Question

Let $S$ be the area bounded by $y=e^{|\cos 4x|},x=0,y=0$ and $x=\pi$, Then

• A. $S=2{\int }_0^{\frac{\pi }{2}}{e}^{\sin t}dt$
• B. $S>\frac{\pi }{2}$
• C. $S<2(2^{\frac{\pi }{2}}-1)$
• D. $S\le 1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A

$S=2{\int }_0^{\frac{\pi }{2}}{e}^{\sin t}dt$

B

$S>\frac{\pi }{2}$

C

$S<2(2^{\frac{\pi }{2}}-1)$

$S=4[{\int }_0^{\frac{\pi }{8}}{e}^{\cos 4x}dx+{\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}{e}^{-\cos 4x}dx]$ [Area is repeated 4 times]

Put $4x=t$

$\Rightarrow S=2{\int }_0^{\frac{\pi }{2}}{e}^{\sin t}dt$

Comparing it with $2{\int }_0^{\frac{\pi }{2}}{e}^{t}dt$, we get,

$\Rightarrow \pi <S<2(e^{\frac{\pi }{2}}-1)$