In the diagram mark the lines $y=a$ and $y=b$ in addition to $y=f(a)$ and $y=f(b)$ and reflect the graph of $y=f(x)$ across the line $y=x$ to obtain the inverse function $y=f^{-1}(x)$. Extending the lines $y=a$ and $y=b$ helps us to find $x=f(a)$ and $x=f(b)$ the bounds for integration. This gives us the yellow area and thus the formula:

${\displaystyle {\int }_{f(a)}^{f(b)}{f}^{-1}(t)\mathrm{dt}+{\int }_a^{b}f(x)\mathrm{dx}=\mathrm{bf}(b)-\mathrm{af}(a).}$

We can find ${\int }_1^4\sqrt{t}\mathrm{dt}$ by direct integration:

${\displaystyle {\int }_1^4\sqrt{t}\mathrm{dt}={\left[\frac{2}{3}{t}^{\frac{3}{2}}\right]}_1^4=\frac{14}{3}.}$

Or we can use the formula with $f(x)=x^2$, $f(a)=a^2=1$ so $a=1$ and $f(b)=b^2=4$ so $b=2$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\int }_1^4\sqrt{t}\mathrm{dt}}& =-{\int }_1^2{x}^2\mathrm{dx}+(2\times 4-1\times 1)\\ \multicolumn{1}{c}{}& =-{\left[\frac{1}{3}{x}^3\right]}_1^2+7\\ \multicolumn{1}{c}{}& =-\frac{7}{3}+7\\ \multicolumn{1}{c}{}& =\frac{14}{3}\end{array}}$

and this is in agreement with the previous result.

In a similar fashion we find ${\int }_0^1{\sin }^{-1}\mathrm{tdt}$ with $f(x)=\sin x$, $f(a)=\sin a=0$ so $a=0$, and $f(b)=\sin b=1$ so $b=\pi /2$ :

${\displaystyle {\int }_0^1{\sin }^{-1}\mathrm{tdt}+{\int }_0^{\pi /2}\sin x\mathrm{dx}=\frac{\pi }{2}.}$

Hence

${\displaystyle {\int }_0^1{\sin }^{-1}\mathrm{tdt}=\frac{\pi }{2}-[-\cos x{]}_0^{\pi /2}=\frac{\pi }{2}-1.}$

Verification by direct integration:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\int }_0^1{\sin }^{-1}\mathrm{tdt}}& =[t{\sin }^{-1}t+\sqrt{1-t^2}{]}_0^1\\ \multicolumn{1}{c}{}& =(1\times {\sin }^{-1}1+\sqrt{1-1^2})-(0\times {\sin }^{-1}0+\sqrt{1-0^2})\\ \multicolumn{1}{c}{}& =(\frac{\pi }{2}+0)-(0+1)=\frac{\pi }{2}-1\end{array}}$

which is agreement with the previous result.

Editors note: this formula provides a method for carrying out integration where the function cannot be integrated directly but the inverse function can be integrated.