Integration Techniques

Antidifferentiation Review

Integration (antidifferentiation) is the reverse process of differentiation. If $F"(x) = f(x)$ Then:

\int f(x)\,dx = F(x) + C

Where $C$ is the constant of integration.

Basic Integration Rules

Rule	Formula
Power rule	$\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)$
Reciprocal	$\displaystyle\int \frac{1}{x},dx = \ln\|x\|
Constant multiple	$\displaystyle\int kf(x)\,dx = k\int f(x)\,dx$
Sum/difference	$\displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$

Standard Integrals

$f(x)$	$\int f(x)\,dx$
$e^x$	$e^x + C$
$a^x$	$\dfrac{a^x}{\ln a} + C$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$
$\csc^2 x$	$-\cot x + C$
$\sec x \tan x$	$\sec x + C$
$\dfrac{1}{\sqrt{1-x^2}}$	$\arcsin x + C$
$\dfrac{1}{1+x^2}$	$\arctan x + C$

The Fundamental Theorem of Calculus

First Fundamental Theorem

If $f$ is continuous on $[a, b]$ and $F'(x) = f(x)$ Then:

\int_a^b f(x)\,dx = F(b) - F(a)

Second Fundamental Theorem

If $f$ is continuous on an interval containing $a$ Then:

\frac{d}`\{dx}`\left[\int_a^x f(t)\,dt\right] = f(x)

Definite Integrals

Properties

For continuous functions $f$ and $g$ on $[a, b]$ :

$\displaystyle\int_a^b kf(x)\,dx = k\int_a^b f(x)\,dx$
$\displaystyle\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$
$\displaystyle\int_a^a f(x)\,dx = 0$
$\displaystyle\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$
$\displaystyle\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$ for any $c$

Even and Odd Functions

If $f$ is even ( $f(-x) = f(x)$ ):

\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx

If $f$ is odd ( $f(-x) = -f(x)$ ):

\int_{-a}^{a} f(x)\,dx = 0

Evaluate $\displaystyle\int_{-2}^{2} x^3\cos(x^2)\,dx$ .

Since $x^3\cos(x^2)$ is odd (odd $\times$ even $=$ odd):

\int_{-2}^{2} x^3\cos(x^2)\,dx = 0

Integration by Substitution

Method

When the integrand contains a function and its derivative (or a composite function), substitute $u = g(x)$ where $g(x)$ is the “inner” function:

\int f(g(x))g'(x)\,dx = \int f(u)\,du

Steps

Choose $u = g(x)$ (the inner function).
Compute $\dfrac{du}{dx} = g'(x)$ So $du = g'(x)\,dx$ .
Rewrite the integral in terms of $u$ .
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Evaluate $\displaystyle\int_0^1 xe^{x^2}\,dx$ .

Let $u = x^2$ So $du = 2x\,dx$ .

When $x = 0$ : $u = 0$ . When $x = 1$ : $u = 1$ .

\int_0^1 xe^{x^2}\,dx = \frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}\big[e^u\big]_0^1 = \frac{1}{2}(e - 1)

Trigonometric Substitutions

Certain integrals can be simplified using trigonometric substitutions:

Expression	Substitution	Identity
$\sqrt{a^2 - x^2}$	$x = a\sin\theta$	$1 - \sin^2\theta = \cos^2\theta$
$\sqrt{a^2 + x^2}$	$x = a\tan\theta$	$1 + \tan^2\theta = \sec^2\theta$
$\sqrt{x^2 - a^2}$	$x = a\sec\theta$	$\sec^2\theta - 1 = \tan^2\theta$

Evaluate $\displaystyle\int \frac{1}{\sqrt{4-x^2}}\,dx$ .

This is a standard integral: $\arcsin\!\left(\dfrac{x}{2}\right) + C$ .

Alternatively, let $x = 2\sin\theta$ So $dx = 2\cos\theta\,d\theta$ :

\int \frac{2\cos\theta}{2\cos\theta}\,d\theta = \int 1\,d\theta = \theta + C = \arcsin\!\left(\frac{x}{2}\right) + C

Integration by Parts

For the product of two functions:

\int u\,dv = uv - \int v\,du

Choosing $u$ and $dv$

Use the LIATE rule (prioritise choosing $u$ from):

Logarithmic functions
Inverse trig functions
Algebraic functions (polynomials)
Trigonometric functions
Exponential functions

Evaluate $\displaystyle\int x^2 e^x\,dx$ .

First application: $u = x^2$$dv = e^x\,dx$ :

\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx

Second application on $\int 2x e^x\,dx$ : $u = 2x$$dv = e^x\,dx$ :

\int 2x e^x\,dx = 2xe^x - \int 2e^x\,dx = 2xe^x - 2e^x

Combining:

\int x^2 e^x\,dx = x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C

Cyclic Integration by Parts

Integrals like $\displaystyle\int e^x \sin x\,dx$ require applying integration by parts twice and solving algebraically.

Evaluate $\displaystyle\int e^x \sin x\,dx$ .

Let $u = e^x$$dv = \sin x\,dx$ . Then $du = e^x\,dx$$v = -\cos x$ .

I = -e^x\cos x + \int e^x\cos x\,dx

Now let $u = e^x$$dv = \cos x\,dx$ . Then $du = e^x\,dx$$v = \sin x$ .

\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - I

Substituting back:

I = -e^x\cos x + e^x\sin x - I

2I = e^x(\sin x - \cos x)

I = \frac{e^x(\sin x - \cos x)}{2} + C

Partial Fractions

When the integrand is a rational function with a denominator that can be factorised, decompose into partial fractions.

Types of Decomposition

Denominator	Partial Fractions
$(ax+b)(cx+d)$	$\dfrac{A}{ax+b} + \dfrac{B}{cx+d}$
$(ax+b)^2$	$\dfrac{A}{ax+b} + \dfrac{B}{(ax+b)^2}$
$(ax+b)(x^2+c)$	$\dfrac{A}{ax+b} + \dfrac{Bx+C}{x^2+c}$

Evaluate $\displaystyle\int \frac{5x+1}{(x+1)(x-2)}\,dx$ .

\frac{5x+1}{(x+1)(x-2)} = \frac{A}{x+1} + \frac{B}{x-2}

5x + 1 = A(x-2) + B(x+1)

$x = -1$ : $-5 + 1 = A(-3) \implies A = \dfrac{4}{3}$ .

$x = 2$ : $10 + 1 = B(3) \implies B = \dfrac{11}{3}$ .

\int \frac{5x+1}{(x+1)(x-2)}\,dx = \frac{4}{3}\ln\|x+1\| + \frac{11}{3}\ln\|x-2\| + C

Area Under Curves

Area Between a Curve and the $x$ -axis

The signed area between $y = f(x)$ and the $x$ -axis from $x = a$ to $x = b$ is:

\mathrm{Area} = \int_a^b f(x)\,dx

If the curve crosses the $x$ -axis, split the integral at the zeros and take the absolute value of each part.

Area Between Two Curves

The area between $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ (where $f(x) \ge g(x)$ ):

\mathrm{Area} = \int_a^b [f(x) - g(x)]\,dx

Find the area enclosed by $y = x^2 - 4$ and the $x$ -axis.

Zeros: $x^2 - 4 = 0 \implies x = \pm 2$ .

\mathrm{Area} = \int_{-2}^{2} |x^2 - 4|\,dx = 2\int_0^2 (4 - x^2)\,dx = 2\left[4x - \frac{x^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3}

Volume of Revolution

Rotation about the $x$ -axis

The volume generated by rotating $y = f(x)$ from $x = a$ to $x = b$ about the $x$ -axis:

V = \pi \int_a^b [f(x)]^2\,dx

Rotation about the $y$ -axis

The volume generated by rotating $x = g(y)$ from $y = c$ to $y = d$ about the $y$ -axis:

V = \pi \int_c^d [g(y)]^2\,dy

Alternatively, using the shell method about the $y$ -axis:

V = 2\pi \int_a^b x |f(x)|\,dx

Find the volume generated by rotating the region between $y = x$ and $y = x^2$ about the $x$ -axis.

Intersection: $x = x^2 \implies x = 0$ or $x = 1$ .

V = \pi\int_0^1 \left(x^2 - x^4\right)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}

Kinematics Applications

Displacement, Velocity, and Acceleration

Velocity is the derivative of displacement: $v = \dfrac{ds}{dt}$
Acceleration is the derivative of velocity: $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$
Displacement from velocity: $s = \displaystyle\int v\,dt$
Velocity from acceleration: $v = \displaystyle\int a\,dt$

Total Distance vs Displacement

\mathrm{Displacement} = \int_{t_1}^{t_2} v(t)\,dt

\mathrm{Total distance} = \int_{t_1}^{t_2} |v(t)|\,dt

A particle moves with velocity $v(t) = t^2 - 4t + 3\mathrm{ m/s}$ for $0 \le t \le 5$ .

(a) Find the displacement.

S = \int_0^5 (t^2 - 4t + 3)\,dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^5 = \frac{125}{3} - 50 + 15 = \frac{40}{3}\mathrm{ m}

(b) Find the total distance travelled.

Find when $v = 0$ : $t^2 - 4t + 3 = 0 \implies (t-1)(t-3) = 0 \implies t = 1, 3$ .

Interval	Sign of $v$	Motion
$0 \lt t \lt 1$	$+$	Forward
$1 \lt t \lt 3$	$-$	Backward
$3 \lt t \lt 5$	$+$	Forward

\mathrm{Distance} = \int_0^1 v\,dt + \left|\int_1^3 v\,dt\right| + \int_3^5 v\,dt

= \frac{4}{3} + \left|-\frac{4}{3}\right| + \frac{20}{3} = \frac{4}{3} + \frac{4}{3} + \frac{20}{3} = \frac{28}{3}\mathrm{ m}

Improper Integrals

Type 1: Infinite Limits

\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx

Type 2: Discontinuous Integrand

\int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx

An improper integral converges if the limit exists (is finite) and diverges otherwise.

Determine whether $\displaystyle\int_1^{\infty} \frac{1}{x}\,dx$ converges.

\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty}[\ln x]_1^b = \lim_{b \to \infty}\ln b = \infty

The integral diverges.

IB Exam-Style Questions

Question 1 (Paper 1 style)

Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x\,dx$ .

Using the identity $\sin^2 x = \dfrac{1 - \cos 2x}{2}$ :

\int_0^{\pi/2} \frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{1}{2}\left(\frac{\pi}{2} - 0\right) = \frac{\pi}{4}

Question 2 (Paper 2 style)

Let $R$ be the region bounded by the curve $y = x(x-2)$ and the $x$ -axis.

(a) Find the area of $R$ .

Zeros: $x = 0$ and $x = 2$ . The curve is below the axis (opens upward).

\mathrm{Area} = \int_0^2 |x(x-2)|\,dx = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}

(b) Find the volume when $R$ is rotated $360\degree$ about the $x$ -axis.

V = \pi\int_0^2 [x(x-2)]^2\,dx = \pi\int_0^2 x^2(x-2)^2\,dx

= \pi\int_0^2 (x^4 - 4x^3 + 4x^2)\,dx = \pi\left[\frac{x^5}{5} - x^4 + \frac{4x^3}{3}\right]_0^2

= \pi\left(\frac{32}{5} - 16 + \frac{32}{3}\right) = \pi \cdot \frac{16}{15} = \frac{16\pi}{15}

Question 3 (Paper 2 style)

Evaluate $\displaystyle\int \frac{2x + 3}{x^2 + 3x + 2}\,dx$ .

Factorise the denominator: $x^2 + 3x + 2 = (x+1)(x+2)$ .

\frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}

2x + 3 = A(x+2) + B(x+1)

$x = -1$ : $1 = A \implies A = 1$ .

$x = -2$ : $-1 = -B \implies B = 1$ .

\int \frac{2x+3}{x^2+3x+2}\,dx = \ln\|x+1\| + \ln\|x+2\| + C = \ln\|(x+1)(x+2)\| + C

Question 4 (Paper 1 style)

Use integration by parts to evaluate $\displaystyle\int_0^1 xe^{2x}\,dx$ .

Let $u = x$$dv = e^{2x}\,dx$ . Then $du = dx$$v = \dfrac{1}{2}e^{2x}$ .

\int_0^1 xe^{2x}\,dx = \left[\frac{1}{2}xe^{2x}\right]_0^1 - \frac{1}{2}\int_0^1 e^{2x}\,dx = \frac{e^2}{2} - \frac{1}{2}\left[\frac{e^{2x}}{2}\right]_0^1

= \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}

Question 5 (Paper 2 style)

A particle moves in a straight line with acceleration $a(t) = 6t - 2\mathrm{ m/s}^2$ . At $t = 0$ The velocity is $4\mathrm{ m/s}$ and the displacement is $0\mathrm{ m}$ .

(a) Find the velocity function.

V(t) = \int (6t-2)\,dt = 3t^2 - 2t + C

$v(0) = 4 \implies C = 4$ .

V(t) = 3t^2 - 2t + 4

(b) Find the displacement function.

S(t) = \int (3t^2 - 2t + 4)\,dt = t^3 - t^2 + 4t + D

$s(0) = 0 \implies D = 0$ .

S(t) = t^3 - t^2 + 4t

(c) Find the total distance travelled in the first 3 seconds.

Check if $v = 0$ : $3t^2 - 2t + 4 = 0$ . Discriminant $= 4 - 48 \lt 0$ So $v \gt 0$ always.

\mathrm{Distance} = \int_0^3 v\,dt = \int_0^3 (3t^2 - 2t + 4)\,dt = \left[t^3 - t^2 + 4t\right]_0^3 = 27 - 9 + 12 = 30\mathrm{ m}

Question 6 (Paper 2 style)

The region bounded by $y = e^x$$y = 1$$x = 0$ And $x = 2$ is rotated about the $x$ -axis. Find The volume.

V = \pi\int_0^2 [(e^x)^2 - 1^2]\,dx = \pi\int_0^2 (e^{2x} - 1)\,dx = \pi\left[\frac{e^{2x}}{2} - x\right]_0^2

= \pi\left(\frac{e^4}{2} - 2 - \frac{1}{2} + 0\right) = \pi\left(\frac{e^4}{2} - \frac{5}{2}\right) = \frac{\pi(e^4 - 5)}{2}

Summary Table of Techniques

Technique	When to Use	Key Idea
Direct	Simple power functions, standard forms	Apply basic rules directly
Substitution	Composite functions, $f(g(x))g'(x)$	Let $u = g(x)$
By parts	Product of different function types	$\int u\,dv = uv - \int v\,du$
Partial fractions	Rational functions, factorisable denominator	Decompose then integrate each term
Trig substitution	$\sqrt{a^2 \pm x^2}$ or $\sqrt{x^2 - a^2}$	Replace with trig function

When facing an integral, first check if it is a standard form. If not, consider substitution (especially if you see a function and its derivative). If it is a product of different function Types, use integration by parts. If it is a rational function, consider partial fractions.

Additional Integration Techniques

Integrals Involving Trigonometric Identities

Many trigonometric integrals require using identities to simplify before integrating.

Powers of Sine and Cosine

Odd power of sine: Factor out one $\sin x$ Convert the rest to cosines using $\sin^2 x = 1 - \cos^2 x$ Then substitute $u = \cos x$ .

Evaluate $\displaystyle\int \cos^4 x\,dx$ .

\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}

= \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8}

\int \cos^4 x\,dx = \frac{3x}{8} + \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C

Products of Sine and Cosine

For $\displaystyle\int \sin mx \cos nx\,dx$ Use the product-to-sum identities:

\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]

\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]

\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]

Evaluate $\displaystyle\int \sin 3x \cos 5x\,dx$ .

\sin 3x \cos 5x = \frac{1}{2}[\sin 8x + \sin(-2x)] = \frac{1}{2}(\sin 8x - \sin 2x)

\int \sin 3x \cos 5x\,dx = \frac{1}{2}\left(-\frac{\cos 8x}{8} + \frac{\cos 2x}{2}\right) + C

\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C

\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + C

\int \frac{1}{x^2 - a^2}\,dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C

Evaluate $\displaystyle\int \frac{1}{4 + 9x^2}\,dx$ .

Rewrite as: $\displaystyle\int \frac{1}{4 + (3x)^2}\,dx$ .

Let $u = 3x$$du = 3\,dx$ :

= \frac{1}{3}\int \frac{1}{4 + u^2}\,du = \frac{1}{3} \cdot \frac{1}{2}\arctan\frac{u}{2} + C = \frac{1}{6}\arctan\frac{3x}{2} + C

Integrals Involving $e^x$ and $\ln x$

Integrals with $e^x$ and Polynomials

Use integration by parts when $e^x$ is multiplied by a polynomial.

Integrals with $\ln x$

Use integration by parts with $u = \ln x$ and $dv = dx$ :

\int \ln x\,dx = x\ln x - x + C

Evaluate $\displaystyle\int x^2 \ln x\,dx$ .

Let $u = \ln x$$dv = x^2\,dx$ . Then $du = \dfrac{1}{x}\,dx$$v = \dfrac{x^3}{3}$ .

\int x^2 \ln x\,dx = \frac{x^3}{3}\ln x - \int \frac{x^3}{3} \cdot \frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx

= \frac{x^3}{3}\ln x - \frac{x^3}{9} + C

Arc Length

The arc length of a curve $y = f(x)$ from $x = a$ to $x = b$ :

L = \int_a^b \sqrt{1 + \left(\frac`\{dy}`, `\{dx}`\right)^2}\,dx

For a parametric curve $(x(t), y(t))$ from $t = t_1$ to $t = t_2$ :

L = \int_{t_1}^{t_2} \sqrt{\left(\frac`\{dx}`, `\{dt}`\right)^2 + \left(\frac`\{dy}`, `\{dt}`\right)^2}\,dt

Find the arc length of $y = \dfrac{x^3}{6} + \dfrac{1}{2x}$ from $x = 1$ to $x = 3$ .

\frac`\{dy}`, `\{dx}` = \frac{x^2}{2} - \frac{1}{2x^2}

\left(\frac`\{dy}`, `\{dx}`\right)^2 = \frac{x^4}{4} - \frac{1}{2} + \frac{1}{4x^4}

1 + \left(\frac`\{dy}`, `\{dx}`\right)^2 = \frac{x^4}{4} + \frac{1}{2} + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2

L = \int_1^3 \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)\,dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^3

= \left(\frac{27}{6} - \frac{1}{6}\right) - \left(\frac{1}{6} - \frac{1}{2}\right) = \frac{26}{6} + \frac{1}{3} = \frac{14}{3}

Mean Value of a Function

The mean value of a function $f(x)$ over $[a, b]$ :

\bar{f} = \frac{1}{b - a}\int_a^b f(x)\,dx

Additional Exam-Style Questions

Question 7 (Paper 2 style)

Evaluate $\displaystyle\int_0^{\pi/4} x\cos 2x\,dx$ .

Let $u = x$$dv = \cos 2x\,dx$ . Then $du = dx$$v = \dfrac{\sin 2x}{2}$ .

\int_0^{\pi/4} x\cos 2x\,dx = \left[\frac{x\sin 2x}{2}\right]_0^{\pi/4} - \frac{1}{2}\int_0^{\pi/4}\sin 2x\,dx

= \left(\frac{\pi/4 \times 1}{2} - 0\right) - \frac{1}{2}\left[-\frac{\cos 2x}{2}\right]_0^{\pi/4}

= \frac{\pi}{8} - \frac{1}{2}\left(-\frac{0}{2} + \frac{1}{2}\right) = \frac{\pi}{8} - \frac{1}{4}

Question 8 (Paper 2 style)

Find the area enclosed by the curves $y = x^3$ and $y = \sqrt{x}$ .

Intersection: $x^3 = \sqrt{x} \implies x^6 = x \implies x(x^5 - 1) = 0 \implies x = 0$ or $x = 1$ .

For $0 \le x \le 1$ : $\sqrt{x} \ge x^3$ .

\mathrm{Area} = \int_0^1 (\sqrt{x} - x^3)\,dx = \left[\frac{2x^{3/2}}{3} - \frac{x^4}{4}\right]_0^1 = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}

Question 9 (Paper 1 style)

Evaluate $\displaystyle\int \frac{x}{x^2 + 2x + 5}\,dx$ .

Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$ .

Let $u = x^2 + 2x + 5$$du = (2x + 2)\,dx$ .

We need to split: $\dfrac{x}{x^2+2x+5} = \dfrac{(2x+2)/2 - 1}{x^2+2x+5} = \dfrac{1}{2} \cdot \dfrac{2x+2}{x^2+2x+5} - \dfrac{1}{x^2+2x+5}$ .

\int \frac{x}{x^2+2x+5}\,dx = \frac{1}{2}\ln(x^2+2x+5) - \int \frac{1}{(x+1)^2 + 4}\,dx

= \frac{1}{2}\ln(x^2+2x+5) - \frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C

Question 10 (Paper 2 style)

The region bounded by $y = \ln x$$y = 0$ And $x = e$ is rotated $360\degree$ about the $x$ -axis. Find the volume.

V = \pi\int_1^e (\ln x)^2\,dx

Using integration by parts with $u = (\ln x)^2$$dv = dx$ :

= \pi\left[x(\ln x)^2\right]_1^e - 2\pi\int_1^e \ln x\,dx

= \pi(e - 0) - 2\pi[x\ln x - x]_1^e = \pi e - 2\pi(e - e + 1) = \pi e - 2\pi = \pi(e - 2)

For the A-Level treatment of this topic, see Integration.

the IB specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Integration with other IB mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Intuition

Integration is the art of undoing differentiation, and each technique is a different strategy for dismantling complexity. Substitution is the chain rule in reverse, if you spot a function inside another function, you peel away the outer layer by changing variables. Integration by parts splits a product into something you can integrate and something that disappears, guided by the LIATE hierarchy which ranks functions by how much they simplify when differentiated. Partial fractions拆 apart rational functions into pieces that each integrate to logarithms or arctangents. The Fundamental Theorem of Calculus is the bridge connecting the two halves of calculus, area and slope are inverse operations.

Common Pitfalls

Forgetting the chain rule when integrating composite functions. Look for an inner function and its derivative.
Confusing definite and indefinite integrals. Definite integrals give a numerical value; indefinite integrals give a family of functions.
Incorrectly applying limits of integration after a substitution. Remember to change the limits to the new variable.
Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’. The former requires using previous work.
Cancelling terms instead of factors, $\frac{ab + ac}{a}$ simplifies to $b + c$ , not $bc$ .

Summary

Indefinite integral always includes $+C$ ; definite integral evaluates at bounds
Area between curves: $\int |f(x) - g(x)|\,dx$
Volume of revolution: $V = \pi\int y^2\,dx$
Techniques: substitution, integration by parts (LIATE), partial fractions

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

Differential Calculus covers differentiation techniques that integration reverses, forming the fundamental theorem of calculus.
Integral Calculus applies these integration techniques to calculate areas, volumes, and accumulated quantities.
Series and Differential Equations connects integration methods to infinite series and solutions of differential equations.

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