Question

The value of $\underset{-4}{\overset{4}{\int }}(x^3\sec x+3)\sqrt{16-x^2}dx$ is

• A. $8\pi$
• B. $16\pi$
• C. $24\pi$
• D. $48\pi$

Answer 1

The correct option is C $24\pi$

$I=\underset{-4}{\overset{4}{\int }}(x^3\sec x+3)\sqrt{16-x^2}dx=\underset{-4}{\overset{4}{\int }}{x}^3\sec x\sqrt{16-x^2}dx+3\underset{-4}{\overset{4}{\int }}\sqrt{16-x^2}dx=\underset{-4}{\overset{4}{\int }}f\left(x\right)dx+3\underset{-4}{\overset{4}{\int }}g\left(x\right)dx$
$\because f\left(x\right)$ is an odd function.
$\therefore \underset{-4}{\overset{4}{\int }}f\left(x\right)dx=0$

And $g\left(x\right)$ is half the area of circle $x^2+y^2=16$
$\therefore \underset{-4}{\overset{4}{\int }}g\left(x\right)dx=\frac{\pi \times 4^2}{2}=8\pi \Rightarrow I=0+3\times 8\pi =24\pi$

Alternate solution:
$3\underset{-4}{\overset{4}{\int }}\sqrt{16-x^2}dx=3{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_{-4}^4=24\pi$