Inverse integrals

Inverse integrals

Excerpt: Students are asked to think about inverse functions in order to find a definite integral involving $\ln x$. Can they find a method to calculate certain integrals without knowing how to integrate the function directly?

Warm-up

Take a look at these graphs.

Say what you see…

Things to think about

• Is there any information you can add to the graphs?

• What links are there between the two diagrams?

• What shapes can you see in the diagrams? Can you work out their areas?

Warm-up ideas

Take a look at these graphs.

Say what you see…

Here are some things you might have noticed.

• There are two different graphs: each has a function and two $x$-coordinates given. There is an area shaded on each graph but they are in different places. One is between the curve and the $y$-axis and the other between the curve and the $x$-axis.

• Neither curve has any $y$-coordinates labelled so we can add these to the diagram. For $y = e^x$, the $y$-coordinates are $2$ and $3$. For $y = \ln x$ they are $\ln 2$ and $\ln 3$. So the same numbers appear in both graphs except the $x$ and $y$-coordinates have swapped over.

• The swapping of coordinates might remind us that there is a relationship between $e^x$ and $\ln x$. They are inverse functions of each other, which we can think about graphically as a reflection in $y = x$.

This symmetry around $y = x$ is not immediately obvious from the diagrams, as the scaling in the $x$ and $y$ directions is different on each sketch. It is important not to make assumptions about shape or symmetry based on sketch graphs.

• There are two rectangles on each graph with areas of $2 \times \ln 2$ and $3 \times \ln 3$ respectively. These areas are the same on both graphs as the $x$ and $y$ coordinates are reflected in $y=x$.

What can we say about the curved regions in the diagrams?

• The symmetry of the graphs implies that the two shaded areas are actually identical as they are reflected across $y=x$. They could be represented by the integral $\displaystyle{\int_2^{3} \ln x \, dx}$ or the integral $\displaystyle{\int_2^{3} \ln y \, dy}.$

Can you see which area is represented by which integral?

• We can also find the area of other regions bounded by the curves. For example, the area A can be represented by the integral

$\displaystyle{\int_{\ln 2}^{\ln 3} e^x \, dx}.$

As inverse functions have symmetry around $y = x$, we know that area A is the same as area B, shaded in the diagram below.

How would we represent B as an integral?

Problem

Use the diagrams to help you find $\displaystyle{\int_2^3 \ln x \, dx}$.

Use a similar method to find $\displaystyle{\int_{\frac{1}{2}}^{\frac{\sqrt3}{2}} \arcsin x \, dx}$.

Solution

This solution will be based on ideas from the warm-up, so you may wish to look at the Warm-up ideas section if you haven’t already done so.

The integral represents the area of the shaded region under the curve $y = \ln x$. However, we also know that because these graphs are inverse functions, the shaded region between $e^x$ and the $y$-axis has the same area.

If we don’t know how to integrate $\ln x$ directly, then we need to use other areas that we do know how to find. We have already calculated the areas of the large rectangle, $3\ln 3$, and the smaller rectangle, $2\ln 2$.

Therefore the L-shaped shaded region has area $3\ln 3 - 2\ln 2$.

The area between the curve, the

$y$

-axis,

$\ln 2$

and

$\ln 3$

is given by the integral

$$\begin{align*} \displaystyle{\int_{\ln 2}^{\ln 3} e^y \, dy} &= \left[e^y \right]_{\ln 2}^{\ln 3} \\ &= 3 - 2 \\ &= 1. \end{align*}$$

Putting all these together means the area represented by

$\displaystyle{\int_2^3 \ln x \, dx}$

can be found by

$$\begin{align*} \int_2^3 \ln x \, dx &= 3\ln 3 - 2\ln 2 - \int_{\ln 2}^{\ln 3} e^y \, dy \\ &= 3\ln 3 - 2\ln 2 - 1 \end{align*}$$

The answer can be written in several different forms. If you combined the logarithms you might have ended up with $\ln \left(\frac{27}{4}\right) - 1$ or $\ln \left(\frac{27}{4e}\right)$.

Remember $\arcsin x$ is the same as $\sin^{-1} x$.

We could have sketched the graph of $y= \sin x$ instead and thought about the area between the curve and the $y$-axis.

In this case we find

$$\begin{align*} \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \arcsin x \, dx &= \frac{\pi}{3}\cdot\frac{\sqrt{3}}{2} - \frac{\pi}{6}\cdot\frac{1}{2} - \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin y \, dy \\ &= \frac{\sqrt{3}\pi}{6} - \frac{\pi}{12} + \frac{1-\sqrt{3}}{2} \end{align*}$$

We have seen that we can find the definite integral of any function if it has an inverse function that is easy to integrate.

Is it also possible to work out the indefinite integral? What happens if you think about the integral $\displaystyle{\int_a^x \ln t \, dt}?$