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If $\int f(x)dx =g(x)$ then $\int f^{-1}(x)dx $ is equal to

(1) $g^{-1}(x)$

(2) $xf^{-1}(x)-g(f^{-1}(x))$

(3) $xf^{-1}(x)-g^{-1}(x)$

(4) $f^{-1}(x)$

My approach is as follows: Let $f(x)=y$, therefore $f^{-1}(y)=x$, $\int f^{-1}(f(x))dx =g(f(x))$

On differentiating we get $x=g'(f(x))f'(x)$

After this step, I am not able to proceed.

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    $\begingroup$ Why not try an example to see whether any of those formulas work? $\int x^2\,dx=x^3/3+C, $\int\sqrt x\,dx=(2/3)x^{3/2}+C$, do any of the formulas work in this case? $\endgroup$ Jul 11, 2020 at 6:51

4 Answers 4

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There's a nice visual computation of the antiderivative of an inverse function: $$F(x) := \int_0^x f^{-1}(t) dt$$ is an antiderivative for $f^{-1}(x)$, and for $x = a$, $F(a)$ is equal to the green area in the picture below: area under the curve of the inverse function

If we could figure out the blue area, we would be set, because

\begin{align*} \text{ (green area) } &= \text{ (rectangle area) } - \text{ (blue area) } \\ F(a) &= af^{-1}(a) - \text{ (blue area) }.\ \end{align*}

But if we reflect this picture across the line $y = x$, we see the blue area is just the antiderivative of $f$, namely $g(x) := \int_0^x f(t) dt$, evaluated at $f^{-1}(a)$:

antiderivative of the original function

So we get $$\text{ (blue area) } = g(f^{-1}(a)),$$

and plugging this in, we get that $F(a)$ is equal to the second of the four choices:

\begin{align*} F(a) &= af^{-1}(a) - \text{ (blue area) } \\ &= af^{-1}(a) - g(f^{-1}(a)). \ \end{align*}


† To make the picture look nice, we assumed $f(0) = f^{-1}(0) = 0$ without loss of generality.

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  • $\begingroup$ Very nice explaination $\endgroup$ Jul 12, 2020 at 3:09
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Ignoring the constant of integration the answer is (2):$$\int f^{-1}(x)dx=\int yf'(y)dy=yf(y)-\int f(y)dy$$ (where I have used integration by parts). Hence $$\int f^{-1}(x)dx=f^{-1}(x)x-g(y)=xf^{-1}(x)-g(f^{-1}(x))$$.

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Given $y=f(x) \implies x=f^{-1}(y)$ Then $$\int f^{-1}(x) dx= \int f^{-1}(y) dy= \int x dy= \int x \frac{dy}{dx} dx=\int xf'(x) dx $$ $$=xf(x)-\int f(x)dx=xf(x)-g(x)+C.$$ Lastly, we have done integration by parts.

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  • $\begingroup$ It’s a very clear answer. You have made every step so clear. $\endgroup$ Jul 12, 2020 at 3:46
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All the other answers have already given you the finest methods of solving the problem. I want to say that as you’re given the options for the integral $\int f^{-1} (x) dx$, it would be a nice thing to differentiate the options to see if we get $f^{-1}(x)$ (it should follow from The Fundamental Theorem of Calculus ).

Let’s try the first option: $$ If ~~~~ \int f^{-1}(x) dx = g^{-1} (x) \\ then~~~~~ f^{-1} (x)= \frac{d}{dx} g^{-1} (x) \\ f^{-1} (x)= \left( \frac{d~g(x)}{dx} \right)^{-1}\\ f^{-1} (x)= \frac{1}{f(x)}$$ (in the third step I once again used the FTC for the function$f(x)$) the last equality is not true in general, therefore this option is not valid.

Let’s try the second option: $$ If~~~~ \int f^{-1}(x) dx = xf^{-1} (x) - g\left( f^{-1}(x)\right) \\ then~~~~ f^{-1}(x)= \frac{d}{dx} \left[xf^{-1} (x) - g\left( f^{-1}(x)\right) \right] \\ f^{-1} (x)= f^{-1}(x) + x \frac{d~f^{-1}(x)}{dx} - f\left(f^{-1} (x)\right) \frac{d~f^{-1}(x)}{dx}\\ f^{-1}(x) = f^{-1}(x) $$ Hence, the second option is correct.

Hope it adds something to this thread!

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