Rich example

## Solution

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 $F(a)$?