Question

Let $f$ be a one-to-one continuous function such that $f(2)=3$ and $f(5)=7$. Given $\underset{2}{\overset{5}{\int }}f(x)dx=17$, then the value of the definite integral $\underset{3}{\overset{7}{\int }}{f}^{-1}(x)dx$ equals

• A. 10
• B. 11
• C. 12
• D. 13

Answer 1

Answer: (C)

$y=f(x)\Rightarrow x=f^{-1}(y)$ and $dy=f^{\prime }(x)dx$ now $I=\underset{3}{\overset{7}{\int }}{f}^{-1}(x)dx=\underset{3}{\overset{7}{\int }}{f}^{-1}(y)dy=\underset{2}{\overset{5}{\int }}x{f}^{\prime }(x)dx;($ when $y=3$ then $x=2$ and $y=7$ then $x=5$ ) hence $I=\underset{2}{\overset{5}{\int }}x{f}^{\prime }(x)dx$. Integrating by parts gives, $I={xf(x)|}_2^5-\underset{2}{\overset{5}{\int }}f(x)dx$ $I=5\cdot 7-2\cdot 3-17=35-6-17=12$