Question

The area bounded by the curves $y={\left(x+1\right)}^2$, $y={\left(x+1\right)}^2$ and $y=\frac{1}{4}$ is

• A. $4$ sq. units
• B. $\frac{1}{6}$ sq. units
• C. $\frac{4}{3}$ sq. units
• D. $\frac{1}{3}$ sq. units

Answer 1

Answer: (D)

We have, $y={\left(x+1\right)}^2$ $...\left(i\right)$, parabola with vertex $\left(-1,0\right)$
and $y={\left(x-1\right)}^2$ $...\left(ii\right)$, parabola with vertex $\left(1,0\right)$.
Solving $\left(i\right)$ and $\left(ii\right)$, we get $x=0$ and $y=1$

Required area $=A=2{\int }_{\frac{1}{4}}^1\left(1-\sqrt{y}\right)dy$
$=2{\left[y-\frac{2}{3}{y}^{3/2}\right]}_{1/4}^1$
$=2\left[\frac{3}{4}-\frac{2}{3}\left(1-\frac{1}{8}\right)\right]$
$=\frac{3}{2}-\frac{7}{6}=\frac{1}{3}$ sq. units.