_ - / 4 ^ / 2 e ^ -x x d x= - Vidyadip

MCQ

$\int_{-\pi / 4}^{\pi / 2} e ^{-x} \sin x d x=$

✓
$-\frac{1}{2} e ^{\frac{-\pi}{2}}$
B
$-\frac{\sqrt{2}}{2} e ^{\frac{-\pi}{4}}$
C
$-\sqrt{2}\left( e ^{\frac{-\pi}{4}}+ e ^{\frac{-\pi}{4}}\right)$
D
$0$

✓

Answer

Correct option: A.

$-\frac{1}{2} e ^{\frac{-\pi}{2}}$

(A)
Let $I =\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} e ^{-x} \sin x d x$
$\therefore \quad I =\left[- e ^{-x} \sin x- e ^{-x} \cos x\right]_{-\pi / 4}^{\pi / 2}-\int_{\frac{-\pi}{4}}^{\frac{\pi}{2}} e ^{-x} \sin x d x$
$\begin{array}{l}\Rightarrow 2 I =\left[ e ^{-x}(-\sin x-\cos x)\right]_{-\pi / 4}^{\pi / 2} \\ \Rightarrow I =\frac{1}{2}\left[ e ^{\frac{-\pi}{2}}(-1-0)-\left\{ e ^{\frac{\pi}{4}}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\right\}\right] \\ \Rightarrow I =-\frac{1}{2} e ^{\frac{-\pi}{2}}\end{array}$

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