Definite integral of an inverse function

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A function of x, f(x) maps x to y . The inverse of this function is $f^{-1} (y)$ which gives back x. f(x) is continuous on the interval [0,2) and is always increasing ( $f'(x) >0)$ and f(0)= 1 and f(2)=33 .

If $\displaystyle\int^{2}_{0} f(x) dx =\dfrac{38}{3}$ find $\displaystyle\int^{33}_{1} f^{-1}(y) dy$

$f(x)=y$

$f^{-1} (y) =x$

The answer is 53.333.

1 solution

Amal Hari
Oct 18, 2019

Since the function is always increasing on the interval given we do not need to concern with one y mapping to two different x within the interval.

Geometrically looking definite integral is the area under the curve from 0 to 2.

consider a rectangle of height f(2) and width 2 , from the question f(2)=33

then the area of the rectangle is = 66

and

$\displaystyle\int^{f(2)}_{f(0)} f^{-1}(y) dy = 2\times f(2) -\displaystyle \int^{2}_{0} f(x) dx$

$\displaystyle\int^{f(2)}_{f(0)} f^{-1}(y) dy = 66 -\dfrac{38}{3} =\dfrac{160}{3} =53.333$

Definite integral of an inverse function

The answer is 53.333.

1 solution

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