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

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, \(2\ln 2\).

shaded area of rectangle height 3 length l n 3 minus the rectangle height 2 length l n 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 \(\ln \left(\frac{27}{4e}\right)\).

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

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

graph of arcsin x with area between graph, x axis, x = one half and x = root 3 over 2 shaded.

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 \[\displaystyle{\int_a^x \ln t \, dt}?\]