MCQ
$\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} = $
- ✓$ - \frac{1}{2}{e^{ - \pi /2}}$
- B$ - \frac{{\sqrt 2 }}{2}{e^{ - \pi /4}}$
- C$ - \sqrt 2 ({e^{ - \pi /4}} + {e^{ - \pi /4}})$
- D$0$
✓
Answer
Correct option: A.
$ - \frac{1}{2}{e^{ - \pi /2}}$
a
(a) $\int_{ - \pi /4}^{\pi /2} {{e^{ - x}}\sin x\,dx} $
(a) $\int_{ - \pi /4}^{\pi /2} {{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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