Integration of Inverse Functions

God does not care about our mathematical difficulties. He integrates empirically.— Albert Einstein

$\text{Inverse Integral Set-up}$

$\text{Integration of inverse functions can be computed by means of a formula that expresses}$ $\text{the}$ $\text{indefinite integral of the inverse}$ $f^{-1}$ $\text{of a continuous and}$ invertible function $f$, $\text{in terms of}$ $f^{-1}$ $\text{and}$ $\text{an indefinite integration of}$ $f$.

$\text{Statement of The Theorem}$

$\text{Suppose},$ ${\text{I}}_1\text{and}{\text{I}}_2$ $\text{be two intervals of}$ $\mathbb{R}$. $\text{Assuming that}$ $f:I_1\to I_2$ $\text{is a continuous}$ $\text{and invertible function. It follows from the}$ intermediate value theorem $\text{that}$ $f$ $\text{is strictly monotone. Since}$ $f$ $\text{and the inverse function}$ $f^{-1}:I_2\to I_1$ $\text{are continuous,}$ $\text{they}$ $\text{have}$ $\text{indefinite integration by the}$ fundamental theorem of calculus.

$\int f^{-1}(y)dy\text{is the typical notation of inverse integral. We will solve this integral using integration}$ $\text{by parts. Also, we have to use substitution.}$

$\text{So, We have,}$ $$I=\int f^{-1}(y)dy$$ $\text{Let us assume,}x=f^{-1}(y)\text{where}y=f(x)$ $\therefore y=f(x)⟹dy=d(f(x))=df(x)$

$\text{Here we will write our integrand in terms of}x$. $\text{Therefore,}$

$$I=\int f^{-1}(f(x))df(x)=\int xdf(x)$$

$\text{Here, we will apply the integration by parts to solve the above integral.}$$$\int xdf(x)=x\int df(x)-\int [{\displaystyle \frac{d}{dx}}(x)\int df(x)]dx$$

$$=xf(x)-\int f(x)dx$$

$$=xf(x)-F(x)+C$$

$$=y{f}^{-1}(y)-F(f^{-1}(y))+C$$

$$\therefore \boxed{{\displaystyle \int f^{-1}(y)dy=y{f}^{-1}(y)-F\circ f^{-1}(y)+C}}\dots (\ast )$$

$\text{Problem 1}$

$\text{Approch 1 :}$

$\text{If}$ $f$ $\text{be a decreasing and continuous function on}$ $[a,b]$, $\text{then prove that}$

$${\int }_{f(b)}^{f(a)}{f}^{-1}(y)dy=af(a)-bf(b)+{\int }_a^{b}f(x)dx$$

$\text{We shall express}$ ${\int }_{f(b)}^{f(a)}{f}^{-1}(y)dy$ $\text{ in terms of }$ $x$ $\text{ instead of}$ $y$.

$${\int }_{f(b)}^{f(a)}{f}^{-1}(x)dx$$

$\text{Using the fact that},$ $$\int f^{-1}(y)dy=y{f}^{-1}(y)-F(f^{-1}(y))+C$$

$\text{where}$ $F$ $\text{is the indefinite integral of}$ $f$, $\text{we can obtain},$

$${\int }_{f(b)}^{f(a)}{f}^{-1}(x)dx={\left[x{f}^{-1}(x)-F(f^{-1}(x))\right]}_{f(b)}^{f(a)}$$

$$=f(a)a-F(a)-(f(b)b-F(b))$$

$$\boxed{{\displaystyle {\int }_{f(a)}^{f(b)}{f}^{-1}(x)dx=F(a)-F(b)-af(a)+bf(b)}}$$

$\text{Now},$

$${\int }_a^{b}f(x)dx+{\int }_{f(a)}^{f(b)}{f}^{-1}(x)dx$$

$$=F(b)-F(a)+F(a)-F(b)-af(a)+bf(b)$$

$$\therefore {\int }_a^{b}f(x)dx+{\int }_{f(a)}^{f(b)}{f}^{-1}(x)dx=-af(a)+bf(b)$$

$$⟹-{\int }_{f(a)}^{f(b)}{f}^{-1}(x)dx=af(a)-bf(b)+{\int }_a^{b}f(x)dx$$

$$⟹\boxed{{\displaystyle {\int }_{f(b)}^{f(a)}{f}^{-1}(x)dx=af(a)-bf(b)+{\int }_a^{b}f(x)dx}}$$ $\text{Hence, we are done.}$

$\text{We can also do this problem by integration by parts. For the convention of reder, it is shown below.}$

$\text{Approach 2:}$

$\text{Let’s define}$ $f^{-1}(y)=t\iff y=f(t)\iff dy=f^{\prime }(t)dt$ $\text{then we get},$

$${\int }_{f(b)}^{f(a)}{f}^{-1}(y)dy={\int }_b^{a}t{f}^{{}^{\prime }}(t)dt$$

$\text{Then using integration by parts we get},$

$${\int }_b^{a}t{f}^{{}^{\prime }}(t)dt=tf(t){|}_b^a-{\int }_b^{a}f(t)dt=af(a)-bf(b)-{\int }_b^{a}f(x)dx$$

$\text{Changing variable t to x we get that,}$

$${\int }_{f(b)}^{f(a)}{f}^{-1}(y)dy-{\int }_a^{b}f(x)dx=af(a)-bf(b)-{\int }_b^{a}f(x)dx-{\int }_a^{b}f(x)dx=af(a)-bf(b)$$

$$⟹\boxed{{\displaystyle {\int }_{f(b)}^{f(a)}{f}^{-1}(y)dy=af(a)-bf(b)+{\int }_a^{b}f(x)dx}}$$

$\text{Problem 2}$

$f:[0,1]\to [0,1]$ $\text{is continuous f(0)=0,f(1)=1}$. ${\int }_0^{1}f(x)dx=\frac{1}{3}$ $\text{and f is invertible. Find the value of}$

$${\int }_0^1{f}^{-1}(y)dy$$

$\text{There are two ways of doing this problem. Let us explore the geometrical method first.}$ $\text{This method}$ $\text{will provide you a visual understanding.}$

$\text{We know that f(0) = 0, f(1) = 1 and f is invertible which means it always passes the horizontal line}$ $\text{test.}$

$\text{Note that,}$ $\text{An invertible function}$ $f:[0,0]\to [1,1]$ $\text{is either strictly increasing or strictly decreasing.}$ $\text{So, in this case, it is strictly increasing. Because if it is not strictly increasing then at some point,}$ $\text{it goes up and it goes down and in that case it would not pass the horizontal line test. That’s why in}$ $\text{this case, it’s strictly increasing. Moreover, we know that }$ ${\int }_0^{1}f(x)dx=\frac{1}{3}$

$\therefore$ $\text{OBC}={\displaystyle \frac{1}{3}}.$ $\text{As the curve OB i.e., the function}$ $y=f(x)$ $\text{is invertible then we can solve for}$ $x$ $\text{and we get}$ $x=f^{-1}(y).$ $\text{But our objective is to figure out}$ ${\int }_0^1{f}^{-1}(y)dy.$

$\text{If you tilt your head and the interesting thing is this is not the graph of}f\text{but in terms of}y$ $\text{this is the graph of}{f}^{-1}(y).$ $\text{Notice that area of OAB is precisely}$ $\text{ABCO}-\text{OBC}.$

$\therefore \text{Our answer would be ( area of the ABCO}-\text{area of OBC)}=(1-\frac{1}{3})=\frac{2}{3}.$

$\text{Another solution of this problem can be written as,}$

$${\int }_0^1{f}^{-1}(y)dy$$

$\text{Let us substitute}y=f(x)⟹dy=f^{{}^{\prime }}(x)dx.$ $\text{Notice that,}$ $f(0)=0=f(x)⟹x=0.$ $\text{Similarly,}f(1)=1=f(x)⟹x=1.$ $\text{Therefore we can substitute above integral as,}$

$${\int }_0^1{f}^{-1}(y)dy=[xf(x){]}_0^1-{\int }_0^{1}f(x)dx$$

$$=f(1)-{\int }_0^{1}f(x)dx=1-{\displaystyle \frac{1}{3}}={\displaystyle \frac{2}{3}}$$

$$\therefore \boxed{{\displaystyle {\int }_0^1{f}^{-1}(y)dy={\displaystyle \frac{2}{3}}}}$$

$\text{Problem 3}$

$\text{Let us assume A}$ $(x_0,y_0),$ $\text{B}(\sqrt{\frac{y_0}{2}},\frac{y_0}{2})$ $\text{P}(\sqrt{\frac{y_0}{2}},y_0).$ $\text{Moreover it’s given that}$ $y=2x^2\text{or},x=\sqrt{\frac{y}{2}}.$ $\text{The area(OPB),}$

$${\int }_0^{\sqrt{\frac{y_0}{2}}}(2x^2-x^2)dx={\int }_0^{\sqrt{\frac{y_0}{2}}}{x}^{2}dx$$

$$=[{\displaystyle \frac{1}{3}}{x}^3{]}_0^{\sqrt{\frac{y_0}{2}}}$$

$$={\displaystyle \frac{y_0^{\frac{3}{2}}}{6\sqrt{2}}}$$ $\text{The area of OPA will be figuring out with respect to y-axis.}$ $\text{The area (OPA)},$

$${\int }_0^{y_0}(\sqrt{{\displaystyle \frac{y}{2}}}-f^{-1}(y))dy$$

$$={\int }_0^{y_0}\sqrt{{\displaystyle \frac{y}{2}}}dy-{\int }_0^{y_0}{f}^{-1}(y)dy$$

$$=\frac{2}{3\sqrt{2}}{y}_0^{\frac{3}{2}}-{\int }_0^{y_0}{f}^{-1}(y)dy$$

$\text{As, the areas of OAP and OBP are equal. Therefore, }$

$$\frac{2}{3\sqrt{2}}{y}_0^{\frac{3}{2}}-{\int }_0^{y_0}{f}^{-1}(y)dy={\displaystyle \frac{y_0^{\frac{3}{2}}}{6\sqrt{2}}}$$

$\text{Differentiate both side with respect to}{y}_0$

$$⟹\sqrt{{\displaystyle \frac{y_0}{2}}}-f^{-1}(y_0)={\displaystyle \frac{1}{4}}\sqrt{{\displaystyle \frac{y_0}{2}}}$$

$$⟹f^{-1}(y_0)={\displaystyle \frac{3}{4}}\sqrt{{\displaystyle \frac{y_0}{2}}}$$

$$\therefore y_0=f({\displaystyle \frac{3}{4}}\sqrt{{\displaystyle \frac{y_0}{2}}})$$

$\text{Let us assume}{\displaystyle \frac{3}{4}}\sqrt{{\displaystyle \frac{y_0}{2}}}=x⟹y_0=\frac{32x^2}{9},\text{hence}\boxed{{\displaystyle f(x)=\frac{32x^2}{9}}}$

$\text{See Also}$

• Strictly Increasing/Decreasing Functions

• Inverse Function

• Horizontal line test

$\text{References}$

1. Key, E. (Mar 1994). “Disks, Shells, and Integrals of Inverse Functions”. The College Mathematics Journal. 25 (2): 136–138.

2. Bensimhoun, Michael (2013). “On the antiderivative of inverse functions”. arXiv:1312.3839 [math.HO].

3. Parker, F. D. (Jun–Jul 1955). “Integrals of inverse functions”. The American Mathematical Monthly. 62 (6): 439–440. doi:10.2307/2307006. JSTOR 2307006.