Question

The area between the curves $x^2=4y$ and line $x+2=4y$ is

• A. $\frac{9}{8}$
• B. $\frac{9}{4}$
• C. $\frac{9}{2}$
• D. $\frac{5}{4}$

Answer 1

The correct option is A $\frac{9}{8}$

For intersection points, $x+2=4y$

$\Rightarrow y=\frac{x+2}{4}$

$\Rightarrow x^2=4\left(\frac{x+2}{4}\right)$

$\Rightarrow x^2-x-2=0$

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

$\Rightarrow x=2$ or $x=-1$

$\Rightarrow y=\frac{x^2}{4}=1$ or $\frac{1}{4}$

$\Rightarrow$ P.O.I $=(2,1),(-1,\frac{1}{4})$

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

$=\frac{1}{4}{\int }_1^{2}x+2-x^{2}dx$

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

$=\frac{1}{4}[2+4-\frac{8}{3}-(\frac{1}{2}-2+\frac{1}{3})]$

$=\frac{1}{4}[6-8/3+2-5/6]=\frac{1}{4}[8-21/6]=2-7/8=9/8$.

$\Rightarrow$ (A).