Integral Calculus Question 96
Question: Which of the following is incorrect?
Options:
A) $ \int_{a+c}^{b+c}{f(x)dx=\int_a^{b}{f(x+c)dx}} $
B) $ \int_ac^{bc}{f(x)dx=c\int_a^{b}{f(cx)dx}} $
C) $ \int_{-a}^{a}{f(x)dx=\frac{1}{2}\int_{-a}^{a}{f(x)+f(-x)dx}} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ I=\int\limits_{a+c}^{b+c}{f(x)dx.} $ Putting $ x=t+c $ or $ dx=dt, $ we get $ I=\int_a^{b}{f(t+c)dt=\int_a^{b}{f(x+c)dx=\int_ac^{bc}{f(x)dx}}} $ Putting $ x=tc $ or $ dx=c,dt $ , we get $ I=c\int_a^{b}{f(ct)dt=c\int_a^{b}{f(cx)dx}} $ $ f(x)=\frac{1}{2}(f(x)+f(-x)+f(x)-f(-x)) $
$ \therefore \int_{-a}^{a}{f(x)dx=\frac{1}{2}\int_{-a}^{a}{(f(x)+f(-x)+f(x)-f(-x))dx}} $ $ =\frac{1}{2}\int_{-a}^{a}{(f(x)+f(-x))dx+\frac{1}{2}\int_{-a}^{a}{(f(x)-f(-x))}dx} $ $ =\frac{1}{2}\int_{-a}^{a}{(f(x)+f(-x)dx} $ as $ f(x)+f(-x) $ is even and $ f(x)-f(-x) $ is odd.
BETA
