## Fundamental theorem of calculus and the area of the unit circle

The area of the unit circle can be estimated from below and from above by rectangles and octagons as illustrated in the following picture: This idea can of course be continued to obtain better and better estimates. But how to calculate the area precisely? In other words, how to derive the familiar formula for the area of a circle?

Besides learning how to write LaTeX in WordPress, the purpose of this first blog post is to demonstrate how the exact area can be calculated. Mathematics used here should be accessible to anyone familiar with basic trigonometric identities and concepts such as continuity, derivative, and integral.

Recall that the integral $\int_a^b f(t) \,\mathrm{d}t$ of a positive function $f$ is, by definition, the area beneath the graph of $f$ between $a$ and $b$. Loosely speaking, the integral is defined to be the limit of the area of better and better approximating rectangles. The definition is therefore unsuitable for calculating the exact values of integrals in most cases. Assuming $a and $f$ to be a continuous function as in the above picture, we see that $\displaystyle hf(x) \le \int_a^{x+h} f(t) \,\mathrm{d}t - \int_a^x f(t) \,\mathrm{d}t \le hf(x+h)$

and therefore, by the continuity of $f$, $\displaystyle 0 \le \frac{1}{h}\biggl(\int_a^{x+h} f(t) \,\mathrm{d}t - \int_a^x f(t) \,\mathrm{d}t\biggr) - f(x) \le f(x+h) - f(x) \to 0$

as $h \to 0$. So, writing $G(x) = \int_a^x f(t) \,\mathrm{d}t$, we have $\displaystyle f(x) = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = G'(x).$

Let us formulate our observation as a theorem. Note that, for simplicity, we assumed $f$ to be positive and increasing on a neighborhood of $x$. To get rid of these assumptions is left as an exercise.

Theorem 1 (Fundamental Theorem of Calculus). Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. If $G \colon [a,b] \to \mathbb{R}$ is a function such that $\displaystyle G(x) = \int_a^x f(t) \,\mathrm{d}t$

for all $x \in [a,b]$, then $G$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $\displaystyle G'(x) = f(x)$

for all $x \in (a,b)$.

The following corollary introduces us a method to calculate exact values of integrals.

Corollary 2. Let $f \colon [a,b] \to \mathbb{R}$ be a continuous function. If there exists a continuous function $F \colon [a,b] \to \mathbb{R}$ that satisfies $F'(x) = f(x)$ for all $x \in (a,b)$, then $\displaystyle \int_a^b f(t) \,\mathrm{d}t = F(b) - F(a).$

Proof. Let $G(x) = \int_a^x f(t) \,\mathrm{d}t$ for all $x \in [a,b]$. By the fundamental theorem of calculus, also $G'(x) = f(x)$ for all $x \in (a,b)$. Therefore, by the linearity of the derivative operation, $\displaystyle (F-G)'(x) = F'(x) - G'(x) = 0$

for all $x \in (a,b)$. The mean value theorem and the continuity of $F$ and $G$ now imply that there exists a constant $c$ such that $(F-G)(x) = c$ for all $x \in [a,b]$. Choosing $x=a$, we see that $\displaystyle F(a) - c = G(a) = \int_a^a f(t) \,\mathrm{d}t = 0$

and therefore, $c = F(a)$. In other words, $(F-G)(x) = F(a)$ and $G(x) = F(x) - F(a)$ for all $x \in [a,b]$, and so $\displaystyle \int_a^b f(t) \,\mathrm{d}t = G(b) = F(b) - F(a). \;\blacksquare$

Example 3. Let $F(x) = \tfrac{1}{2}(x + \sin x \cos x)$ and observe that $\displaystyle F'(x) = \tfrac{1}{2}(1-\sin^2x+\cos^2x) = \cos^2x.$

Therefore, $\displaystyle \int_0^{\pi/2} \cos^2t \,\mathrm{d}t = F(\pi/2) - F(0) = \pi/4.$

In the previous example, by first defining $F$, it was easy to find out its derivative $f$ and then calculate the integral of $f$. We are, however, almost always interested in calculating the integral of a given function $f$. The challenge thus is to find its antiderivative $F$. For example, if $f(x) = \cos^2x$, then how do we find $F$ for which $F' = f$?

To overcome this problem one often transfers the integral into a form which is easier to compute. The following substitution method is one way to implement this idea.

Theorem 4. Let $\varphi \colon [a,b] \to [c,d]$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that its derivative is continuous. If $f \colon [c,d] \to \mathbb{R}$ is continuous, then $\displaystyle \int_{\varphi(a)}^{\varphi(b)} f(s)\,\mathrm{d}s = \int_a^b f(\varphi(t)) \varphi'(t) \,\mathrm{d}t.$

Proof. By the fundamental theorem of calculus, there exists continuous $F \colon [c,d] \to \mathbb{R}$ such that $F'(x) = f(x)$ for all $x \in (c,d)$. Corollary 2 thus implies $\displaystyle \int_{\varphi(a)}^{\varphi(b)} f(s)\,\mathrm{d}s = F(\varphi(b)) - F(\varphi(a)) = (F \circ \varphi)(b) - (F \circ \varphi)(a).$

Since, by the chain rule, $\displaystyle (F \circ \varphi)'(t) = F'(\varphi(t)) \varphi'(t) = f(\varphi(t)) \varphi'(t),$

Corollary 2 also gives $\displaystyle \int_a^b f(\varphi(t)) \varphi'(t) \,\mathrm{d}t = \int_a^b (F \circ \varphi)'(t) \,\mathrm{d}t = (F \circ \varphi)(b) - (F \circ \varphi)(a).$

This finishes the proof. $\blacksquare$

Example 5. Let $\varphi \colon [0,\pi/2] \to [0,1]$, $\varphi(t) = \sin t$, and $\displaystyle f \colon [0,1] \to \mathbb{R}$, $f(s) = \sqrt{1-s^2}.$

Since $f(\varphi(t)) = \sqrt{1-\sin^2t}$ and $\varphi'(t) = \cos t$, Theorem 4 shows that $\displaystyle \int_0^1 \sqrt{1-s^2} \,\mathrm{d}s = \int_0^{\pi/2} \sqrt{1-\sin^2t} \cos t \,\mathrm{d}t = \int_0^{\pi/2} \cos^2t \,\mathrm{d}t.$

We determined the above integral in Example 3, so we conclude that $\displaystyle \int_0^1 \sqrt{1-s^2} \,\mathrm{d}s = \pi/4.$ Observe that $(x,y)$ is contained in the unit circle if and only if $x^2+y^2 \leq 1$. For $x,y \in [0,1]$, this is equivalent to $y \le \sqrt{1-x^2}$. Therefore, the integral $\int_0^1 \sqrt{1-s^2} \,\mathrm{d}s$ is the area of the intersection of the unit circle and the first quadrant. Since, by symmetry, the area of the unit circle is four times this number, an application of Example 5 shows that the area of the unit circle is $\pi$.

To finish the discussion let us remark that this method can easily be adapted to show that the area of a circle of radius $r>0$ is $\pi r^2$. This is also left as an exercise.