Definite Integration Question 331
Question: Let $ f(x) $ be a function satisfying $ {f}’(x)=f(x) $ with $ f(0)=1 $ and $ g(x) $ be the function satisfying $ f(x)+g(x)=x^{2}. $ The value of integral $ \int_0^{1}{f(x)g(x)dx} $ is equal to
[AIEEE 2003; DCE 2005]
Options:
A) $ \frac{1}{4}(e-7) $
B) $ \frac{1}{4}(e-2) $
C) $ \frac{1}{2}(e-3) $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
We have $ f’(x)=f(x)\Rightarrow \frac{f’(x)}{f(x)}=1 $
therefore $ \log f(x)=x+\log c\Rightarrow f(x)=ce^{x} $
Since $ f(0)=1 $ , therefore $ 1=ce^{0}\Rightarrow c=1 $
Thus $ f(x)=e^{x} $ .
Hence $ g(x)=x^{2}-e^{x} $
$ \therefore \int_0^{1}{f(x)g(x)dx=\int_0^{1}{e^{x}(x^{2}}-e^{x})dx} $
$ =e-\frac{1}{2}e^{2}-\frac{3}{2} $ .
BETA
