Question

Find the area bounded by the curve x2=4y and the line x=4y-2

Answer 1

The area bounded by the curve,x2=4y,and line,x=4y-2,is represented by the

shaded area OBAO.

Let A and B be the points of intersection of the line and parabola.

Coordinates of point A are$(-1,\frac{1}{4}).$

Coordinates of point B are(2,1).

We draw AL and BM perpendicular to x-axis.

It can be observed that,

Area OBAO=Area OBCO+Area OACO...(1)

Then,Area OBCO=Area OMBC-Area OMBO

=$∫_0^2 \frac{x+2}{4} dx-∫_0^2 \frac{x^2}{4}dx$

=$\frac{1}{4}[\frac{x^2}{2}+2x]_0^2-\frac{1}{4}[\frac{x^3}{3}]_0^2$

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

=$\frac{3}{2}-\frac{2}{3}$

=$\frac{5}{6}$

Similarly,Area OACO=Area OLAC-Area OLAO

=$∫_0^{-1} \frac{x+2}{4}dx-∫_0^{-1} \frac{x^2}{4}dx$

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

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

=-$\frac{1}{4}[\frac{1}{2}-2]-\frac{1}{12}$

=$\frac{1}{2}-\frac{1}{8}-\frac{1}{12}$

=$\frac{7}{24}$

Therefore,required area=$(\frac{5}{6}+\frac{7}{24})$= $\frac{9}{8}$ units.