# Chain Rule & Integration by Substitution

Area of a Circle (equation derived with calculus)
Area of a Circle (equation derived with calculus)

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 $$F(a)$$?

You are watching: Chain Rule & Integration by Substitution. Info created by THVinhTuy selection and synthesis along with other related topics.

Rate this post