Q. The area bounded by the curves $ y = (x - 1)^2 $, $y = (x + 1)^2$ and $ y = \frac{1}{4} $ is
IIT JEEIIT JEE 2005Application of Integrals
Solution:
The curves $ y = (x - 1)^2 $, $y = (x + 1)^2$ and $ y = \frac{1}{4} $ are
shown as
where, points of intersection are
$\, \, \, \, \, \, \, (x - 1)^2= \frac{1}{4} \Rightarrow x = \frac{1}{2}$
and $\, \, \, \, \, (x + 1)^2= \frac{1}{4} \Rightarrow x = - \frac{1}{2}$
i.e $\, \, \, \, \, \, \, \, Q \bigg (\frac{1}{2},\frac{1}{4}\bigg)\, \, and\, R \bigg (-\frac{1}{2},\frac{1}{4}\bigg)$
$ \therefore Required\, \, area = 2 \int \limits_0^{1/2} \bigg[(x - 1)^2 - \frac{1}{4}\bigg] dx $
$\, \, \, \, = 2 \bigg[\frac{(x - 1)^3}{3} - \frac{1}{4}x\bigg]_0^{1/2}$
$\, \, \, \, \, \, \, \, = 2 \bigg[-\frac{1}{8.3} - \frac{1}{8} - \bigg(-\frac{1}{3}-0\bigg)\bigg] = \frac{8}{24} = \frac{1}{3}sq units$
