Integral Calculus Question 387
Question: $ \int_{{}}^{{}}{\sqrt{x}{e^{\sqrt{x}}}\ dx=} $
[Karnataka CET 2004]
Options:
A) $ 2\sqrt{x}-{e^{\sqrt{x}}}-4\sqrt{x}\ {e^{\sqrt{x}}}+c $
B) $ (2x-4\sqrt{x}+4){e^{\sqrt{x}}}+c $
C) $ (2x+4\sqrt{x}+4){e^{\sqrt{x}}}+c $
D) $ (1-4\sqrt{x}){e^{\sqrt{x}}}+c $
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Answer:
Correct Answer: B
Solution:
$ I=\int_{{}}^{{}}{\sqrt{x}.{e^{\sqrt{x}}}}dx $ . Let $ x=t^{2}\Rightarrow dx=2t,dt $
$ \therefore I=2\int_{{}}^{{}}{t^{2}},.,e^{t}dt $
Þ $ I=2(t^{2}.e^{t}-2te^{t}+2e^{t}]+c $
Þ $ I=\frac{1}{\sqrt{2}}\log | \tan ( \frac{x}{2}+\frac{3\pi }{8} ), |+c $ i.e., $ I={e^{\sqrt{x}}}[2x-4\sqrt{x}+4]+c $ .
BETA
