Integration

Lessons

Senior 6 · Mathematics

Integration

5 lessons

Lesson 1 of 5

Differentials and the indefinite integral

30 min

Running differentiation backwards

In S4 (Differentiation of Polynomial, Rational and Irrational Functions) you learned to go from a function to its rate of change, and in S6 Unit 2 you added $\ln x$ and $e^{x}$ . Very often the problem comes the other way round. A speed sensor on a Kigali–Musanze bus records the velocity $v (t)$ , and the operator wants the distance; a lab measures the rate at which a drug leaves the blood, and the doctor wants the amount left. In both cases you know the derivative and must recover the function. That reverse process is integration, and this unit builds it step by step. We start with differentials, because they give the $d x$ that appears in every integral.

Definition

Differential: let $y = f (x)$ be differentiable. The differential of $x$ , written $d x$ , is any real increment of $x$ . The differential of $y$ is

dy = f'(x)\,dx .

With this definition $\frac{dy}{dx} = f'(x)$ is a true quotient of two numbers, not just a symbol.

Example: if $y = x^{3} - 5 x$ , then $dy = (3x^2 - 5)\,dx$ . At $x = 2$ with $d x = 0.1$ , $dy = 7 \times 0.1 = 0.7$ .

What $d y$ measures

Move from $x$ to $x + d x$ . The true change in $y$ along the curve is $\Delta y = f(x + dx) - f(x)$ . The differential $dy = f'(x)\,dx$ is the change along the tangent instead. For a small $d x$ the curve hugs its tangent, so

\Delta y \approx dy \qquad\text{and}\qquad f(x + dx) \approx f(x) + f'(x)\,dx .

The tangent at P rises by dy = f′(x) dx over a run of dx; the curve itself rises by Δy. The gap between them (red) shrinks faster than dx as dx → 0.

This linear approximation is how you estimate a value near one you know, and how an error in a measurement spreads into an error in a result. The relative error is $\frac{dy}{y}$ , and the percentage error is $\frac{dy}{y} \times 100\%$ .

Worked example 1: estimates and errors with differentials

(a) Estimate $\sqrt{25.3}$ . (b) The radius of a steel ball bearing is measured as $10\,\mathrm{cm}$ with a possible error of $0.05\,\mathrm{cm}$ . Estimate the possible error in its volume, and the percentage error.

Solution

Take $f(x) = \sqrt{x}$ , a value we know, $x = 25$ , and $d x = 0.3$ . Then $f'(x) = \frac{1}{2\sqrt{x}}$ , so $dy = \frac{1}{2 \times 5} \times 0.3 = 0.03$ , and $\sqrt{25.3} \approx 5 + 0.03 = 5.03 .$ A calculator gives $5.029\,91$ , so the estimate is good to about $10^{-4}$ .
$V = \frac{4}{3}\pi r^3$ , so $dV = 4\pi r^2\,dr = 4\pi (10)^2 (0.05) = 20\pi \approx 62.8\,\mathrm{cm^3}$ . The relative error is $\frac{dV}{V} = \frac{4\pi r^2\,dr}{\frac{4}{3}\pi r^3} = 3\,\frac{dr}{r} = 3 \times \frac{0.05}{10} = 0.015 ,$ a percentage error of $1.5 %$ : three times the $0.5 %$ error in the radius.

Warning

Use radians. To estimate $\sin 31^\circ$ from $\sin 30^\circ$ , the increment is $dx = 1^\circ = \frac{\pi}{180} \approx 0.017\,45$ , not 1. The rule $\frac{d}{dx}\sin x = \cos x$ is true only when $x$ is in radians.

Definition

Antiderivative (primitive): a function $F$ is an antiderivative of $f$ on an interval $I$ if $F^{'} (x) = f (x)$ for every $x$ in $I$ .

Example: $F (x) = x^{3}$ is an antiderivative of $f (x) = 3 x^{2}$ on $\mathbb{R}$ . So are $x^{3} + 7$ and $x^3 - \sqrt{2}$ , since a constant differentiates to 0.

Important

Theorem. If $F$ and $G$ are both antiderivatives of $f$ on an interval $I$ , then $G (x) = F (x) + C$ for some constant $C$ .

Reason: $(G - F)^{'} = f - f = 0$ on $I$ . By the mean value theorem, a function whose derivative is 0 at every point of an interval does not change on it, so $G - F$ is a constant.

Definition

Indefinite integral: the set of all antiderivatives of $f$ ,

\int f(x)\,dx = F(x) + C, \quad\text{where } F'(x) = f(x).

Notation: $\int$ is the integral sign, $f (x)$ is the integrand, $d x$ names the variable of integration, and $C$ is the constant of integration. Finding $\int f(x)\,dx$ is called integrating $f$ .

Example: $\int 2x\,dx = x^2 + C$ .

Every antiderivative of 2x has the form x² + C. Changing C slides the graph up or down without changing its slope at any x: at x = 1 all five tangents are parallel, with slope 2.

Properties

Each rule below is a differentiation rule read backwards, so each is checked by differentiating the right-hand side. For functions with antiderivatives and a constant $k$ :

$\int \left[f(x) \pm g(x)\right]dx = \int f(x)\,dx \pm \int g(x)\,dx$
$\int k f(x)\,dx = k\int f(x)\,dx$
$\frac{d}{dx}\int f(x)\,dx = f(x)$ , and $\int f'(x)\,dx = f(x) + C$ ; in differentials, $\int dF = F + C$ .

There is no product rule of this simple kind: $\int f g\,dx$ is not $\int f\,dx \times \int g\,dx$ . Products need substitution (below) or integration by parts (lesson 3).

Table of standard integrals

$f (x)$	$\int f(x)\,dx$	condition
$x^{n}$	$\frac{x^{n+1}}{n+1} + C$	$n \ne -1$
$\frac{1}{x}$	$\ln\|x\| + C$	$x \ne 0$
$e^{x}$	$e^{x} + C$
$a^{x}$	$\frac{a^x}{\ln a} + C$	$a \gt 0,\ a \ne 1$
$\sin x$	$-\cos x + C$	$x$ in radians
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$	$\cos x \ne 0$
$\csc^2 x$	$-\cot x + C$	$\sin x \ne 0$
$\sec x \tan x$	$\sec x + C$	$\cos x \ne 0$
$\frac{1}{\sqrt{1 - x^2}}$	$\sin^{-1} x + C$	$\|x\| \lt 1$
$\frac{1}{1 + x^2}$	$\tan^{-1} x + C$

Why $\ln|x|$ ? For $x \gt 0$ , $\frac{d}{dx}\ln x = \frac{1}{x}$ . For $x \lt 0$ , $\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}$ as well. The absolute value covers both sides of 0 in one formula.

Worked example 2: using the table

Find (a) $\int (4x^3 - 6x + 5)\,dx$ (b) $\int \frac{(x^2 + 1)^2}{x^2}\,dx$ (c) $\int \left(3\sqrt{x} - \frac{2}{x}\right)dx$ (d) $\int (2e^x + 3\cos x)\,dx$ .

Solution

Term by term: $x^{4} - 3 x^{2} + 5 x + C$ .
First expand and divide, so that every term is a power of $x$ : $\frac{x^4 + 2x^2 + 1}{x^2} = x^2 + 2 + x^{-2}$ . Then $\int (x^2 + 2 + x^{-2})\,dx = \frac{x^3}{3} + 2x - \frac{1}{x} + C .$
$3\sqrt{x} = 3x^{\frac{1}{2}}$ integrates to $3 \times \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = 2x^{\frac{3}{2}}$ . The answer is $2x^{\frac{3}{2}} - 2\ln|x| + C$ .
$2e^x + 3\sin x + C$ .

Check (b): $\frac{d}{dx}\left(\frac{x^3}{3} + 2x - x^{-1}\right) = x^2 + 2 + x^{-2}$ ✓

Worked example 3: fixing the constant

A minibus pulls away from a stop with velocity $v (t) = 3 t - 0.15 t^{2}$ metres per second, $0 \le t \le 10$ , where $t$ is in seconds. Find its distance $s (t)$ from the stop.

Solution Since $s^{'} (t) = v (t)$ ,

s(t) = \int (3t - 0.15t^2)\,dt = 1.5t^2 - 0.05t^3 + C .

The bus starts at the stop, so $s (0) = 0$ , which gives $C = 0$ . Hence $s (t) = 1.5 t^{2} - 0.05 t^{3}$ , and after 10 s it has covered $150 - 50 = 100\,\mathrm{m}$ . A condition such as $s (0) = 0$ is an initial condition; it picks one curve out of the family.

Definition

Integration by substitution (change of variable): if $u = g (x)$ is differentiable and $F^{'} = f$ , then

\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du = F\big(g(x)\big) + C .

Reason: by the chain rule, $\frac{d}{dx}F\big(g(x)\big) = F'\big(g(x)\big)\,g'(x) = f\big(g(x)\big)\,g'(x)$ . In differentials, $u = g (x)$ gives $du = g'(x)\,dx$ : the substitution replaces $g'(x)\,dx$ by $d u$ .

Method. (1) Choose $u$ , usually the "inside" function whose derivative also appears as a factor. (2) Write $du = g'(x)\,dx$ . (3) Rewrite the whole integrand in $u$ ; no $x$ may be left. (4) Integrate in $u$ . (5) Put $u = g (x)$ back.

Two special cases are used so often that they are worth learning as rules:

\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C \ (a \ne 0), \qquad \int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C .

For example, $\int \cos(5x - 1)\,dx = \frac{1}{5}\sin(5x - 1) + C$ , and $\int \frac{2x}{x^2 + 9}\,dx = \ln(x^2 + 9) + C$ (no absolute value is needed, because $x^2 + 9 \gt 0$ ).

Worked example 4: substitution

Find (a) $\int x(x^2 + 3)^5\,dx$ (b) $\int e^{\sin x}\cos x\,dx$ (c) $\int x\sqrt{x - 1}\,dx$ .

Solution

Let $u = x^{2} + 3$ ; then $du = 2x\,dx$ , so $x\,dx = \frac{1}{2}du$ . $\int x(x^2 + 3)^5\,dx = \frac{1}{2}\int u^5\,du = \frac{u^6}{12} + C = \frac{(x^2 + 3)^6}{12} + C .$
Let $u = \sin x$ , $du = \cos x\,dx$ : $\int e^u\,du = e^u + C = e^{\sin x} + C$ .
Here no factor is the derivative of $x - 1$ times a function of it, but the substitution still works if you also rewrite $x$ . Let $u = x - 1$ , so $x = u + 1$ and $d x = d u$ : $\int (u + 1)u^{\frac{1}{2}}\,du = \int \left(u^{\frac{3}{2}} + u^{\frac{1}{2}}\right)du = \frac{2}{5}(x - 1)^{\frac{5}{2}} + \frac{2}{3}(x - 1)^{\frac{3}{2}} + C .$

Warning

Two common slips. (1) $\int (2x + 1)^4\,dx$ is $\frac{(2x + 1)^5}{10} + C$ , not $\frac{(2x + 1)^5}{5} + C$ : the factor $\frac{1}{a} = \frac{1}{2}$ comes from $du = 2\,dx$ . (2) A substitution needs the derivative of $u$ to be present, up to a constant factor. In $\int (x^2 + 3)^5\,dx$ the factor $x$ is missing, so $u = x^{2} + 3$ fails; expand instead.

Tip

An integral can always be checked: differentiate your answer and you must get the integrand back exactly. This takes thirty seconds and catches most sign and factor errors.

3Blue1Brown: The paradox of the derivative, Chapter 2 of Essence of Calculus (about 17 min). Watch how dx and dy are treated as tiny nudges, and how their ratio settles to the derivative as dx shrinks: this is the idea behind differentials.

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Practice

Check each indefinite integral by differentiating. Questions 7 and 8 are harder.

Find $d y$ when (a) $y = x^{4} - 3 x$ (b) $y = \sin 2x$ (c) $y = e^{x^2}$ .
Use differentials to estimate (a) $\sqrt[3]{27.54}$ (b) $\cos 61^\circ$ , starting from $\cos 60^\circ$ .
Find (a) $\int (5x^4 - 2x + 7)\,dx$ (b) $\int (x - 1)(x + 2)\,dx$ (c) $\int \left(\frac{1}{\sqrt{x}} + e^x\right)dx$ .
Find (a) $\int (2\sin x - 3\sec^2 x)\,dx$ (b) $\int \frac{4}{1 + x^2}\,dx$ (c) $\int 5^x\,dx$ .
Find $f (x)$ if $f^{'} (x) = 3 x^{2} - 2 x$ and $f (2) = 1$ .
Use a substitution: (a) $\int 3x^2(x^3 + 1)^4\,dx$ (b) $\int \sin x\cos^3 x\,dx$ (c) $\int x e^{x^2}\,dx$ (d) $\int \frac{\ln x}{x}\,dx$ .
Find $\int \frac{x}{\sqrt{x + 3}}\,dx$ with $u = x + 3$ .
The edge of a cube is measured as $5\,\mathrm{cm}$ with a possible error of $0.02\,\mathrm{cm}$ . Estimate the possible error in the volume, and the percentage error. Explain why the percentage error is three times that of the edge.

Show answersHide answers

(a) $(4x^3 - 3)\,dx$ (b) $2\cos 2x\,dx$ (c) $2x e^{x^2}\,dx$
(a) $3 + \frac{1}{27} \times 0.54 = 3.02$ (true value $3.0199$ ) (b) $0.5 - \frac{\sqrt{3}}{2} \times \frac{\pi}{180} \approx 0.4849$ (true value $0.4848$ )
(a) $x^{5} - x^{2} + 7 x + C$ (b) $\frac{x^3}{3} + \frac{x^2}{2} - 2x + C$ (c) $2\sqrt{x} + e^x + C$
(a) $-2\cos x - 3\tan x + C$ (b) $4\tan^{-1} x + C$ (c) $\frac{5^x}{\ln 5} + C$
$f (x) = x^{3} - x^{2} + C$ ; $8 - 4 + C = 1$ , so $f (x) = x^{3} - x^{2} - 3$ .
(a) $\frac{(x^3 + 1)^5}{5} + C$ (b) $u = \cos x$ : $-\frac{\cos^4 x}{4} + C$ (c) $\frac{1}{2}e^{x^2} + C$ (d) $\frac{(\ln x)^2}{2} + C$
$\int \frac{u - 3}{\sqrt{u}}\,du = \frac{2}{3}(x + 3)^{\frac{3}{2}} - 6\sqrt{x + 3} + C$
$dV = 3s^2\,ds = 3(25)(0.02) = 1.5\,\mathrm{cm^3}$ ; $\frac{1.5}{125} = 1.2\%$ . Since $\frac{dV}{V} = 3\frac{ds}{s}$ , the $0.4 %$ error in the edge is tripled.

Sources and credits

Written by Ganzaa: the explanations, the diagrams (drawn from computed values), the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S6 Core Mathematics (2026–2027); no REB text is reproduced.
Video: The paradox of the derivative | Chapter 2, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1Running differentiation backwards

2What dy dy dy measures

3Properties

4Table of standard integrals

Running differentiation backwards

What $d y$ measures

Properties

Table of standard integrals