Integral Calculus Question 192
Question: If $ \int{f(x)dx=g(x)+c} $ , then $ \int{{f^{-1}}(x)dx} $ is equal to
Options:
A) $ x{f^{-1}}(x)+C $
B) $ f({g^{-1}})(x))+C $
C) $ x{f^{-1}}(x)-g({f^{-1}})(x))+C $
D) $ {g^{-1}}(x)+C $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let $ I=\int{{f^{-1}}(x)dx} $ and $ {f^{-1}}(x)=t\Rightarrow x=f(t)\Rightarrow dx=f’(t)dt $ Put value of $ dx $ and $ {f^{-1}}(x) $ in I, we get $ I=\int{tf’(t)dt} $ Now, integrate it by parts, $ I=tf(t)-\int{f(t)dt} $ Given, $ \int{f(x)dx=g(x)+C} $
$ \therefore I=tf(t)-[g(t)]+C $ Now, by putting value of t, f (t) and g(t) we get, $ I=x{f^{-1}}(x)-g[{f^{-1}}(x)]+C $
BETA
