A-Level Mathematics – Chapter 7: Integration
Integration is the reverse process of differentiation and one of the most powerful tools in mathematics. It enables us to compute areas, volumes, accumulated quantities, and to solve differential equations that model real-world phenomena. This chapter builds from the fundamental rules introduced at AS level through to the sophisticated Year 2 techniques required for the full A-Level.
Specification Note
Content labelled Year 2 is A2-level only and not required for AS Mathematics. The trapezium rule and definite integration appear on both AS and A-Level papers.
7.1 Indefinite Integration
Integration is formally defined as the inverse (or anti-derivative) of differentiation. If $\frac{d}{dx}[F(x)] = f(x)$, then $\int f(x)\,dx = F(x) + c$, where $c$ is the constant of integration.
Definition: Indefinite Integral
The indefinite integral of $f(x)$ with respect to $x$ is written $\displaystyle\int f(x)\,dx$ and represents the most general antiderivative of $f(x)$. The constant of integration $c$ must always be included because differentiating any constant gives zero.
The Power Rule for Integration
The fundamental rule for integrating powers of $x$ is:
$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + c \qquad (n \neq -1)$$In words: increase the power by 1, then divide by the new power. The case $n = -1$ (integrating $\frac{1}{x}$) requires the natural logarithm and is covered below.
Standard Integrals
You are expected to know these standard results — many appear in the formula booklet, but recognising them instantly saves valuable exam time.
| $f(x)$ | $\displaystyle\int f(x)\,dx$ |
|---|---|
| $x^n$ $(n \neq -1)$ | $\dfrac{x^{n+1}}{n+1} + c$ |
| $\dfrac{1}{x}$ | $\ln|x| + c$ |
| $e^x$ | $e^x + c$ |
| $e^{kx}$ | $\dfrac{1}{k}e^{kx} + c$ |
| $\sin x$ | $-\cos x + c$ |
| $\cos x$ | $\sin x + c$ |
| $\sec^2 x$ | $\tan x + c$ |
| $(ax + b)^n$ | $\dfrac{(ax+b)^{n+1}}{a(n+1)} + c$ |
Example 7.1.1 — Basic Power Rule
Find $\displaystyle\int (3x^2 - 4x + 5)\,dx$.
Step 1 Integrate each term separately using $\int x^n\,dx = \frac{x^{n+1}}{n+1} + c$:
$$\int 3x^2\,dx = \frac{3x^3}{3} = x^3 \qquad \int 4x\,dx = \frac{4x^2}{2} = 2x^2 \qquad \int 5\,dx = 5x$$
Step 2 Combine and add a single constant of integration:
$$\int(3x^2 - 4x + 5)\,dx = x^3 - 2x^2 + 5x + c$$
Example 7.1.2 — Negative and Fractional Indices
Find $\displaystyle\int \left(x^{1/2} + \frac{2}{x^3}\right)dx$.
Step 1 Rewrite $\frac{2}{x^3} = 2x^{-3}$.
Step 2 Apply the power rule to each term:
$$\int x^{1/2}\,dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2} \qquad \int 2x^{-3}\,dx = \frac{2x^{-2}}{-2} = -x^{-2}$$
Result $\displaystyle\int\!\left(x^{1/2} + \frac{2}{x^3}\right)dx = \tfrac{2}{3}x^{3/2} - x^{-2} + c$
Example 7.1.3 — Exponential and Trigonometric
Find $\displaystyle\int(e^{2x} + 3\sin x - 4\cos x)\,dx$.
$$= \frac{1}{2}e^{2x} + 3(-\cos x) - 4(\sin x) + c = \frac{1}{2}e^{2x} - 3\cos x - 4\sin x + c$$
Example 7.1.4 — Finding a Particular Integral
Given that $\frac{dy}{dx} = 6x^2 - 2x$ and $y = 5$ when $x = 1$, find $y$ in terms of $x$.
Step 1 Integrate: $y = 2x^3 - x^2 + c$.
Step 2 Substitute the initial condition $x=1$, $y=5$: $5 = 2(1) - 1 + c \Rightarrow c = 4$.
Result $y = 2x^3 - x^2 + 4$.
Exam Tip
Always include $+\,c$ in indefinite integrals. Omitting it is a common mark-losing error. When an initial condition is given, use it immediately to find $c$ and write the particular solution.
7.2 Definite Integration
A definite integral has upper and lower limits and produces a numerical value rather than an expression. It represents the signed area between the curve and the $x$-axis.
Fundamental Theorem of Calculus
If $F(x)$ is an antiderivative of $f(x)$, then:
$$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$$
The constant of integration cancels out in definite integrals, so it need not be written.
Area Under a Curve
The area enclosed between $y = f(x)$, the $x$-axis, and the lines $x = a$ and $x = b$ is:
$$A = \int_a^b f(x)\,dx \quad \text{(when } f(x) \geq 0 \text{ throughout)}$$If $f(x) < 0$ in part of the interval, the integral gives a negative value for that region. To find the total area (always positive), split the integral at any roots and take the modulus of each part.
Figure 7.1 — Shaded area under $y = x^2 + 1$ from $x = 1$ to $x = 3$. The definite integral gives $A = \frac{20}{3} + 2 = \frac{26}{3} \approx 8.67$.
Example 7.2.1 — Evaluating a Definite Integral
Evaluate $\displaystyle\int_1^3 (x^2 + 1)\,dx$.
$$\left[\frac{x^3}{3} + x\right]_1^3 = \left(\frac{27}{3} + 3\right) - \left(\frac{1}{3} + 1\right) = 12 - \frac{4}{3} = \frac{32}{3}$$
Example 7.2.2 — Area Below the Axis
Find the total area enclosed between $y = x^2 - 4$ and the $x$-axis.
Step 1 The curve crosses the $x$-axis where $x^2 - 4 = 0$, i.e.\ $x = \pm 2$.
Step 2 For $-2 \leq x \leq 2$, the curve is below the axis. So the area is:
$$A = \left|\int_{-2}^{2}(x^2-4)\,dx\right| = \left|\left[\frac{x^3}{3} - 4x\right]_{-2}^{2}\right| = \left|\left(\frac{8}{3}-8\right)-\left(\frac{-8}{3}+8\right)\right| = \left|\!-\frac{32}{3}\right| = \frac{32}{3}$$
Area Between Two Curves
The area between $y = f(x)$ (upper) and $y = g(x)$ (lower) from $x = a$ to $x = b$ is:
$$A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx$$Example 7.2.3 — Area Between Two Curves
Find the area enclosed between $y = 4 - x^2$ and $y = x + 2$.
Step 1 Find intersections: $4 - x^2 = x + 2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x+2)(x-1) = 0$, so $x = -2$ and $x = 1$.
Step 2 On $[-2,1]$, $4 - x^2 \geq x + 2$. Area:
$$\int_{-2}^{1}\bigl[(4-x^2)-(x+2)\bigr]\,dx = \int_{-2}^{1}(2 - x - x^2)\,dx = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-2}^{1} = \frac{9}{2}$$
7.3 Integration by Substitution Year 2
Integration by substitution is the chain-rule in reverse. It converts a complicated integral into a simpler one by changing the variable.
Method: Integration by Substitution
To evaluate $\int f(g(x))\,g'(x)\,dx$, let $u = g(x)$. Then $\frac{du}{dx} = g'(x)$, so $du = g'(x)\,dx$, and the integral becomes $\int f(u)\,du$.
For definite integrals, also change the limits: when $x = a$, $u = g(a)$; when $x = b$, $u = g(b)$.
Example 7.3.1 — Polynomial Substitution
Find $\displaystyle\int 2x(x^2 + 3)^4\,dx$.
Let $u = x^2 + 3 \Rightarrow du = 2x\,dx$.
$$\int u^4\,du = \frac{u^5}{5} + c = \frac{(x^2+3)^5}{5} + c$$
Example 7.3.2 — Trigonometric Substitution
Find $\displaystyle\int \sin^3 x \cos x\,dx$.
Let $u = \sin x \Rightarrow du = \cos x\,dx$.
$$\int u^3\,du = \frac{u^4}{4} + c = \frac{\sin^4 x}{4} + c$$
Example 7.3.3 — Exponential Substitution
Find $\displaystyle\int x\,e^{x^2}\,dx$.
Let $u = x^2 \Rightarrow du = 2x\,dx$, so $x\,dx = \frac{1}{2}\,du$.
$$\int \frac{1}{2}e^u\,du = \frac{1}{2}e^u + c = \frac{1}{2}e^{x^2} + c$$
Example 7.3.4 — Definite Integral with Changed Limits
Evaluate $\displaystyle\int_0^1 \frac{x}{(1+x^2)^2}\,dx$.
Let $u = 1 + x^2 \Rightarrow du = 2x\,dx$. When $x=0$, $u=1$; when $x=1$, $u=2$.
$$\int_1^2 \frac{1}{2u^2}\,du = \left[-\frac{1}{2u}\right]_1^2 = -\frac{1}{4} + \frac{1}{2} = \frac{1}{4}$$
7.4 Integration by Parts Year 2
Integration by parts is derived from the product rule for differentiation. It is used when the integrand is a product of two functions.
Formula: Integration by Parts
$$\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$$
Equivalently written as $\displaystyle\int u\,dv = uv - \int v\,du$.
LIATE rule — choose $u$ in this order of preference: Logarithms, Inverse trig, Algebraic, Trigonometric, Exponential.
Example 7.4.1 — Polynomial × Exponential
Find $\displaystyle\int x\,e^x\,dx$.
Let $u = x$ and $\frac{dv}{dx} = e^x$, so $\frac{du}{dx} = 1$ and $v = e^x$.
$$\int x\,e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + c = e^x(x-1) + c$$
Example 7.4.2 — Polynomial × Trigonometric
Find $\displaystyle\int x\cos x\,dx$.
Let $u = x$, $\frac{dv}{dx} = \cos x$, so $\frac{du}{dx} = 1$, $v = \sin x$.
$$\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + c$$
Example 7.4.3 — Logarithm (Special Case)
Find $\displaystyle\int \ln x\,dx$.
Write as $\int \ln x \cdot 1\,dx$. Let $u = \ln x$, $\frac{dv}{dx} = 1$, so $\frac{du}{dx} = \frac{1}{x}$, $v = x$.
$$\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + c$$
Example 7.4.4 — Applying Twice
Find $\displaystyle\int x^2 e^x\,dx$.
First application: $u = x^2$, $v = e^x$: $\int x^2 e^x\,dx = x^2 e^x - 2\int xe^x\,dx$.
Second application (from Example 7.4.1): $\int xe^x\,dx = e^x(x-1) + c$.
$$\int x^2 e^x\,dx = x^2 e^x - 2e^x(x-1) + c = e^x(x^2 - 2x + 2) + c$$
7.5 Partial Fractions and Integration Year 2
When integrating a rational function $\frac{f(x)}{g(x)}$, it is often necessary to decompose it into partial fractions first, after which each term can be integrated using standard results.
Key Result
$$\int \frac{1}{ax+b}\,dx = \frac{1}{a}\ln|ax+b| + c$$
This follows from the substitution $u = ax + b$ and is used directly after decomposing into partial fractions.
Example 7.5.1 — Distinct Linear Factors
Find $\displaystyle\int \frac{5x - 1}{(x+1)(2x-1)}\,dx$.
Step 1 Decompose: $\frac{5x-1}{(x+1)(2x-1)} = \frac{A}{x+1} + \frac{B}{2x-1}$.
Comparing numerators: $5x - 1 = A(2x-1) + B(x+1)$. Setting $x = -1$: $A = 2$; $x = \frac{1}{2}$: $B = \frac{3}{2} \cdot \frac{1}{1} = \frac{3}{2}$.
Step 2 Integrate: $\displaystyle\int\!\left(\frac{2}{x+1} + \frac{3/2}{2x-1}\right)dx = 2\ln|x+1| + \frac{3}{4}\ln|2x-1| + c$.
Example 7.5.2 — Repeated Linear Factor
Find $\displaystyle\int \frac{3x}{(x-1)^2}\,dx$.
Step 1 Decompose: $\frac{3x}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$. This gives $A = 3$, $B = 3$.
Step 2 Integrate: $3\ln|x-1| - \frac{3}{x-1} + c$.
7.6 Trapezium Rule
When an integral cannot be evaluated analytically, numerical methods provide approximate answers. The trapezium rule approximates the area under a curve by dividing it into trapezoids.
Trapezium Rule Formula
With $n$ strips of equal width $h = \frac{b-a}{n}$ and ordinates $y_0, y_1, \ldots, y_n$:
$$\int_a^b f(x)\,dx \approx \frac{h}{2}\bigl(y_0 + 2y_1 + 2y_2 + \cdots + 2y_{n-1} + y_n\bigr)$$
Equivalently: $\frac{h}{2}\bigl(\text{first} + \text{last} + 2\times\text{sum of middles}\bigr)$.
Figure 7.2 — Trapezium rule approximation to $\int_0^2 e^{x/2}\,dx$ using 4 strips. The exact value is $2(e-1) \approx 3.436$; the trapezium estimate slightly overestimates since $e^{x/2}$ is convex.
Example 7.6.1 — Applying the Trapezium Rule
Use the trapezium rule with 4 strips to estimate $\displaystyle\int_0^2 e^{x/2}\,dx$.
Setup $h = 0.5$; ordinates at $x = 0, 0.5, 1, 1.5, 2$:
$y_0 = e^0 = 1$, $y_1 = e^{0.25} \approx 1.2840$, $y_2 = e^{0.5} \approx 1.6487$, $y_3 = e^{0.75} \approx 2.1170$, $y_4 = e^1 \approx 2.7183$.
$$\approx \frac{0.5}{2}\bigl(1 + 2(1.2840 + 1.6487 + 2.1170) + 2.7183\bigr) = 0.25 \times 14.1167 \approx 3.529$$
The exact value is $2(e-1) \approx 3.436$; percentage error $\approx 2.7\%$.
Example 7.6.2 — Error and Improvement
Explain how to reduce the error in the trapezium rule, and state whether the rule over- or under-estimates for a convex function.
Increasing the number of strips $n$ reduces the error, which is $O(h^2)$ — doubling $n$ roughly quarters the error. For a convex function ($f''(x) > 0$), the straight-line tops of the trapezoids lie above the curve, so the rule over-estimates. For a concave function ($f''(x) < 0$), it under-estimates.
7.7 Differential Equations Year 2
A differential equation relates a function to its derivatives. At A-Level, you are required to solve first-order separable differential equations — those in which the variables $x$ and $y$ can be placed on opposite sides of the equation.
Method: Separation of Variables
For an equation of the form $\dfrac{dy}{dx} = f(x)\,g(y)$:
- Separate: $\dfrac{1}{g(y)}\,dy = f(x)\,dx$
- Integrate both sides: $\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$
- Include a single constant of integration $c$
- Apply initial conditions (if given) to find the particular solution
Example 7.7.1 — Basic Separation
Solve $\dfrac{dy}{dx} = 2xy$.
$$\frac{1}{y}\,dy = 2x\,dx \Rightarrow \int \frac{1}{y}\,dy = \int 2x\,dx \Rightarrow \ln|y| = x^2 + c$$
Exponentiating: $|y| = e^c \cdot e^{x^2}$, so $y = Ae^{x^2}$ where $A$ is an arbitrary constant.
Example 7.7.2 — Initial Value Problem
Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ given $y = 3$ when $x = 0$.
$$y\,dy = x\,dx \Rightarrow \frac{y^2}{2} = \frac{x^2}{2} + c$$
Applying $y=3$, $x=0$: $\frac{9}{2} = 0 + c$, so $c = \frac{9}{2}$. The particular solution is $y^2 = x^2 + 9$, i.e.\ $y = \sqrt{x^2+9}$ (positive root since $y=3>0$).
Example 7.7.3 — Exponential Growth/Decay
The rate of decay of a radioactive substance is proportional to the amount $N$ present. Write down and solve the differential equation, given $N = N_0$ at $t = 0$.
$$\frac{dN}{dt} = -kN \quad (k > 0) \Rightarrow \frac{dN}{N} = -k\,dt \Rightarrow \ln N = -kt + c$$
Applying initial condition: $N = N_0 e^{-kt}$.
Example 7.7.4 — Mixed Variables
Solve $\dfrac{dy}{dx} = \dfrac{y^2 + 1}{x}$, given $y = 1$ when $x = 1$.
$$\frac{dy}{y^2+1} = \frac{dx}{x} \Rightarrow \arctan y = \ln|x| + c$$
At $x=1$, $y=1$: $\arctan 1 = 0 + c$, so $c = \frac{\pi}{4}$. Hence $y = \tan\!\left(\ln x + \dfrac{\pi}{4}\right)$.
Exam Tip
When forming a differential equation from a word problem, pay attention to the sign: "decreasing at a rate proportional to" gives a negative sign. Also check whether the question asks for a general or particular solution — if initial conditions are given, you must find $c$.
Practice Problems
Problem 1
Find $\displaystyle\int\!\left(4x^3 - \frac{3}{x^2} + \sqrt{x}\right)dx$.
Show solution
$$x^4 + 3x^{-1} + \frac{2}{3}x^{3/2} + c = x^4 + \frac{3}{x} + \frac{2}{3}x^{3/2} + c$$
Problem 2
Evaluate $\displaystyle\int_0^{\pi/2} (2\sin x + \cos x)\,dx$.
Show solution
Problem 3
Find the area enclosed between $y = x^3 - x$ and the $x$-axis.
Show solution
$\int_0^1(x^3-x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_0^1 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.
Total area $= 2 \times \frac{1}{4} = \frac{1}{2}$.
Problem 4 Year 2
Use the substitution $u = 3x + 1$ to find $\displaystyle\int_0^1 \sqrt{3x+1}\,dx$.
Show solution
$$\int_1^4 \sqrt{u}\cdot\frac{du}{3} = \frac{1}{3}\left[\frac{2u^{3/2}}{3}\right]_1^4 = \frac{2}{9}(8-1) = \frac{14}{9}$$
Problem 5 Year 2
Find $\displaystyle\int x^2\ln x\,dx$.
Show solution
$$\frac{x^3}{3}\ln x - \int \frac{x^3}{3}\cdot\frac{1}{x}\,dx = \frac{x^3}{3}\ln x - \frac{1}{3}\int x^2\,dx = \frac{x^3}{3}\ln x - \frac{x^3}{9} + c$$
Problem 6 Year 2
Express $\dfrac{4}{(x-1)(x+3)}$ in partial fractions and hence find $\displaystyle\int_2^5 \frac{4}{(x-1)(x+3)}\,dx$.
Show solution
$$\int_2^5\!\left(\frac{1}{x-1} - \frac{1}{x+3}\right)dx = \big[\ln|x-1| - \ln|x+3|\big]_2^5 = \ln\frac{4}{8} - \ln\frac{1}{5} = \ln\frac{4}{8} + \ln 5 = \ln\frac{5}{2}$$
Problem 7
Use the trapezium rule with 5 ordinates (4 strips) to estimate $\displaystyle\int_1^3 \frac{1}{\ln x}\,dx$. Give your answer to 3 decimal places.
Show solution
$y_0$: undefined at $x=1$ ($\ln 1=0$) — use $x$ values from $x=1$ as $x\to1^+$ gives $\infty$. In practice this integral requires $x>1$; use $x = 1.0001$ to obtain $y_0 \approx 2.309$ or note the question may intend $x$ from a suitable lower limit. Ordinates: $y(1.5)=\frac{1}{\ln 1.5}\approx 2.466$, $y(2)=\frac{1}{\ln 2}\approx 1.443$, $y(2.5)=\frac{1}{\ln 2.5}\approx 1.093$, $y(3)=\frac{1}{\ln 3}\approx 0.910$.
Approximation $\approx \frac{0.5}{2}(2.466 + 2(1.443 + 1.093) + 0.910) \approx \frac{0.5}{2}(8.448) \approx 2.112$.
Problem 8 Year 2
Solve the differential equation $\dfrac{dy}{dx} = \dfrac{3x^2}{y}$, given that $y = 2$ when $x = 1$.
Show solution
Apply $y=2$, $x=1$: $2 = 1 + c \Rightarrow c = 1$.
Particular solution: $y^2 = 2x^3 + 2$, so $y = \sqrt{2x^3 + 2}$ (positive since $y=2>0$).
Problem 9 Year 2
Find $\displaystyle\int e^x\sin x\,dx$.
Show solution
First: $u=\sin x$, $v=e^x$: $I = e^x\sin x - \int e^x\cos x\,dx$.
Second: $u=\cos x$, $v=e^x$: $\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I$.
Substituting back: $I = e^x\sin x - e^x\cos x - I \Rightarrow 2I = e^x(\sin x - \cos x)$.
$$I = \frac{e^x(\sin x - \cos x)}{2} + c$$
Problem 10
Find the area of the region enclosed between $y = \sin x$ and $y = \cos x$ for $x \in \left[\frac{\pi}{4}, \frac{5\pi}{4}\right]$.
Show solution
$$\int_{\pi/4}^{5\pi/4}(\sin x - \cos x)\,dx = \big[-\cos x - \sin x\big]_{\pi/4}^{5\pi/4}$$ $$= \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} = 2\sqrt{2}$$