Definite Integration Question 149
Question: Let $ \int_0^{1}{f(x)dx=1,} $
$ \int_0^{1}{xf(x)dx=a} $ and $ \int_0^{1}{x^{2}f(x)dx=a^{2},} $ then the value of $ \int_0^{1}{{{(x-a)}^{2}}f(x)dx=} $
[IIT 1990]
Options:
A) 0
B) $ a^{2} $
C) $ a^{2}-1 $
D) $ a^{2}-2a+2 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_0^{1}{{{(x-a)}^{2}}f(x)dx=\int_0^{1}{x^{2}f(x)dx+a^{2}\int_0^{1}{f(x)dx}}}-\int_0^{1}{2axf(x)dx} $
= $ a^{2}+a^{2}-2a\times a=0 $ .
BETA
