Integral Calculus Question 104
Question: If $ \int_1^{2}{{e^{x^{2}}}dx=a} $ , then $ \int_e^{e^{4}}{\sqrt{\ln x}}dx $ is equal to
Options:
A) $ 2e^{4}-2e-a $
B) $ 2e^{4}-e-a $
C) $ 2e^{4}-e-2a $
D) $ e^{4}-e-a $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ I_1=\int_e^{e^{4}}{\sqrt{\ln x}dx} $ Putting $ t=\sqrt{\ln x}, $ i.e., $ dt=\frac{dx}{2x\sqrt{\ln x}}, $ we get $ dx=2t{e^{t^{2}}}dt $ Or $ \int_e^{e^{4}}{\sqrt{\ln x}}dx=\int_1^{2}{2t^{2}{e^{t^{2}}}dt} $ $ =t{e^{t^{2}}}|_1^{2}-\int_1^{2}{{e^{t^{2}}}dt} $ $ =2e^{4}-e-a $
BETA
