Integral Calculus Question 16
Question: $ \int_{{}}^{{}}{{e^{x\log a}}.\ e^{x}\ dx} $ is equal to
[Kerala (Engg.) 2005]
Options:
A) $ {{(ae)}^{x}}+c $
B) $ \frac{{{(ae)}^{x}}}{\log (ae)}+c $
C) $ \frac{e^{x}}{1+\log a}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{{e^{x\log a}}e^{x}dx}=\int_{{}}^{{}}{{e^{\log a^{x}}}.,e^{x}dx}=\int_{{}}^{{}}{a^{x}e^{x}dx} $ $ =\int_{{}}^{{}}{{{(ae)}^{x}}dx}=\frac{{{(ae)}^{x}}}{\log (ae)}+C. $
BETA
