Definite Integration Question 393
Question: $ \int _{-\pi /4}^{\pi /2}{{e^{-x}}\sin xdx}= $
[CEE 1993]
Options:
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
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Answer:
Correct Answer: A
Solution:
$ \int _{-\pi /4}^{\pi /4}{{e^{-x}}\sin xdx}=[ \frac{{e^{-x}}}{2}(-\sin x-\cos x) ] _{-\pi /4}^{\pi /2} $
$ =\frac{1}{2}[{e^{-x}}(-\sin x-\cos x)] _{-\pi /4}^{\pi /2} $
$ =\frac{1}{2}[ {e^{-\pi /2}}(-1-0)-{ {e^{\pi /4}}( \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} ) } ]=-\frac{{e^{-\pi /2}}}{2} $ .
BETA
