Integral Calculus Question 401
Question: $ \int_{{}}^{{}}{{e^{\sqrt{x}}}\ dx} $ is equal to
[MP PET 1998]
Options:
A) $ {e^{\sqrt{x}}}+A $
B) $ \frac{1}{2}{e^{\sqrt{x}}}+A $
C) $ 2(\sqrt{x}-1){e^{\sqrt{x}}}+A $
D) $ 2(\sqrt{x}+1){e^{\sqrt{x}}}+A $ (A is an arbitrary constant)
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int_{{}}^{{}}{{e^{\sqrt{x}}}.,dx} $ . Put $ \sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}},dx=dt\Rightarrow dx=2t,dt $ \ $ I=\int_{{}}^{{}}{e^{t}.,2t,dt}=2,[t,.,e^{t}-e^{t}]+A=2,[\sqrt{x},.,{e^{\sqrt{x}}}-{e^{\sqrt{x}}}]+A $
Þ $ I=2(\sqrt{x}-1),.,{e^{\sqrt{x}}}+A $ .
BETA
