A) \[-\frac{1}{2}{{e}^{-\pi /2}}\]
B) \[-\frac{\sqrt{2}}{2}{{e}^{-\pi /4}}\]
C) \[-\sqrt{2}({{e}^{-\pi /4}}+{{e}^{-\pi /4}})\]
D) 0
Correct Answer: A
Solution :
\[\int_{-\pi /4}^{\pi /4}{{{e}^{-x}}\sin x\,dx}=\left[ \frac{{{e}^{-x}}}{2}(-\sin x-\cos x) \right]_{-\pi /4}^{\pi /2}\] \[=\frac{1}{2}[{{e}^{-x}}(-\sin x-\cos x)]_{-\pi /4}^{\pi /2}\] \[=\frac{1}{2}\left[ {{e}^{-\pi /2}}(-1-0)-\left\{ {{e}^{\pi /4}}\left( \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \right) \right\} \right]=-\frac{{{e}^{-\pi /2}}}{2}\].
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