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

$e^{|\cos 4x|}=\{\begin{array}{cc}{e}^{\cos 4x}& \text{if}0\le 4x\le \frac{\pi }{2}\\ e^{-\cos 4x}& \text{if}\frac{\pi }{2}\le 4x\le \pi \end{array}$
$S=4[{\int }_0^{\frac{\pi }{8}}{e}^{\cos 4x}dx+{\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}{e}^{-\cos 4x}dx]$
$\text{Let}4x=t$
$S=4[\frac{1}{4}{\int }_0^{\frac{\pi }{2}}{e}^{\cos t}dt+\frac{1}{4}{\int }_{\frac{\pi }{2}}^{\pi }{e}^{-\cos t}dt]$
$\text{Let}t=\pi -x\text{for the second integral}$
$S=[{\int }_0^{\frac{\pi }{2}}{e}^{\cos t}dt+{\int }_{\frac{\pi }{2}}^0{e}^{\cos x}(-dx\left)\right]$
$S=2{\int }_0^{\frac{\pi }{2}}{e}^{\cos t}dt=2{\int }_0^{\frac{\pi }{2}}{e}^{\sin t}dt$
$\text{Also,}$
$2\times {\int }_0^{\frac{\pi }{2}}{e}^{0}dt<2{\int }_0^{\frac{\pi }{2}}{e}^{\sin t}dt<2{\int }_0^{\frac{\pi }{2}}{e}^{t}dt$
$\Rightarrow \pi <S<2(e^{\frac{\pi }{2}}-1)$