \left(\because y^2=4 a x\right)$
$=2 \times 2 \sqrt{a} \int_o^a \sqrt{x} d x$
$=4 a \sqrt{a}\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)_0^a$
$=4 a \sqrt{a} \times \frac{2}{3} \times a^{\frac{3}{2}}$
$=\frac{8}{3} a^2 \text {}$

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Question 32 Marks

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

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Question 42 Marks

Find the area of the region bounded by $y=|x|$, $x =-3, x =1$ and the $x-$axis.

Answer

racing $y=|x|, x=-3$ and $x=1$,
Required area $=($ Area $\text{OBAO} )+($ Area $\text{ODCO )}$
$=\int_{-3}^0 y d x+\int_0^1 y d x$
$=\int_{-3}^0(-x) d x+\int_0^1 x d x$

$=\left(\frac{-x^2}{2}\right)_{-3}^0+\left(\frac{x^2}{2}\right)_0^1$
$=-\frac{1}{2}\left(0-(-3)^2\right)+\frac{1}{2}\left(1^2-0\right)$
$=-\frac{1}{2}(-9)+\frac{1}{2}(1-0)$
$=\frac{9}{2}+\frac{1}{2}=\frac{10}{2}=5$ square units Ans.

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Question 52 Marks

Find the area bounded by the curve $y=x^3-6 x^2+$ $8 x, x=a, x=b$ and the $x$-axis.

Answer

$\text { Sol. Required area }
=\int_a^b y d x$
$=\int_a^b\left(x^3-6 x^2+8 x\right) d x$
$=\left(\frac{x^4}{4}-\frac{6 x^3}{3}+\frac{8 x^2}{2}\right)_a^b$
$=\left(\frac{x^4}{4}-2 x^3+4 x^2\right)_a^b$
$\therefore \text { Area }=\frac{1}{4}\left(b^4-a^4\right)-2\left(b^3-a^3\right)+4\left(b^2-a^2\right)$

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Question 62 Marks

Find the area of the region bounded by the curve $y=\sin x$ and the $x$-axis where $0 \leq x \leq 2 \pi$.

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Question 72 Marks

Find the area of the region enclosed by the narabola $v^2=4 x$ and the line $x=3$.

Answer

$=2 \int_0^3 y d x$
$=2 \int_0^3 \sqrt{4 x} d x$
$=4 \int_0^3 \sqrt{x} d x$
$=4 \times\left(\frac{2 x^{\frac{3}{2}}}{3}\right)_0^3$
$=\frac{8}{3} \times\left((3)^{\frac{3}{2}}-0\right)$
$=\frac{8}{3} \times 3 \sqrt{3}$
$=8 \sqrt{3}$ square units.

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Question 82 Marks

Find the area of the region enclosed by $x-$axis, the curve $y=\sin ^3 x \cos x$ and the ordinates $x=0$ and $x=\frac{\pi}{2}$.

Answer

The required area bounded with $x-$axis is
$=\int_a^b y d x$
Hence required area $=\int_0^{\pi / 2} \sin ^3 x \cos x d x$
$\because y=\sin ^3 x \cos x$
Let $\sin x=t$
$\therefore \cos x d x=d t$
when $x=0 \text { then } t=0$
when $x=\frac{\pi}{2}$ then $t=1$
$\int_0^1 t^3 d t=\left(\frac{t^4}{4}\right)_0^1=\frac{1}{4}$ square unit.

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Maths 2 Marks Questions

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