Integral Calculus Question 349
Question: $ \int_{{}}^{{}}{x^{3}{e^{x^{2}}}dx=} $
[MNR 1980]
Options:
A) $ \frac{1}{2}(x^{2}+1){e^{x^{2}}}+c $
B) $ (x^{2}+1){e^{x^{2}}}+c $
C) $ \frac{1}{2}(x^{2}-1){e^{x^{2}}}+c $
D) $ (x^{2}-1){e^{x^{2}}}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ x^{2}=t\Rightarrow 2x,dx=dt, $ then $ \int_{{}}^{{}}{x^{3}{e^{x^{2}}}dx}=\frac{1}{2}\int_{{}}^{{}}{te^{t}dt} $ $ =\frac{1}{2}[ te^{t}-e^{t} ]+c $ $ =\frac{1}{2}{e^{x^{2}}}(x^{2}-1)+c. $
BETA
