How does your answer to \(\displaystyle{\int_a^b \sqrt{1-x^2} \, dx}\) relate to the following diagram?

Firstly we might notice that the diagram has an area shaded between lines which could be \(x=a\) and \(x = b\), so this could be represented by an integral between \(a\) and \(b\). The curve appears to be a semicircle with radius \(1\). A unit circle has equation \(x^2 + y^2 = 1\), but if this is rearranged to find \(y\) then we get \(y = \pm\sqrt{1-x^2}\). The positive root, gives us a semicircle above the \(x\)-axis as all the \(y\)-values must be positive. Therefore the shaded area is represented by the integral \[\displaystyle{\int_a^b \sqrt{1-x^2} \, dx},\] which we have already worked out using the substitution \(x = \cos \theta\) to get \[\displaystyle{\int_a^b \sqrt{1-x^2} \, dx} = \frac{1}{2}b\sqrt{1-b^2} - \frac{1}{2}\arccos b - \frac{1}{2}a\sqrt{1-a^2} + \frac{1}{2}\arccos a.\]

#### Finding the area with geometry

As it is part of a semicircle, we can break the area down into more familiar shapes. We can find the area of the triangle and the sector shaded below.

The green triangle has area \(\frac{1}{2}b\sqrt{1-b^2}\) and the orange sector has area \(\frac{1}{2}(\arccos a - \arccos b)\), (to see why, look at the suggestion section). We can also find the area of the triangle with base \(a\) which is \(\frac{1}{2}a\sqrt{1-a^2}\).

The total area is found by adding the orange sector and the green triangle with base \(b\), then subtracting the triangle with base \(a\). This gives us \[\frac{1}{2}b\sqrt{1-b^2} + \frac{1}{2}(\arccos a - \arccos b) - \frac{1}{2}a\sqrt{1-a^2}.\]

The geometric pieces can be thought about in a number of ways. You might have started from the answer of the integral which can be written in the form \[\displaystyle{\int_a^b \sqrt{1-x^2} \, dx} = F(b) - F(a),\] where \(F(b) = \frac{1}{2}b\sqrt{1-b^2} - \frac{1}{2}\arccos b\) and \(F(a) = \frac{1}{2}a\sqrt{1-a^2} - \frac{1}{2}\arccos a.\)

Can you see which areas are represented by \(F(b)\) and \(F(a)\)?