Category: Exercises

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

January 29, 2026
The Negative Integral
Big Idea

Integrating rational functions can get quite challenging unless we split it up into a sum of “smaller” rational functions whose integrals are relatively more trivial to compute.

Questions

Question 1. Evaluate the integral $\displaystyle \int \frac{1}{x^4 - 1}\, \mathrm dx$ .
(Click for Solution)

Solution. Factorise $x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^2+1)$ , so that

$\begin{aligned} \frac 1{ (x-1)(x+1)(x^2+1)} &= \frac A{x-1} + \frac B{x+1} + \frac{Cx+D}{x^2+1}. \end{aligned}$

Using the cover-up rule to make the following calculations:
- For $x - 1=0$ , set $x = 1$ so that $\displaystyle A = \frac 1{(1+1)(1^2+1)} = 1/4$ .
- For $x = -1$ , set $x = -1$ so that $\displaystyle B = \frac 1{ (-1-1)((-1)^2+1)}=-1/4$ .
- For $x^2+1 = 0$ , set $x = i$ so that $\displaystyle Ci+D = \frac 1{(i-1)(i+1)} = -1/2$ .
In particular, $C = 0$ and $D = -1/2$ , so that

$\begin{aligned} \int \frac{1}{x^4-1}\, \mathrm dx &= \int \frac 1{ (x-1)(x+1)(x^2+1)}\, \mathrm dx \\ &= \frac 14 \cdot \int \frac 1{x-1}\, \mathrm dx - \frac 14 \cdot \int \frac 1{x+1}\, \mathrm dx - \frac 12 \cdot \int \frac{1}{x^2+1}\, \mathrm dx \\ &= \frac 14 \ln|x-1| - \frac 14 \ln|x+1| - \frac 12 \tan^{-1}(x)+C. \end{aligned}$

Alternate Solution. We observe that

$\begin{aligned} \int \frac{1}{x - 1}\, \mathrm dx &= \int \frac{(x-1)'}{x - 1}\, \mathrm dx =\ln|x-1| + C_1, \\ \int \frac{x}{x^4 - 1}\, \mathrm dx &= \frac 12 \int \frac{(x^2)'}{(x^2)^2 - 1}\, \mathrm dx = \frac 12 \ln \left| \frac{x^2 - 1}{x^2 + 1} \right| + C_2, \\ \int \frac{x^3}{x^4 - 1}\, \mathrm dx &= \frac 14 \int \frac{(x^4-1)'}{x^4-1}\, \mathrm dx = \frac 14 \ln |x^4-1| + C_3. \end{aligned}$

Making the substitution $u = x^2$ , we decompose

$\displaystyle \frac{x^2}{x^4-1} = \frac{u}{(u-1)(u+1)} = \frac A{u-1} + \frac B{u+1}.$

Using the cover-up rule,

$\displaystyle A = \frac{1}{1+1} = \frac 12,\quad B = \frac{-1}{-1-1} = \frac 12.$

Therefore,

$\begin{aligned} \int \frac{x^2}{x^4-1} \mathrm dx &= \frac 12 \int \frac 1{x^2-1} \mathrm dx + \frac 12 \int \frac 1{x^2+1}\, \mathrm dx \\ &= \frac 12 \cdot \frac 12 \ln \left| \frac{x-1}{x+1} \right| + \frac 12 \tan^{-1}(x) + C_4. \end{aligned}$

Finally, we make the clever observation that

$\begin{aligned} (x^4 - 1) &= (x^2 -1)(x^2 + 1) \\ &= (x-1)(x+1)(x^2+1) \\ &= (x-1)(x^3 + x^2 + x + 1), \end{aligned}$

so that

$\begin{aligned} \int \frac 1{x-1}\, \mathrm dx &= \int \frac{x^3 + x^2 + x + 1}{x^4-1}\, \mathrm dx \\ &= \int \frac{x^3}{x^4-1}\, \mathrm dx + \int \frac{x^2}{x^4-1}\, \mathrm dx + \int \frac{x}{x^4-1}\, \mathrm dx + \int \frac{1}{x^4-1}\, \mathrm dx. \end{aligned}$

Therefore,

$\begin{aligned} \int \frac{1}{x^4-1}\, \mathrm dx &= \int \frac 1{x-1}\, \mathrm dx- \int \frac{x^3}{x^4-1}\, \mathrm dx - \int \frac{x^2}{x^4-1}\, \mathrm dx - \int \frac{x}{x^4-1}\, \mathrm dx \\ &= \ln|x-1| - \frac 14 \ln|x^4-1| - \frac 14 \ln \left| \frac{x-1}{x+1} \right| - \frac 12 \tan^{-1}(x) - \frac 12 \ln \left| \frac{x^2 - 1}{x^2 + 1} \right| + C. \end{aligned}$
Question 2. Evaluate the integral $\displaystyle \int \frac{1}{x^4 + 4}\, \mathrm dx$ .

(Click for Solution)

Solution. First make the clever factorisation

$\begin{aligned} x^4 + 4 &= x^4 + 4x^2 + 4 - 4x^2 \\ &= (x^2 + 2)^2 - (2x)^2 \\ &= (x^2 + 2x + 2)(x^2 - 2x +2) \\ &= ((x+1)^2 + 1)((x-1)^2 + 1)\end{aligned}$

Then use the partial fraction decomposition

$\displaystyle \frac{1}{((x+1)^2 + 1)((x-1)^2 + 1)} = \frac{A(x+1)+B}{(x+1)^2 + 1} + \frac{C(x-1)+D}{(x-1)^2 + 1}.$

Now the roots of $(x + 1)^2 + 1$ are $-1 \pm i$ . Therefore, using $x = -1 + i$ in the cover-up rule,

$\displaystyle Ai + B = \frac{1}{(-1 + i - 1)^2 + 1} = \frac 18 + \frac 18i \quad \Rightarrow \quad A = B = \frac 18.$

Likewise, the roots of $x^2 - 2x + 2$ are $1 \pm i$ . Therefore, using $x = 1 + i$ in the cover-up rule,

$\displaystyle Ci + D = \frac{1}{(1 + i + 1)^2 + 1} = \frac 18 - \frac 18i \quad \Rightarrow \quad C = -\frac 18,\quad D = \frac 18.$

Therefore,

$\begin{aligned} \int \frac{1}{x^4 + 4}\, \mathrm dx &= \frac 1{16} \int \frac{2(x+1)}{(x+1)^2 + 1} \, \mathrm dx + \frac 18 \int \frac{1}{(x+1)^2 + 1} \, \mathrm dx \\ &\phantom{==} - \frac 1{16} \int \frac{2(x-1)}{(x-1)^2 + 1} \, \mathrm dx + \frac 18 \int \frac{1}{(x-1)^2 + 1}\, \mathrm dx \\ &= \frac 1{16} \ln((x+1)^2 + 1) + \frac 18 \tan^{-1}(x+1) \\ &\phantom{==} - \frac 1{16} \ln((x-1)^2+1) + \frac 18 \tan^{-1}(x-1) + C. \end{aligned}$

—Joel Kindiak, 4 Sept 25, 2239H
January 15, 2026
Reduction Formulae

Big Idea

To integrate general products of functions, we use integration by parts, a.k.a. what I call the IS-ID technique:

$\displaystyle \int u\, \mathrm dv = \underbrace{ \phantom- v \phantom- }_{\mathrm I}\, \underbrace{ \phantom- u \phantom- }_{\mathrm S} - \int \underbrace{ \phantom- v \phantom- }_{\mathrm I}\, \underbrace{ \phantom- \mathrm du \phantom- }_{\mathrm D}.$

The function $\mathrm I$ is taken from the easier-to-integrate function $\mathrm dv$ .

Questions

Question 1. For any non-negative integer $n$ , define

$\displaystyle I_n := \int_0^{\pi} \sin^n(x)\, \mathrm dx.$

Evaluate $I_0$ and $I_1$ . Hence, evaluate $I_n$ in terms of $n$ .

(Click for Solution)

Solution. To evaluate the first two integrals is straightforward:

$\begin{aligned} I_0 &= \int_0^{\pi} \sin^0(x)\, \mathrm dx = \int_0^{\pi}\, \mathrm dx = \pi, \\ I_1 &= \int_0^{\pi} \sin(x)\, \mathrm dx = \left[ - \cos(x) \right]_0^{\pi} \\ &= (-(-1) - (-1)) = 2. \end{aligned}$

For the general case, assume $n \geq 2$ , so that

$\begin{aligned} I_n &= \int_0^{\pi} \sin^n(x)\, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \cdot \sin^2(x) \, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \cdot (1-\cos^2(x)) \, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \, \mathrm dx - \int_0^{\pi} \sin^{n-2}(x) \cos^2(x) \, \mathrm dx \\ &= I_{n-2} - \int_0^{\pi} \sin^{n-2}(x) \cos^2(x) \, \mathrm dx. \end{aligned}$

To evaluate the integral,

$\begin{aligned} &\int_0^{\pi} \sin^{n-2}(x) \cos^2(x) \, \mathrm dx \\ &= \int_0^{\pi} \sin^{n-2}(x) \cos(x) \cdot \cos(x) \, \mathrm dx \\ &= \left[ \underbrace{ \phantom- \frac{\sin^{n-1}(x)}{n-1} \phantom- }_{\mathrm I}\, \underbrace{ \phantom- \cos(x) \phantom- }_{\mathrm S} \right]_0^{\pi} - \int_0^{\pi} \underbrace{ \phantom- \frac{\sin^{n-1}(x)}{n-1} \phantom- }_{\mathrm I}\, \underbrace{ \phantom- (-\sin(x))\, \mathrm dx \phantom- }_{\mathrm D} \\ &= (0 - 0) + \frac 1{n-1} \int_0^{\pi} \sin^n(x)\, \mathrm dx = \frac 1{n-1} \cdot I_{n}. \end{aligned}$

Therefore,

$\displaystyle I_n = I_{n-2} - \frac 1{n-1} \cdot I_{n} \quad \Rightarrow \quad I_n = \frac {n-1}n \cdot I_{n-2}.$

For example,

$I_2 = \frac 12 \cdot I_0,\quad I_3 = \frac 23 \cdot I_1,\quad I_4 = \frac 34 \cdot \frac 12 \cdot I_0, \quad I_5 = \frac 45 \cdot \frac 23 \cdot I_1.$

More generally, if $n = 2k + l$ where $l = 0, 1$ ,

$\displaystyle I_n = I_{2k+l} = \frac{(n-1)(n-3) \cdot \cdots \cdot (n-(2k-1)) }{n(n-2) \cdot \cdots \cdot (n-2k)} \cdot I_l,$

where $I_l = \pi$ if $l = 0$ and $I_1 = 2$ if $l = 1$ .

Question 2. For any positive integer $n$ , define

$\displaystyle I_n := \int_{-1}^1 \frac{ x^n }{ \sqrt{2x+3} }\, \mathrm dx.$

Evaluate $I_0$ . Then evaluate $I_n$ in terms of $I_{n-1}$ . Finally, evaluate $I_1, I_2$ .

(Click for Solution)

Solution. Setting $n = 0$ ,

$\begin{aligned} I_0 &= \int_{-1}^{1} \frac 1{\sqrt{2x+3}}\, \mathrm dx = \int_{-1}^{1} (2x+3)^{-1/2}\, \mathrm dx \\ &= \left[ \frac 12 \cdot \frac{(2x+3)^{1/2}}{1/2} \right]_{-1}^1 = \left[ \sqrt{2x+3} \right]_{-1}^1 = -1 + \sqrt 5. \end{aligned}$

Writing $x^n = x \cdot x^{n-1}$ ,

$\begin{aligned} I_n &= \int_{-1}^1 \frac{ x^n }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \int_{-1}^1 x^n \cdot \frac{ 1 }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \frac 12 \int_{-1}^1 x^{n-1} \cdot \frac{ (2x+3-3) }{ \sqrt{2x+3} }\, \mathrm dx \\ &= \frac 12 \int_{-1}^1 x^{n-1} \sqrt{2x+3}\, \mathrm dx - \frac 32 \int_{-1}^1 x^{n-1} \cdot \frac 1{\sqrt{2x+3}}\, \mathrm dx \\ &= \frac 12 \left( \left[\underbrace{ \phantom- \frac{x^n}{n} \phantom- }_{\mathrm I}\cdot \underbrace{ \phantom- \sqrt{2x+3} \phantom- }_{\mathrm S} \right]_{-1}^1 - \int_{-1}^1 \underbrace{ \phantom- \frac{x^n}{n} \phantom- }_{\mathrm I}\cdot \underbrace{ \phantom- \frac{1}{2\sqrt{2x+3}} \cdot 2 \, \mathrm dx\phantom- }_{\mathrm D}\right) - \frac 32 \cdot I_{n-1} \\ &= \frac 12 \left( \frac 1n \cdot (\sqrt 5 - (-1)^n) - \frac 1n \cdot I_n \right) - \frac 32 \cdot I_{n-1} \\ &=\frac 1{2n} \cdot (\sqrt 5 - (-1)^n) - \frac 1{2n} \cdot I_n - \frac 32 \cdot I_{n-1}. \end{aligned}$

Doing a little bit more algebruh,

$\begin{aligned} I_n &= \frac {1}{2n+1} \cdot (\sqrt 5 - (-1)^n) - \frac {3n}{2n+1} \cdot I_{n-1}.\end{aligned}$

Finally,

$\begin{aligned} I_1 &= \frac 13 \cdot (\sqrt 5 -(-1)) - \frac 33 \cdot I_0 = \frac 43-\frac 23 \sqrt 5 ,\\ I_2 &= \frac {1}{5} \cdot (\sqrt 5 - 1) - \frac {6}{5} \cdot I_{1} = - \frac {9}{5} + \sqrt 5 . \end{aligned}$

—Joel Kindiak, 4 Sept 25, 1727H

January 8, 2026
Real-Analytic Inequalities

Fix $p > 1, q > 1$ such that $\displaystyle \frac 1p + \frac 1q = 1$ .

Problem 1. Verify that $p+q = pq$ .

(Click for Solution)

Solution. By algebruh,

$\displaystyle 1 = \frac 1p + \frac 1q = \frac{p+q}{pq}\quad \Rightarrow \quad p+q = pq.$

Problem 2. Show that for $x \geq 0$ , $\displaystyle x \leq \frac{x^p}{p} + \frac 1q$ with equality if and only if $x = 1$ .

(Click for Solution)

Solution. Define the function $f(x) := x^p/p + 1/q - x$ . By observation,

$\displaystyle f(1) = \frac 1p + \frac 1q - 1 = 0.$

Furthermore,

$f'(x) = x^{p-1} - 1 \quad \Rightarrow \quad f''(x) = (p-1)x^{p-2} > 0.$

By the second derivative test, solving for $f'(x) = 0$ yields a global minimum:

$\displaystyle x^{p-1} - 1 = 0 \quad \Rightarrow \quad x = 1.$

Therefore, $f(x) \geq f(1) = 0$ for any $x \geq 0$ , as required.

Problem 3. Use Problem 2 to prove Young’s inequality: for any $a \geq 0, b \geq 0$ ,

$\displaystyle ab \leq \frac{a^p}{p} + \frac{ b^q }{q}$

with equality if and only if $a^p = b^q$ .

(Click for Solution)

Solution. Setting $x = a/b^{q-1}$ ,

$\begin{aligned} \frac a{b^{q-1}} &\leq \frac{\left( \frac a{b^{q-1}} \right)^p}{p} + \frac 1q \\ \frac a{b^{q-1}} \cdot b^q &\leq \frac{1}{p} \cdot \frac{a^p}{b^{p(q-1)}} \cdot b^q + \frac {b^q}q \\ ab &\leq \frac{a^p}{p} \cdot b^{p+q-pq} + \frac {b^q}q \\ ab &\leq \frac{a^p}{p} + \frac {b^q}q . \end{aligned}$

Furthermore, since $p(q-1) = pq - p = q$ , equality holds if and only if

$\displaystyle x = 1 \quad \iff \quad a = b^{q-1} \quad \iff \quad a^p = b^{p(q-1)} \quad \iff \quad a^p = b^q.$

Problem 4. Use Young’s inequality to prove Hölder’s inequality for two-dimensional vectors: given non-negative numbers $a_1,a_2,b_1,b_2$ ,

$\displaystyle a_1 b_1 + a_2 b_2 \leq \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q}.$

(Click for Solution)

Solution. If $\left( a_1^p + a_2^p \right)^{1/p} = 0$ , then $0 \leq a_1^p + a_2^p = 0$ implies $a_1 = a_2 = 0$ , and the inequality holds trivially. Without loss of generality, suppose the right-hand side is non-zero. It suffices to prove that

$\displaystyle \frac{ a_1 b_1 + a_2 b_2 }{ \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q} } \leq 1.$

By one application of Young’s inequality,

$\begin{aligned} \frac{ a_1 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_1 }{\left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right)^{p/p} } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right)^{q/q} } \\ &\leq \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right) }. \end{aligned}$

By a second application of Young’s inequality,

$\begin{aligned} \frac{ a_2 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_2 }{\left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) }. \end{aligned}$

Summing the inequalities,

$\begin{aligned} \frac{ a_1 b_1 + a_2 b_2 }{ \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{ a_1 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_1 }{\left( b_1^q + b_2^q \right)^{1/q} } + \frac{ a_2 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_2 }{\left( b_1^q + b_2^q \right)^{1/q} } \\ &= \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right) } + \frac{a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) } \\ &= \frac{a_1^p + a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q + b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) } = \frac 1p + \frac 1q = 1.\end{aligned}$

Problem 5. Use Hölder’s inequality to prove Minkowski’s inequality for two-dimensional vectors: given real numbers $a_1,a_2,b_1,b_2$ ,

$\displaystyle (|a_1 + b_1|^p + |a_2 + b_2|^p)^{1/p} \leq (|a_1|^p + |a_2|^p )^{1/p} + (|b_1|^p + |b_2|^p )^{1/p}.$

(Click for Solution)

Solution. By the vanilla triangle inequality,

$\begin{aligned} &|a_1 + b_1|^p + |a_2 + b_2|^p \\ &= |a_1 + b_1| |a_1 + b_1|^{p-1} + |a_2 + b_2| |a_2 + b_2|^{p-1} \\ &\leq (|a_1| + |b_1|) |a_1 + b_1|^{p-1} +(|a_2| + |b_2|) (a_2 + b_2)^{p-1} \\ &= |a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} + |b_1| |a_1 + b_1|^{p-1} + |b_2| |a_2 + b_2|^{p-1}. \end{aligned}$

By Hölder’s inequality,

$\begin{aligned} & |a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} \\ &\leq (|a_1|^p + |a_2|^p)^{1/p} \cdot (|a_1 + b_1|^{q(p-1) }+ |a_2 + b_2|^{q(p-1)})^{1/q} \\ &\leq (|a_1|^p + |a_2|^p)^{1/p} \cdot (|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}, \end{aligned}$

so that

$\displaystyle \frac{|a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} }{(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|a_1|^p + |a_2|^p)^{1/p}.$

Similarly,

$\displaystyle \frac{|b_1| |a_1 + b_1|^{p-1} + |b_2| |a_2 + b_2|^{p-1} }{(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|b_1|^p + |b_2|^p)^{1/p}.$

Combining the displays,

$\displaystyle \frac{ |a_1 + b_1|^p + |a_2 + b_2|^p} {(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|a_1|^p + |a_2|^p)^{1/p} + (|b_1|^p + |b_2|^p)^{1/p}.$

The result follows by the observation

$\displaystyle \frac{ |a_1 + b_1|^p + |a_2 + b_2|^p} {(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} = (|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/p}.$

Remark 1. Denoting

$\left\| \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} \right\|_p := (|a_1|^p + |a_2|^p )^{1/p}$

and defining $\mathbf a := \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$ , Hölder’s inequality reduces to

$\| \mathbf a \cdot \mathbf b \|_1 \leq \| \mathbf a \|_p \| \mathbf b \|_q,$

and Minkowski’s inequality reduces to

$\| \mathbf a + \mathbf b \|_p \leq \| \mathbf a \|_p + \| \mathbf b \|_p.$

Setting $p = q = 2$ reduces to the usual Cauchy-Schwarz inequality and triangle inequality for two-dimensional vectors. Furthermore, these results hold for $n$ -dimensional vectors, and even “infinite-dimensional” (sufficiently nice) functions.

January 5, 2026
Substitution Magic

Big Idea

The substitution $u = g(x)$ allows us to simplify integrals:

$\displaystyle \int f'(g(x)) \cdot g'(x)\, \mathrm dx = \int f(u)\, \mathrm du.$

In particular, we recover two useful integration forms:

$\displaystyle \int f'(ax+b)\, \mathrm dx = \frac 1a \cdot f'(ax+b) +C,\quad \int \frac{f'(x)}{f(x)}\, \mathrm dx = \ln|f(x)| +C.$

Questions

Question 1. Using $u = -x^2$ , evaluate the integral $\displaystyle \int xe^{-x^2}\, \mathrm dx$ .

Deduce the value of $\displaystyle \int_0^1 xe^{-x^2}\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $u = -x^2 \Rightarrow -\frac 12 \mathrm du = x\, \mathrm dx$ ,

$\begin{aligned} \int xe^{-x^2}\, \mathrm dx &= \int e^{-x^2} \cdot x\, \mathrm dx\\ &= \int e^u\left( -\frac 12 \right) \, \mathrm du \\ &= -\frac 12 \int e^u\, \mathrm du \\ &= -\frac 12 \cdot e^u + C \\ &= -\frac 12 \cdot e^{-x^2} + C. \end{aligned}$

Hence,

$\begin{aligned} \int_0^1 xe^{-x^2}\, \mathrm dx &= \left[ -\frac 12 \cdot e^{-x^2} \right]_0^1 \\ &= \left( -\frac 12 \cdot e^{-1^2} \right) - \left( -\frac 12 \cdot e^{-0^2} \right) \\ &= \frac 12(1-e^{-1}). \end{aligned}$

Question 2. Evaluate the integral

$\displaystyle \int \sin(\cos^2(x)) \cos(1 - \sin^2(x)) \sin(2x)\, \mathrm dx.$

(Click for Solution)

Solution. Make the substitution

$u = \cos^2(x) = 1-\sin^2(x) = \frac 12(1 + \cos 2x)$

so that

$\begin{aligned} \frac{ \mathrm du }{\mathrm dx} &= \frac 12 (0 + (-\sin(2x)) \cdot 2) \\ &= - \sin (2x) \\ \quad (-1)\, \mathrm du &= \sin(2x)\, \mathrm dx. \end{aligned}$

Hence,

$\begin{aligned} &\int \sin(\cos^2(x)) \cos(1 - \sin^2(x)) \sin(2x)\, \mathrm dx \\ &= \int \sin(\cos^2(x)) \cos(\cos^2(x)) \sin(2x)\, \mathrm dx\\ &= \int \sin(u) \cos(u) \cdot (-1)\, \mathrm du \\ &= -\frac 12 \int 2 \sin(u) \cos(u) \, \mathrm du \\ &= -\frac 12 \int \sin(2u) \, \mathrm du \\ &= -\frac 12 \cdot \frac 12 \cdot (-\cos(2u)) + C \\ &= \frac 14 \cdot \cos(2u) + C \\ &= \frac 14 \cdot \cos(2 \cos^2(x)) + C.\end{aligned}$

Question 3. Evaluate the integral $\displaystyle \int x^x(\ln x + 1)\, \mathrm dx$ .

(Click for Solution)

Solution. Using the laws of exponents $a^b = e^{b \ln a}$ ,

$\displaystyle \int x^x(\ln x + 1)\, \mathrm dx = \int e^{x \ln x }(\ln x + 1)\, \mathrm dx.$

Then make the substitution $u = x \ln x$ so that

$\begin{aligned} \frac{\mathrm d u}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx} (x \ln x) \\ &= \ln x +1 \\ \mathrm du &= (\ln x + 1)\, \mathrm dx.\end{aligned}$

Since $e^u = e^{x \ln x} = x^x$ ,

$\displaystyle \int x^x(\ln x + 1)\, \mathrm dx = \int e^u\, \mathrm du = e^u + C = x^x + C.$

—Joel Kindiak, 4 Sept 25, 1644H

January 1, 2026
Tedious Integration

Big Idea

We regard integration, at the pre-university level, as the reverse process of differentiation:

$\displaystyle F(x) = \int f(x)\, \mathrm dx \quad \iff \quad \frac{\mathrm d}{\mathrm dx} (F(x)) = f(x).$

We call $f(x)$ the integrand. Integration is linear in the following sense:

$\displaystyle \int (af(x) + bg(x))\, \mathrm dx = a \int f(x)\, \mathrm dx + b \int g(x)\, \mathrm dx.$

The integrals of powers requires a bit more care: if $n \neq -1$ then nothing weird happens:

$\displaystyle \int x^n\, \mathrm dx = \frac{x^{n+1}}{n+1} +C.$

However, if $n = -1$ , then we cannot divide by zero, and instead recover the logarithm:

$\displaystyle \int x^{-1}\, \mathrm dx = \int \frac 1x \, \mathrm dx = \ln|x| + C.$

Other commonly used integrals are free for use. Furthermore, in the context of pre-university mathematics, the definite integral is an application of the vanilla integral:

$\displaystyle \int_a^b f(x)\, \mathrm dx := F(b) - F(a) \equiv [F(x)]_a^b.$

Questions

You may not use integration by substitution in any of these problems.

Question 1. Evaluate $\displaystyle \int (3^{2x} + \sqrt{25x} - \ln(x^7))\, \mathrm dx$ .

(Click for Solution)

Solution. Simplifying the integrand then integrating term-wise,

$\begin{aligned} \int (3^{2x} + \sqrt{25x} - \ln(x^7))\, \mathrm dx &= \int (9^x + 5x^{1/2} - 7 \ln(x))\, \mathrm dx \\ &= \int 9^x\, \mathrm dx + 5 \int x^{1/2}\, \mathrm dx - 7 \int \ln(x)\, \mathrm dx \\ &= \frac{9^x}{\ln(9)} + 5 \cdot \frac {x^{3/2}}{3/2} - 7(x \ln(x) - x) + C \\ &= \frac {1}{\ln(9)} \cdot 9^x+ \frac{10}3 \cdot x^{3/2} - 7x \ln(x) + 7x + C. \end{aligned}$

Question 2. Evaluate $\displaystyle \int (2+3x)^4\, \mathrm dx$ .

(Click for Solution)

Solution. We first slowly expand the integrand:

$\begin{aligned} (2+3x)^4 &= ( (2+3x)^2 )^2 \\ &= ( 4 + 12x + 9x^2 )^2 \\ &= 16 + 96x + 216x^2 + 216x^3 + 81x^4. \end{aligned}$

Using linearity to integrate term-wise,

$\begin{aligned} \int (2+3x)^4 \, \mathrm dx &= \int (16 + 96x + 216x^2 + 216x^3 + 81x^4)\, \mathrm dx \\ &= \int 16\, \mathrm dx + 96 \int x^1\, \mathrm dx + 216 \int x^2\, \mathrm dx + 216 \int x^3\, \mathrm dx + 81 \int x^4\, \mathrm dx \\ &= 16x + 96 \cdot \frac{x^2}{2} + 216 \cdot \frac{x^3}{3} + 216 \cdot \frac{x^4}{4} + 81 \cdot \frac{x^5}{5} + C \\ &= 16x + 48x^2 + 72x^3 + 54 x^4 + \frac{81}{5} \cdot x^5 + C. \end{aligned}$

Question 3. Evaluate $\displaystyle \int \left( \frac 2x - 7 \sqrt x \right)^2\, \mathrm dx$ .

(Click for Solution)

Solution. We first slowly expand the integrand:

$\begin{aligned} \left( \frac 2x - 7 \sqrt x \right)^2 &= \left( \frac 2x \right)^2 - 2 \cdot \frac 2x \cdot 7\sqrt x + (7 \sqrt x)^2 \\ &= \frac 4{x^2} - 28x^{-1/2} + 49x \\ &= 4x^{-2}- 28x^{-1/2} + 49x^1. \end{aligned}$

Using linearity to integrate term-wise,

$\begin{aligned} \int \left( \frac 2x - 7 \sqrt x \right)^2\, \mathrm dx &= \int (4x^{-2}- 28x^{-1/2} + 49x^1)\, \mathrm dx \\ &= 4 \int x^{-2}\, \mathrm dx - 28 \int x^{-1/2}\, \mathrm dx + 49 \int x^1\, \mathrm dx \\ &= 4 \cdot \frac{x^{-1}}{-1} - 28 \cdot \frac{x^{1/2}}{1/2} + 49 \cdot \frac{x^2}{2} + C \\ &= - 4x^{-1}- 56 x^{1/2} + \frac{49}{2} \cdot x^2 + C. \end{aligned}$

Remark 1. In general, when feasible, final answers ought to take the form

$\displaystyle c_1 \cdot f_1(x) + c_2 \cdot f_2(x) + \cdots + c_n \cdot f_n(x) + C,$

where $c_1,\dots, c_n$ are numerical constants, $f_1(x),\dots, f_n(x)$ are expressions in terms of $x$ , and $C$ denotes the arbitrary constant of integration (if and only if needed), for easy recognition.

—Joel Kindiak, 4 Sept 25, 1334H

December 4, 2025
Optimised Geometry
Big Idea

To maximise (or minimise) a quantity $y\equiv y(x)$ , we apply the zero derivative condition

$\displaystyle \frac{\mathrm dy}{\mathrm dx} \equiv y'(x) = 0,$

and check that the corresponding value $x = c$ yields to a maximum (resp. minimum) via the second derivative test:
- a local maximum is obtained at $c$ when $y''(c) <0$ ,
- a local minimum is obtained at $c$ when $y''(c) > 0$ .
Recall that we denote

$\displaystyle y'' \equiv y''(x) := (y')'(x) \equiv \frac{\mathrm d}{\mathrm dx}(y'(x)) = \frac{\mathrm d}{\mathrm dx}\left(\frac{\mathrm dy}{\mathrm dx}\right) =: \frac{\mathrm d^2 y}{\mathrm dx^2}$

for brevity.

Questions

Question 1. Evaluate the area of the largest rectangle contained inside the ellipse graphed below.

(Click for Solution)

Solution. Given $x, y > 0$ , the base of the rectangle is $2x$ and its height is $2y$ , yielding a total area of $A = 4xy$ . Differentiating with respect to $x$ ,

$\displaystyle \frac{\mathrm dA}{\mathrm dx} = 4y + 4x \cdot \frac{\mathrm dy}{dx}.$

On the other hand, given the ellipse $\displaystyle \frac{x^2}{9} + \frac{y^2}{4} = 1$ , differentiate with respect to $x$ on both sides to obtain

$\displaystyle \frac{2x}{9} + \frac{2y}{4} \cdot \frac{\mathrm dy}{\mathrm dx} = 0 \quad \Rightarrow \quad \frac{\mathrm dy}{\mathrm dx} = -\frac{4x}{9y}.$

Therefore,

$\displaystyle \frac{\mathrm dA}{\mathrm dx} = 4y + 4x \cdot \frac{\mathrm dy}{dx} = 4y - \frac{16x^2}{9y}.$

By the zero-derivative condition, if $A$ is maximised, then $\displaystyle \frac{\mathrm dA}{\mathrm dx} = 0$ :

$\displaystyle 4y - \frac{16x^2}{9y} = 0 \quad \Rightarrow \quad y^2 = \frac{4}{9}x^2.$

Plugging back into the equation of the ellipse,

$\begin{aligned} \frac{x^2}{9} + \frac{1}{4} \cdot \frac{4}{9}x^2 &= 1 \\ x^2 &= 18\\ x &= \sqrt{18} \\ &= 3\sqrt{2}, \end{aligned}$

since $x > 0$ , yielding $y = \frac 23 x = \frac 23 \cdot 3\sqrt 2 = 2\sqrt 2$ . Thus, the area of the corresponding rectangle is $A = 4 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = 12\, \text{units}^2$ .

To apply second derivative test, we differentiate $A$ again:

$\begin{aligned} \frac{\mathrm d^2 A}{\mathrm dx^2} &= 4 \cdot \frac{\mathrm dy}{\mathrm dx} - \frac{16 \cdot 2x \cdot 9y - 16x^2 \cdot \frac{\mathrm dy}{\mathrm dx}}{81y^2} \\ &= 4 \cdot 0 - \frac{16 \cdot 2x \cdot 9y - 16x^2 \cdot 0}{81y^2} \\ &= - \frac{32x }{9y} < 0,\end{aligned}$

so $A$ is indeed maximised when $x = 3\sqrt 2$ and $y = 2\sqrt 2$ .

Question 2. The diagram below shows a movie theatre with a screen $9\, \text{m}$ tall, whose base is $4\, \text{m}$ above the ground.

The permissible viewing area is $50\, \text{m}$ long with an incline of up to $6\, \text{m}$ . Determine the position that a movie-goer $P$ should sit in order to maximise his viewing angle $\theta$ , justifying your answer.

(Click for Solution)

Solution. Let $x$ denote the horizontal distance of the movie-goer from the screen. His height $y$ is then determined using similar triangles:

$\displaystyle \frac{y}{x} = \frac6{50} = \frac 3{25} \quad \Rightarrow \quad y = \frac{3}{25}x.$

By observation,

$\begin{aligned} \theta &= \tan^{-1} \left( \frac{13-y}{x} \right) - \tan^{-1} \left( \frac{4-y}{x} \right) \\ &= \tan^{-1} \left( \frac{13}{x} - \frac 3{25} \right) - \tan^{-1} \left( \frac{4}{x} - \frac 3{25} \right). \end{aligned}$

We remark that if $y > 4$ , then $4-y < 0$ and the formula still works. Differentiating with respect to $x$ ,

$\begin{aligned} \frac{\mathrm d\theta}{\mathrm dx} &= \frac{\mathrm d}{\mathrm dx}\tan^{-1} \left( \frac{13}{x} - \frac 3{25} \right) - \frac{\mathrm d}{\mathrm dx}\tan^{-1} \left( \frac{4}{x} - \frac 3{25} \right) \\ &= \frac{1}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} \left( -\frac{13}{x^2} \right) - \frac{1}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} \left( -\frac{4}{x^2} \right) \\ &= -\frac 1{x^2} \left( \frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} - \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} \right), \\ x^2 \frac{\mathrm d\theta}{\mathrm dx} &= \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} -\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2}. \end{aligned}$

By the zero-derivative condition, if $\theta$ is maximised, then $\frac{\mathrm d\theta}{\mathrm dx} = 0$ :

$\begin{aligned}x^2 \cdot 0 &= \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} -\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2}. \end{aligned}$

We need to (painfully) solve this equation:

$\begin{aligned} \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} &=\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} \\ 4 \cdot \left( 1 + \left( \frac{13}{x} - \frac 3{25} \right)^2 \right) &=13 \cdot \left( 1 + \left( \frac{4}{x} - \frac 3{25} \right)^2 \right) \\ 4 \cdot \left( 1 + \frac{169}{x^2} -\frac{78}{25x} + \frac{9}{625} \right) &=13 \cdot \left( 1 + \frac{16}{x^2} -\frac{24}{25x} + \frac{9}{625} \right) \\ 4 + \frac{676}{x^2} -\frac{312}{25x} + \frac{36}{625} &= 13 + \frac{208}{x^2} -\frac{312}{25x}+ \frac{117}{625} \\ \frac{468}{x^2} &= \frac{5706}{625} \\ x^2 &= \frac{16\, 250}{317} \\ x &= \sqrt{\frac{16\, 250}{317}} \approx 7.160\, \text{m}, \end{aligned}$

since $x > 0$ .

To apply second derivative test, we differentiate again:

$\begin{aligned} 2x \frac{\mathrm d\theta}{\mathrm dx} + x^2 \frac{\mathrm d^2\theta}{\mathrm dx^2} &= \frac{\mathrm d}{\mathrm dx} \left( \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} \right) -\frac{\mathrm d}{\mathrm dx} \left( \frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2} \right) \\ &= -\frac {4 \cdot 2\left( \frac{4}{x} - \frac 3{25} \right) \cdot -\frac{4}{x^2}}{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} + \frac {13 \cdot 2\left( \frac{13}{x} - \frac 3{25} \right) \cdot -\frac{13}{x^2}}{\left(1 + \left( \frac{13}{x} - \frac 3{25} \right)^2\right)^2} \\ &= \frac 2{x^2} \cdot \left( \frac {16 \cdot \left( \frac{4}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} - \frac {169 \cdot \left( \frac{13}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{13}{x} - \frac 3{25} \right)^2\right)^2} \right).\end{aligned}$

When $\displaystyle \frac{\mathrm d\theta}{\mathrm dx} = 0$ , $\displaystyle \frac{4}{1 + \left( \frac{4}{x} - \frac 3{25} \right)^2} =\frac{13}{1 + \left( \frac{13}{x} - \frac 3{25} \right)^2}$ , so that

$\begin{aligned} 2x \cdot 0 + x^2 \frac{\mathrm d^2\theta}{\mathrm dx^2} &= \frac 2{x^2} \cdot \left( \frac {16 \cdot \left( \frac{4}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} - \frac {16 \cdot \left( \frac{13}{x} - \frac 3{25} \right) }{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} \right) \\ &= \frac 2{x^2} \cdot \frac {16}{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} \cdot \left( \left( \frac{4}{x} - \frac 3{25} \right) - \left( \frac{13}{x} - \frac 3{25} \right) \right) \\ &= \frac 2{x^2} \cdot \frac {16}{\left(1 + \left( \frac{4}{x} - \frac 3{25} \right)^2\right)^2} \cdot -\frac 9x < 0, \end{aligned}$

so indeed $\theta$ will attain its maximum at $x = \sqrt{\frac{16\, 250}{317}}$ .

Question 3. Determine the smallest perimeter of a rectangle with area $1$ .

(Click for Solution)

Solution. Let $x, y$ denote the base and height of the rectangle respectively. Since $xy = 1$ implies $y = 1/x$ , the perimeter $P$ of the rectangle is given by

$\displaystyle P(x) = 2x + \frac 2x.$

Differentiating twice,

$\displaystyle P'(x) = 2 - \frac 2{x^2},\quad P''(x) = \frac 2{x^2} > 0.$

Solving $P'(x) = 0$ yields $x = 1$ so that $y = 1$ . Since $P''(1) > 0$ automatically, $x = 1$ yields a (local) minimum for $P$ . Hence, the rectangle has a minimum perimeter of $P(1) = 4$ , which corresponds to the rectangle being a square with base length $1$ .

Remark 1. The result remains true even if we consider rectangles of other areas; the rectangle with area $A$ whose perimeter is minimum must be a square with side length $\sqrt{A}$ .

Question 4. Given that $0 < x < 1$ , maximise $x^2(1-x)^3$ .

(Click for Solution)

Solution. Define $f(x) := x^2(1-x)^3$ . Differentiating twice,

$\begin{aligned} f'(x) &= 2x \cdot (1-x)^3 + x^2 \cdot 3(1-x)^2 \cdot (-1) \\ &= 2x \cdot (1-x)^3 - x^2 \cdot 3(1-x)^2 \\ &= x \cdot (1-x)^2 \cdot (2-5x), \\ f''(x) &= (1-x)^2\cdot (2-5x) + x \cdot 2(1-x) \cdot(-1) \cdot (2-5x) \\ &\phantom{==} + x \cdot (1-x)^2 \cdot (-5) \\ &= (1-x)^2\cdot (2-5x) - 2x(1-x) \cdot (2-5x) - 5x \cdot (1-x)^2. \end{aligned}$

Solving $f'(x) = 0$ , since $0 < x < 1$ , $2-5x = 0 \Rightarrow x = 2/5$ . Substituting into $f''$ ,

$f''(2/5) = 0 - 0 - 5 \cdot 2/5 \cdot (1-2/5)^2 < 0,$

so that $x = 2/5$ yields a local (global) maximum. Hence, $f$ has a maximum value of

$\displaystyle \left(\frac 25 \right)^2 \left(1 - \frac 25 \right)^3 = \frac{2^2 \cdot 3^3}{5^5} = \frac{108}{3125} = 0.03456.$

Remark 2. Using the same technique, we can prove that for any $0 < s < r$ and $s < x < r$ ,

$\displaystyle 0 < (x-s)^m (r-x)^n \leq m^m n^n \cdot \frac{(r-s)^{m+n}}{(m+n)^{m+n}}.$

with equality at $x = (rm+sn)/(m+n)$ . Setting $(s,r,m,n) = (0,1,2,3)$ yields Question 4.

Question 5. Given that $x > 0$ , find the value of $x$ that minimises $\displaystyle 6x^3 + 7/x^5$ .

(Click for Solution)

Solution. Define $f(x) := 6x^3 + 7/x^5 = 6x^3 + 7x^{-5}$ . Differentiating twice,

$\begin{aligned} f'(x) &= 18x^2 - 35x^{-6}, \\ f''(x) &= 36x + 210x^{-7} > 0. \end{aligned}$

Therefore, $f$ is minimised when $f'(x) = 0$ :

$\displaystyle 18x^2 - 35x^{-6} = 0 \quad \Rightarrow \quad x^8 = \frac{35}{18} \quad \Rightarrow \quad x = \sqrt[8]{ \frac{ 35 }{ 18 } },$

since $x > 0$ .

Remark 3. In general, the value of $x > 0$ that minimises $Ax^m + B/x^n$ is given by $x^{m+n} = (Bn)/(Am)$ .

—Joel Kindiak, 4 Sept 25, 1213H
November 27, 2025
Nontrivial Rates of Change

Big Idea

The rate of change of a quantity $x(t)$ in terms of time $t$ is $\displaystyle \frac{\mathrm dx}{\mathrm dt}$ . Furthermore, given $y \equiv y(x)$ , the chain rule yields

$\displaystyle \frac{\mathrm dy}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Questions

Question 1. The point $P(x, y)$ above the $y$ -axis at time $t$ is defined by the equations

$x = \cos (t), \quad y = \sin (t),\quad 0 \leq t < \pi.$

Define $A(-1, 0)$ . Let $Q$ denote the intersection of $AP$ with the $y$ -axis.

Evaluate the rate of change of the $y$ -coordinate of $Q$ at the point the gradient of the tangent to the curve at $P$ is $-1/\sqrt{3}$ .

(Click for Solution)

Solution. Differentiating with respect to $t$ ,

$\displaystyle \frac{\mathrm dx}{\mathrm dt} = -\sin(t),\quad \frac{\mathrm dy}{\mathrm dt} = \cos(t).$

By the chain rule,

$\begin{aligned} \frac{\mathrm dy}{\mathrm dt} &= \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt} \\ \cos(t) &= -\frac 1{\sqrt 3} \cdot -\sin(t)\\ \tan(t) &= \sqrt{3} \\ t &= \pi/3. \end{aligned}$

Let $y_Q$ denote the $y$ -coordinate of $Q$ . The equation of $AP$ is given by

$\displaystyle \frac{ y - 0 }{ x- (-1) } = \frac{\sin(t) - 0}{\cos(t) - (-1)}\quad\Rightarrow \quad y = \frac{\sin(t)}{1 + \cos(t)} \cdot (x+1).$

At the $y$ -axis, $x = 0$ , so that

$\displaystyle y_Q = \frac{\sin(t)}{1 + \cos(t)}.$

Applying the quotient rule,

$\begin{aligned} \frac{\mathrm dy_Q}{\mathrm dt} &= \frac{\mathrm d}{\mathrm dt} \left( \frac{\sin(t)}{1 + \cos(t)} \right) \\ &= \frac{\displaystyle (1 + \cos(t)) \cdot \frac{\mathrm d}{\mathrm dt}(\sin(t)) - \sin(t) \cdot \frac{\mathrm d}{\mathrm dt}(1 + \cos(t))}{( 1 + \cos(t) )^2} \\ &= \frac{(1 + \cos(t)) \cdot \cos(t) - \sin(t) \cdot (0 + (-\sin(t))}{( 1 + \cos(t) )^2} \\ &= \frac{(\cos(t) + \cos^2(t)) + \sin^2(t) }{( 1 + \cos(t) )^2} \\ &= \frac{\cos(t) + (\cos^2(t) + \sin^2(t)) }{( 1 + \cos(t) )^2} \\ &= \frac{ \cos(t) + 1 }{( 1 + \cos(t) )^2} = \frac{ 1 }{ 1 + \cos(t) }. \end{aligned}$

Therefore, at $t = \pi/3$ , since $\cos(t) = 1/2$ ,

$\begin{aligned} \frac{\mathrm dy_Q}{\mathrm dt} &= \frac{ 1 }{ 1 + 1/2 } = \frac 23\, \text{units}/\text{s}. \end{aligned}$

Alternate Solution. Using double-angle formulae,

$\displaystyle y_Q = \frac{\sin(t)}{1 + \cos(t)} = \frac{2 \sin(t/2) \cos (t/2)}{2 \cos^2(t/2)} = \tan(t/2).$

Differentiating,

$\displaystyle \frac{ \mathrm dy_Q }{ \mathrm dt } = \frac 12 \sec^2(t/2).$

The rest of the question follows as per usual.

Question 2. A man of height $1.6\, \text{m}$ metres is currently $300\, \text{m}$ away from a pole of height $3.6\, \text{m}$ . He runs in a straight line towards the pole at a speed of $3\, \text{m}/\text{s}$ . Let $\theta$ denote the angle of elevation from the man to the top of the pole.

Evaluate the rate of change of $\theta$ when the man is at the pole.

(Click for Solution)

Solution. The man’s distance from the pole at time $t$ is given by $(300 - 3t)$ metres. By considering the height difference between the pole and the man,

$\displaystyle \tan(\theta(t)) = \frac{3.6-1.6}{300 - 3t}\quad \Rightarrow \quad \theta(t) = \tan^{-1}\left( \frac{2}{3(100- t)} \right).$

Applying the chain rule,

$\begin{aligned} \frac{\mathrm d\theta}{\mathrm dt} &= \frac{ \mathrm d }{ \mathrm dt } \left( \tan^{-1}\left( \frac{2}{3(100- t)} \right) \right) \\ &= \frac 1{1 + \left( \frac{2}{3(100- t)} \right)^2} \cdot \frac{\mathrm d}{\mathrm dt}\left( \frac{2}{3(100- t)} \right) \\ &= \frac 1{1 + \frac{4}{9(100- t)^2}} \cdot \frac 23 \cdot \frac{\mathrm d}{\mathrm dt}\left( (100-t)^{-1} \right) \\ &= \frac {9(100- t)^2}{4 + 9(100- t)^2} \cdot \frac 23 \cdot (-1)(100-t)^{-2} \cdot \frac{\mathrm d}{\mathrm dt}(100 - t) \\ &= \frac {9(100- t)^2}{4 + 9(100- t)^2} \cdot \frac 23 \cdot (-1)(100-t)^{-2} \cdot (-1) \\ &= \frac {6}{4 + 9(100- t)^2}. \end{aligned}$

At the pole, $t = 100$ , so that

$\displaystyle \frac{\mathrm d\theta}{\mathrm dt} = \frac 6{4 + 9(100-100)^2} = 1.5\, \text{rad}/\text{s}.$

—Joel Kindiak, 4 Sept 25, 1149H

November 20, 2025
Exploiting the Trifecta

Big Idea

Three powerful results in differentiation handle more exotic kinds of differentiation, the product rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(uv) = v \cdot \frac{\mathrm du}{\mathrm dx} + u \cdot \frac{\mathrm dv}{\mathrm dx},$

the quotient rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\frac uv \right) = \frac{\displaystyle v \cdot \frac{\mathrm du}{\mathrm dx} - u \cdot \frac{\mathrm dv}{\mathrm dx}}{v^2},$

and the chain rule: if $y = f(u)$ , then

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(u)) = f'(u) \cdot \frac{\mathrm du}{\mathrm dx} \quad \iff \quad \frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm du} \cdot \frac{\mathrm du}{\mathrm dx}.$

Questions

Question 1. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} (2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8))$ .

(Click for Solution)

Solution. We first simplify the function

$\begin{aligned} 2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8) &= 4^x \cdot 4x^{1/2} \cdot 8 \ln(x) \\ &= 32 \cdot x^{1/2} \cdot 4^x \cdot \ln(x).\end{aligned}$

Apply the product rule once to obtain

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^{1/2} \cdot 4^x) &= 4^x \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \frac{\mathrm d}{\mathrm dx}(4^x).\end{aligned}$

Apply the product rule on the original function to obtain

$\begin{aligned} & \frac{\mathrm d}{\mathrm dx} (2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8)) \\ &= 32 \cdot \frac{\mathrm d}{\mathrm dx} ((x^{1/2} \cdot 4^x) \cdot \ln(x))\\ &= 32 \cdot \left( \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2} \cdot 4^x) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( \ln(x) \cdot \left( 4^x \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \frac{\mathrm d}{\mathrm dx}(4^x) \right) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( 4^x \cdot \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(4^x) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( 4^x \cdot \ln(x) \cdot \frac 12 x^{-1/2} + x^{1/2} \cdot \ln(x) \cdot 4^x \ln (4) + x^{1/2} \cdot 4^x \cdot \frac 1x \right) \\ &= 16 \cdot 4^x \cdot x^{-1/2} \cdot ( \ln(x) + 2x \cdot \ln(x) \cdot \ln (4) + 2 ).\end{aligned}$

Question 2. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\sin \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \right)$ .

(Click for Solution)

Solution. Noting that $\sin^{-1} \neq \csc = 1/{\sin}$ , applying the chain rule repeatedly and carefully,

$\begin{aligned} &\frac{\mathrm d}{\mathrm dx} \left(\sin \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \frac{\mathrm d}{\mathrm dx} \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( \frac{\mathrm d}{\mathrm dx} (\csc(x^2)) - \frac{\mathrm d}{\mathrm dx} (\sin^{-1}(x^2)) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( -\csc(x^2) \cot(x^2) \cdot \frac{\mathrm d}{\mathrm dx}(x^2) - \frac 1{\sqrt{1 - (x^3)^2}} \cdot \frac{\mathrm d}{\mathrm dx}(x^3) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} -\sin^{-1}(x^3) \right) \cdot \left( -\csc(x^2) \cot(x^2) \cdot 2x - \frac 1{\sqrt{1 - x^6}} \cdot 3x^2 \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( -2x\csc(x^2) \cot(x^2) - \frac {3x^2}{\sqrt{1 - x^6}} \right). \end{aligned}$

Question 3. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x)$ .

(Click for Solution)

Solution. Using the laws of exponents, the chain rule, the product rule, linearity, and common derivatives,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^x) &= \frac{\mathrm d}{\mathrm dx} (e^{x \ln(x)})\\ &= e^{x \ln(x)} \cdot \frac{\mathrm d}{\mathrm dx}( x \ln(x) ) \\ &= x^x \cdot \left( \frac{\mathrm d}{\mathrm dx}(x) \cdot \ln(x) + x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= x^x \cdot \left( 1 \cdot \ln(x) + x \cdot \frac{1}{x} \right) \\ &= x^x \cdot (\ln(x) + x). \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1905H

November 13, 2025
Tedious Differentiation

Big Idea

The derivative of a function $f$ at a point $a$ , denoted $f'(c)$ , intuitively measures the gradient of the tangent to $y = f(x)$ at $(c, f(c))$ . For any general $x$ , we make the notation

$\displaystyle f'(x) \equiv \frac{\mathrm d}{\mathrm dx}(f(x)) \equiv \frac{\mathrm dy}{\mathrm dx}.$

For instance, given any real number $n$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = nx^{n-1}.$

Several other derivatives of commonly used functions exist. Furthermore, differentiation is linear in the following sense:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(af(x) + bg(x)) = a \cdot \frac{\mathrm d}{\mathrm dx}(f(x)) + b \cdot \frac{\mathrm d}{\mathrm dx}(g(x)).$

Questions

You may not use the chain rule in any of these problems.

Question 1. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(2 - 3 x + 4\sqrt x - \frac 5x\right)$ .

(Click for Solution)

Solution. Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\left(2 - 3 x + 4\sqrt x - \frac 5x\right) &= \frac{\mathrm d}{\mathrm dx}(2) - 3 \cdot \frac{\mathrm d}{\mathrm dx}(x) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(\sqrt x) - 5 \cdot \frac{\mathrm d}{\mathrm dx} \left( \frac 1x \right) \\ &= \frac{\mathrm d}{\mathrm dx}(2) - 3 \cdot \frac{\mathrm d}{\mathrm dx}(x^1) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) - 5 \cdot \frac{\mathrm d}{\mathrm dx} \left( x^{-1}\right) \\ &= 0 - 3 \cdot 1x^0 + 4 \cdot \frac 12 x^{-1/2} - 5 \cdot (-1)x^{-2} \\ &= -3 + 2 x^{-1/2} + 5x^{-2} \\ &= -3 + \frac 2{\sqrt x} + \frac 5{x^2}.\end{aligned}$

Question 2. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} (2^{2x} + \sqrt{16x} - \ln(x^8))$ .

(Click for Solution)

Solution. Simplifying then using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}( 2^{2x} + \sqrt{16x} - \ln(x^8) ) &= \frac{\mathrm d}{\mathrm dx} ( (2^2)^x + \sqrt{16} \cdot \sqrt{x} - 8 \cdot \ln(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}( 4^x + 4x^{1/2} - 8 \ln(x) ) \\ &= \frac{\mathrm d}{\mathrm dx}(4^x) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) - 8 \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \\ &= 4^x \ln(4) + 4 \cdot \frac 12 x^{-1/2} - 8 \cdot \frac 1x \\ &= 4^x \ln(4) + 2 x^{-1/2} - \frac 8x \\ &= 4^x \ln(4) + \frac 2{\sqrt x} - \frac 8x. \end{aligned}$

Question 3. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} ((5+2x)^4)$ .

(Click for Solution)

Solution. We first slowly expand the function

$\begin{aligned} (5+2x)^4 &= ((5+2x)^2)^2 \\ &= (25 + 20x + 4x^2)^2 \\ &= 625 + 1000x + 600x^2 + 160x^3 + 16x^4. \end{aligned}$

Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} ((5+2x)^4) &= ((5+2x)^2)^2 \\ &= \frac{\mathrm d}{\mathrm dx} (625 + 1000x + 600x^2 + 160x^3 + 16x^4) \\ &= \frac{\mathrm d}{\mathrm dx}(625) + 1000 \cdot \frac{\mathrm d}{\mathrm dx}(x) + 600 \cdot \frac{\mathrm d}{\mathrm dx}( x^2 ) + 160 \cdot \frac{\mathrm d}{\mathrm dx}(x^3) + 16 \cdot \frac{\mathrm d}{\mathrm dx}( x^4) \\ &= 0 + 1000 \cdot 1 + 600 \cdot 2x + 160 \cdot 3x^2 + 16 \cdot 4x^3 \\ &= 1000 + 1200x + 480x^2 + 64x^3.\end{aligned}$

Question 4. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \left( \frac 2x-5 \sqrt x \right)^2 \right)$ .

(Click for Solution)

Solution. We first expand and simplify the function

$\begin{aligned} \left( \frac 2x-5 \sqrt x \right)^2 &= \left( \frac 2x \right)^2 - 2 \cdot \frac 2x \cdot 5\sqrt x + (5 \sqrt x)^2 \\ &= \frac{4}{x^2} - 20 \cdot x^{-1} \cdot x^{1/2} + 25 x \\ &= 4x^{-2} - 20 x^{-1/2} + 25 x. \end{aligned}$

Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( \left( \frac 2x-5 \sqrt x \right)^2 \right) &= \frac{\mathrm d}{\mathrm dx} (4x^{-2} - 20 x^{-1/2} + 25 x) \\ &= 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{-2}) - 20 \cdot \frac{\mathrm d}{\mathrm dx} (x^{-1/2}) + 25 \cdot \frac{\mathrm d}{\mathrm dx}(x) \\ &= 4 \cdot (-2)x^{-3} - 20 \cdot \left(-\frac 12\right)x^{-3/2} + 25 \cdot 1x^0 \\ &= -8x^{-3} + 10x^{-3/2}+25 \\ &= -\frac 8{x^3} + \frac{10}{\sqrt{x^3}}+25. \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1826H

October 30, 2025