Question

The area bounded by the graphs of functions $f\left(x\right)=x^4-2x^2$ and $g\left(x\right)=2x^2$ is

• A. $\frac{121}{15}$
• B. $\frac{124}{15}$
• C. $\frac{128}{15}$
• D. $\frac{131}{15}$

Answer 1

The correct option is C $\frac{128}{15}$


$f\left(x\right)=x^4-2x^2=x^2(x^2-2)g\left(x\right)=2x^2$
Points of intersection of the curves,
$x^4-2x^2=2x^2\Rightarrow x^4-4x^2=0\Rightarrow x^2(x^2-4)=0\Rightarrow x=0,2,-2$
Area is given by
$A=\underset{-2}{\overset{2}{\int }}\left(f\right(x)-g(x\left)\right)dx\Rightarrow A=\underset{-2}{\overset{2}{\int }}(4x^2-x^4)dx$
$\Rightarrow A=2{\int }_0^2(4x^2-x^4)dx$
$\Rightarrow A=2[\frac{4x^3}{3}-\frac{x^5}{5}{]}_0^2$
$\Rightarrow A=\frac{128}{15}$