Definite Integration Question 335
Question: $ \int_0^{1}{f(1-x)dx} $ has the same value as the integral
[SCRA 1990]
Options:
A) $ \int_0^{1}{f(x)dx} $
B) $ \int_0^{1}{f(-x)dx} $
C) $ \int_0^{1}{f(x-1)dx} $
D) $ \int _{-1}^{1}{f(x)dx} $
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ 1-x=t\Rightarrow -dx=dt $ . Also as $ x=0 $ to 1, $ t=1 $ to 0 Therefore, $ \int_0^{1}{f(1-x)dx=\int_1^{0}{f(t)(-dt)}}=\int_0^{1}{f(t)dt=\int_0^{1}{f(x)dx}} $ .
BETA
