Question

Find the area bounded by the curves $y=-2x^2\text{and}y=2x^2-1$

• A. $\frac{1}{3}sq.units$
• B. $\frac{2}{3}sq.units$
• C. $2sq.units$
• D. $\frac{1}{2}sq.units$

Answer 1

The correct option is B $\frac{2}{3}sq.units$


Point of intersection of the two curve,
$-2x^2=2x^2-1\Rightarrow x=\pm \frac{1}{2}$
$I=2\times \underset{0}{\overset{\frac{1}{2}}{\int }}\left[\right(2x^2-1)-(-2x^2\left)\right]dx\Rightarrow I=2{[\frac{4x^3}{3}-x]}_0^{\frac{1}{2}}\Rightarrow I=-\frac{2}{3}$

Hence the required area will be $=\frac{2}{3}sq.units$