Question

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

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{3}$


Point of intersection of $y=(x-1{)}^2$ and $y=\frac{1}{4}$ is $(\frac{1}{2},\frac{1}{4})$ and $(\frac{3}{2},\frac{1}{4})$

Required area
$=2\underset{0}{\overset{1/2}{\int }}\left(\right(x-1{)}^2-\frac{1}{4})dx=2{[\frac{(x-1{)}^3}{3}-\frac{1}{4}x]}_0^{1/2}=2[-\frac{1}{24}-\frac{1}{8}+\frac{1}{3}]=\frac{1}{3}$