Definite Integration Question 496
Question: The value of $ \int _{-1}^{1}{\frac{\sin x-x^{2}}{3-|x|}dx} $ is
[Roorkee 1995]
Options:
A) 0
B) $ 2\int_0^{1}{\frac{\sin x}{3-|x|}dx} $
C) $ 2\int_0^{1}{\frac{-x^{2}}{3-|x|}}dx $
D) $ 2\int_0^{1}{\frac{\sin x-x^{2}}{3-|x|}dx} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int _{-1}^{1}{\frac{\sin x-x^{2}}{3-|x|}}dx=\int _{-1}^{1}{\frac{\sin x}{3-|x|}}dx-\int _{-1}^{1}{\frac{x^{2}}{3-|x|}}dx $
Here, $ f(x)=\frac{\sin x}{3-|x|} $ is an odd function but $ f(x)=\frac{x^{2}}{3-|x|} $ is an even function
$ \therefore I=-\int _{-1}^{1}{\frac{x^{2}}{3-|x|}}dx=-2\int_0^{1}{\frac{x^{2}}{3-|x|}}dx $
$ =2\int_0^{1}{\frac{-x^{2}}{3-|x|}}dx $ .
BETA
