Question

Find the area bounded by the curve $x^2=4y$ and the line x = 4y - 2.

Answer 1

Givenn curve is $x^2=4y$ ...(i)

Which represents an upward parabola with vertex (0, 0) and axis along Y - axis and the equation of straight line x = 4y - 2 ...(ii)

For intersection point put the value of 4y from Eq. (i) in Eq. (ii), we get

$x^2=x+2$

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

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

$\Rightarrow x=2-1$

When x = 2, then from Eq. (ii) we get

4y = 2 + 2

$\Rightarrow y=1$

When x = - 1, then from Eq. (ii), we get

$4y=2-1=1\Rightarrow y=\frac{1}{4}$

$\therefore$ The line meets the parabola at the points $B(-1,\frac{1}{4})$ and A (2, 1).

Required area = (Area under the line x = 4y -2) - (Area under the parabola $x^2=4y$)

$={\int }_{-1}^2\left(\frac{x+2}{4}\right)dx-{\int }_{-1}^2\frac{x^2}{4}dx(FromEq.(ii),y=\frac{x+2}{4}andfromEq.(i),y=\frac{x^2}{4})=\frac{1}{4}[\frac{x^2}{2}+2x{]}_{-1}^2-\frac{1}{4}[\frac{x^3}{3}{]}_{-1}^2=\frac{1}{4}\{\frac{2^2}{2}+2\times 2-(\frac{1}{2}-2\left)\right\}-\frac{1}{12}[2^3-(-1{)}^3]=\frac{1}{4}(6+\frac{3}{2})-\frac{1}{12}\times 9=\frac{15}{8}-\frac{3}{4}=\frac{9}{8}squnit$
Therefore, required area is $\frac{9}{8}$ sq unit