Integral Calculus Question 243
Question: What is the value of $ \int_1^{2}{e^{x}( \frac{1}{x}-\frac{1}{x^{2}} )dx} $ ?
Options:
A) $ e( \frac{e}{2}-1 ) $
B) $ e(e-1) $
C) $ e-\frac{1}{e} $
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ I=\int_1^{2}{e^{x}( \frac{1}{x}-\frac{1}{x^{2}} )dx} $ $ =\int\limits_1^{2}{e^{x}(f(x)+f’(x))dx} $ where $ f(x)=\frac{1}{x} $ $ =e^{x}f. (x) |_1^{2} $
$ \therefore I=. \frac{e^{x}}{x} |_1^{2}=\frac{e^{2}}{2}-e=e( \frac{e}{2}-1 ) $
BETA
