Definite Integration Question 199
Question: If $ y=f(x) $ makes +ve intercept of 2 and 0 unit on x and y axes and encloses an area of 3/4 square unit with the axes then $ \int\limits_0^{2}{xf’(x)dx} $ is
Options:
A) 3/2
B) 1
C) 5/4
D) -3/4
Show Answer
Answer:
Correct Answer: D
Solution:
[d] We have $ \int\limits_0^{2}{f(x)dx=\frac{3}{4}}; $ Now, $ \int\limits_0^{2}{xf’(x)dx=x\int\limits_0^{2}{f’(x)dx}-\int\limits_0^{2}{f(x)dx}} $
$ =[xf(x)]_0^{2}-\frac{3}{4}=2f(2)-\frac{3}{4} $
$ =0-\frac{3}{4}(\because f(2)=0)=-\frac{3}{4}. $
BETA
