Classic Applied Integration

Big Idea

The main use of integration is to determine the area under the graph of a function: given any function $f$ continuous on $[a, b]$ ,

$\displaystyle \lim_{n \to \infty} \sum_{k=1}^n f(x_k)\, \mathrm \Delta x_k = \text{(area under graph)} = \int_a^b f(x)\, \mathrm dx,$

where $\Delta x_k = x_k - x_{k-1}$ in the left-hand side. The diagram below illustrates the equation above in the simple case $[a,b] = [0,1]$ .

For example, by considering the area of a circle,

$\displaystyle \int_0^1 \sqrt{1-x^2}\, \mathrm dx = \frac \pi 4.$

This base application is usually adapted to formulate many other geometric definitions in terms of integrals.

Questions

Question 1. Evaluate the area enclosed by the ellipse below.

(Click for Solution)

Solution. By symmetry, the required area $A$ is given by the integral

$\displaystyle A = 4 \int_0^3 y\, \mathrm dx.$

Making $y$ the subject, since $y \geq 0$ in the first quadrant,

$\displaystyle y = 2 \sqrt{ 1 - \frac{x^2}{9} },$

so that the required integral is

$\displaystyle A = 8 \int_0^3 \sqrt{ 1 - \frac{x^2}{9} }\, \mathrm dx.$

Make the substitution $x = 3u \Rightarrow \mathrm dx = 3\, \mathrm du$ so that

$\displaystyle A = 8 \int_0^1 \sqrt{ 1 - u^2 } \cdot 3\, \mathrm du = 24 \int_0^1 \sqrt{ 1 - u^2 }\, \mathrm du.$

The integral on the right-hand side corresponds to the area of the quadrant of a unit circle, which equals $\pi/4$ . Therefore,

$\displaystyle A = 24 \int_0^1 \sqrt{ 1 - u^2 }\, \mathrm du = 24 \cdot \frac{\pi}{4} = 6\pi\, \mathrm{units}^2.$

Remark 1. The same technique can be used to prove that

$\displaystyle \int_0^a f(x/a)\, \mathrm dx = a \int_0^{1} f(x)\, \mathrm dx.$

In particular, an ellipse with equation $x^2/a^2 + y^2/b^2 = 1$ has area $\pi ab$ .

Question 2. Evaluate the integral $\displaystyle \int_0^1 \frac{x^4}{x^4 + (1-x)^4}\, \mathrm dx$ .

(Click for Solution)

Solution. Make the substitution $u = 1-x$ so that $\mathrm dx = (-1)\, \mathrm du$ , yielding

$\begin{aligned} I &:= \int_0^1 \frac{x^4}{x^4 + (1-x)^4}\, \mathrm dx \\ &= \int_1^0 \frac{(1-u)^4}{(1-u)^4 + u^4} \cdot (-1)\, \mathrm du \\ &= \int_0^1 \frac{(1-u)^4}{(1-u)^4 + u^4}\, \mathrm du \\ &= \int_0^1 \frac{(1-x)^4}{(1-x)^4 + x^4}\, \mathrm dx, \end{aligned}$

where the last equality follows from simply replacing the dummy variable. Consequently,

$\begin{aligned} I + I &= \int_0^1 \frac{x^4}{x^4 + (1-x)^4}\, \mathrm dx + \int_0^1 \frac{(1-x)^4}{(1-x)^4 + x^4}\, \mathrm dx \\ 2I &= \int_0^1 \frac{x^4 + (1-x)^4}{(1-x)^4 + x^4}\, \mathrm dx = \int_0^1 1\, \mathrm dx = 1. \end{aligned}$

Hence,

$\displaystyle \int_0^1 \frac{x^4}{x^4 + (1-x)^4}\, \mathrm dx = I = \frac 12.$

Remark 2. The same technique can be used to prove that

$\displaystyle \int_0^{2a} \frac{f(x)}{f(x) + f(2a-x)}\, \mathrm dx = a.$

Question 3. Evaluate the limit $\displaystyle \lim_{n \to \infty} \frac{\sqrt[3]{1} + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n\sqrt[3]{n}}$ .

(Click for Solution)

Solution. Using the Big Idea, setting $x_k = k/n$ so that $\Delta_k = x_k-x_{k-1} = 1/n$ ,

$\begin{aligned} \lim_{n \to \infty} \frac{\sqrt[3]{1} + \sqrt[3]{2} + \cdots + \sqrt[3]{n}}{n\sqrt[3]{n}} &= \lim_{n \to \infty} \sum_{k=1}^n \frac{\sqrt[3]{k}}{n\sqrt[3]{n}} = \lim_{n \to \infty} \sum_{k=1}^n \sqrt[3]{\frac kn } \cdot \frac 1n \\ &= \lim_{n \to \infty} \sum_{k=1}^n \sqrt[3]{x_k} \cdot \Delta x_k = \int_0^1 \sqrt[3]{x}\, \mathrm dx \\ &= \int_0^1 x^{1/3}\, \mathrm dx = \left[ \frac{x^{4/3}}{4/3} \right]_0^1 = \frac 34. \end{aligned}$

—Joel Kindiak, 5 Sept 25, 1358H

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