Question

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

Answer 1

The given equations of curves are $C_1:=x^2=4y$ and $C_2:=x=4y-2$

Now, substitute the value of $4y$ from $C_2$ to $C_1$ to get

$x^2=x+2$

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

$\therefore x=2\text{or}x=-1$

For $x=2,y=1$ and for $x=-1,y=\frac{1}{4}$.

Let $A(2,1)$ and $B(-1,\frac{1}{4})$ represent the points above. These are the points of intersection of these curves as they lie on both the curves.

Now, rearrange $C_1$ and $C_2$ as $C_1:=y=\frac{x^2}{4}$ and $C_2:=y=\frac{x+2}{4}$.

$\therefore$ area under the curve $=\left|{\int }_B^A(C_1-C_2)dx\right|$

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

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

$=\frac{1}{4}\left|(\frac{x^3}{3}-\frac{x^2}{2}-2x){|}_{-1}^2\right|$

$=\frac{1}{4}\left|(\frac{8}{3}-\frac{4}{2}-4-(\frac{-1}{3}-\frac{1}{2}+2))\right|$

$=\frac{1}{4}\left|(\frac{9}{3}-\frac{3}{2}-6)\right|$

$=\frac{9}{8}$ sq. units

Hence, the area under the curve is $\frac{9}{8}$ sq. units.