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The area (in sq. units) of the region bounded by the curve $x^{2}=4y$ and the straight line $x=4y-2$ is :
Option 1)

$\frac{3}{4}$
Option 2)

$\frac{7}{8}$
Option 3)

$\frac{5}{4}$
Option 4)

$\frac{9}{8}$

Answers (1)

best_answer

Area between two curves -

If we have two functions $f\left ( x \right )\: and\:g \left ( x \right )$ .Area between two curves are

$\int_{a}^{b}\left [ g\left ( x \right )-f\left ( x \right ) \right ]dx$

- wherein

$x^{2}=4y$

$x=4y-2$

$4y=x+2$

Point of intersection,

$x^{2}=x+2$

$\Rightarrow x^{2}-x+2=0$

$\Rightarrow \left ( x-2 \right )\left ( x+1 \right )=0$

$Area=\int_{-1}^{2}\left [ \left ( \frac{x+2}{4} \right )-\left ( \frac{x^{4}}{4} \right ) \right ]dx$

$=\frac{1}{4}\int_{-1}^{2}\left ( x+2-x^{2} \right )dx$

$=\frac{1}{4}\left [ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right ]_{-1}^{2}$

$=\frac{1}{4}\times \frac{27}{6}=\frac{9}{8}$

Option 1)

$\frac{3}{4}$

Option 2)

$\frac{7}{8}$

Option 3)

$\frac{5}{4}$

Option 4)

$\frac{9}{8}$

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