Integral Calculus Question 35
Question: $ \int_{{}}^{{}}{x^{5}.{e^{x^{2}}}dx=} $
Options:
A) $ \frac{1}{2}x^{4}{e^{x^{2}}}-x^{2}{e^{x^{2}}}+{e^{x^{2}}}+c $
B) $ \frac{1}{2}x^{4}{e^{x^{2}}}+x^{2}{e^{x^{2}}}+{e^{x^{2}}}+c $
C) $ \frac{1}{2}x^{4}{e^{x^{2}}}-x^{2}{e^{x^{2}}}-{e^{x^{2}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ x^{2}=t\Rightarrow 2x,dx=dt, $ then $ \int_{{}}^{{}}{x^{5}{e^{x^{2}}}dx}=\frac{1}{2}\int_{{}}^{{}}{t^{2}e^{t}dt}=\frac{1}{2}[ e^{t}t^{2}-2\int_{{}}^{{}}{te^{t}dt} ]+c $ $ =\frac{t^{2}e^{t}}{2}-[ te^{t}-e^{t} ]+c=\frac{1}{2}x^{4}{e^{x^{2}}}-x^{2}{e^{x^{2}}}+{e^{x^{2}}}+c. $
BETA
