Question

Area bounded by the curve $x^2=4y$ and the angle straight line $x=4y-2$

• A. $\left(\frac{8}{9}\right)sq.units$
• B. $\left(\frac{9}{8}\right)sq.units$
• C. $\left(\frac{4}{3}\right)sq.units$
• D. none of these

Answer 1

The correct option is B $\left(\frac{9}{8}\right)sq.units$

$x^2=4y$

$y=\frac{x^2}{4}$

Equating both the equations, we get

$\frac{x^2}{4}=\frac{x+2}{4}$

$x^2-x-2=0$

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

Hence $x=2$ & $x=-1$

If $x=2$, $y=1$ & if $x=1$, $y=\frac{1}{4}$ ………………$x=4y-2$………$y=\frac{x+2}{4}$

Hence points of intersection are $(2,1)$ & $(-1,\frac{1}{4})$

The curve moves from $-1$ to $0$ & from $0$ to $2$

Area $=\left[{\int }_{-1}^0\left(\frac{x+2}{4}\right)dx-{\int }_{-1}^0\left(\frac{x^2}{4}\right)dx\right]+\left[{\int }_0^2\left(\frac{x+2}{4}\right)dx\right]$

$={[\frac{1}{4}(\frac{x^2}{2}+2x)-\frac{1}{4}\left(\frac{x^3}{3}\right)]}_{-1}^0+{[\frac{1}{4}(\frac{x^2}{2}+2x)-\frac{1}{4}\left(\frac{x^3}{3}\right)]}_0^2$

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

$=\frac{7}{24}+\frac{10}{12}=\frac{27}{24}=\frac{9}{8}$ sq. units.