Notes · MathematicsUK · A-Levels

Integration

Integration is the reverse of differentiation and the tool for accumulating change. The A-Level covers indefinite integrals and the standard results, definite integrals as areas, the techniques of substitution and integration by parts and integration using partial fractions, and the forming and solving of differential equations by separating the variables.

4 sections~9 min reading time3 competenciesLevel Standard 2 · Advanced 2

T·0999 / 20

Exam profile

AO1 · Integrate standard functions and evaluate definite integralsAO2 · Justify area and limit arguments and interpret the constant of integrationAO3 · Form and solve differential-equation models and interpret them

Operators:integratefindevaluateshow thatsolvedetermine

basic level

AS-Level covers integration of powers of $x$ , indefinite and definite integrals, and areas under curves.

higher level

The full A-Level adds integration of $e^x$ , $\tfrac{1}{x}$ and trigonometric functions, substitution, integration by parts, partial fractions and differential equations.

Depth

Text

Contents · 4 sections

Integration

Indefinite integration and standard results#

StandardDfE GCE Mathematics H1AQA 7357 H

The family of antiderivatives

Fig. 1

\displaystyle\int x\,dx = \tfrac{1}{2}x^2 + c

: every value of

c

gives a curve with the same gradient function.

Integration is the reverse of differentiation: to integrate

f(x)

is to find a function whose derivative is

f(x)

. Because differentiating a constant gives zero, any two antiderivatives differ by a constant, so an indefinite integral always carries an arbitrary constant of integration:

\displaystyle\int f(x)\,dx = F(x) + c

. Omitting the

+c

is one of the most common and most easily avoided errors.

Reversing the power rule gives the fundamental result

\displaystyle\int x^{n}\,dx = \dfrac{x^{n+1}}{n + 1} + c

, valid for all rational

n

except

n = -1

(where the answer is

\ln|x|

instead). You raise the power by one and divide by the new power — exactly undoing 'multiply by the power, drop it by one'. Roots and reciprocals are integrated by first writing them as powers.

The other standard integrals mirror the standard derivatives:

\displaystyle\int e^{x}\,dx = e^{x} + c

\displaystyle\int \dfrac{1}{x}\,dx = \ln|x| + c

\displaystyle\int \cos x\,dx = \sin x + c

and

\displaystyle\int \sin x\,dx = -\cos x + c

. Learning these as the reverse of the derivatives you already know is far more reliable than memorising them separately.

Integration is linear, so constant multiples come out and sums integrate term by term. When the integrand is a function of a linear expression, such as

e^{2x + 1}

(3x - 1)^4

, you integrate as usual and divide by the derivative of the inner bracket — a reversal of the chain rule that handles the majority of routine integrals.

\int x^{n}\,dx = \frac{x^{n+1}}{n + 1} + c \quad (n \neq -1)

Reverse power rule

Raise the power by one and divide by the new power; the exception $n = -1$ integrates to $\ln|x|$ .

\int \frac{1}{x}\,dx = \ln|x| + c

The exceptional case

The reciprocal integrates to the natural logarithm of the modulus.

Worked example

Integrating a sum of powers

Find $\displaystyle\int \left(6x^2 + \dfrac{2}{x^3}\right)\,dx$ .

Write in index form
$\displaystyle\int (6x^2 + 2x^{-3})\,dx$ .
Integrate term by term
$\dfrac{6x^3}{3} + \dfrac{2x^{-2}}{-2} + c = 2x^3 - x^{-2} + c$ .
$2x^3 - x^{-2} + c$
Tidy
$= 2x^3 - \dfrac{1}{x^2} + c$ .

Result: $\displaystyle\int \left(6x^2 + \dfrac{2}{x^3}\right)dx = 2x^3 - \dfrac{1}{x^2} + c$ .

Typical mistakes

Forgetting the constant of integration.
Applying $\dfrac{x^{n+1}}{n+1}$ when $n = -1$ , which divides by zero.

Active revision

Find $\displaystyle\int \left(3x^2 - \dfrac{4}{x^2} + \sqrt{x}\right)\,dx$ .

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Definite integrals and areas#

StandardDfE GCE Mathematics H2-H3AQA 7357 H

Area under a curve

Fig. 2

\displaystyle\int_0^2 x^2\,dx = \left[\tfrac{x^3}{3}\right]_0^2 = \tfrac{8}{3}

, the shaded area.

A definite integral

\displaystyle\int_a^b f(x)\,dx

has numerical value

F(b) - F(a)

, where

F

is any antiderivative of

f

; the constant of integration cancels in the subtraction, so it is omitted. This is the fundamental theorem of calculus: differentiation and integration are inverse processes, and the accumulated change of

f

over

[a, b]

is the difference in its antiderivative at the endpoints.

Geometrically,

\displaystyle\int_a^b f(x)\,dx

is the signed area between the curve and the

x

-axis from

x = a

x = b

. Area below the axis counts as negative, so if a curve crosses the axis in the interval you must split the integral at the crossing points and take the modulus of each part to find the total physical area. Ignoring this sign convention is a classic source of wrong answers.

The area between two curves

y = f(x)

(upper) and

y = g(x)

(lower) is

\displaystyle\int_a^b \big(f(x) - g(x)\big)\,dx

, where

a

and

b

are the

x

-coordinates of the points of intersection. You find the limits by solving

f(x) = g(x)

, then integrate the difference 'top minus bottom'. This handles regions enclosed between a line and a curve, or between two curves.

Definite integrals obey useful properties: reversing the limits changes the sign,

\displaystyle\int_b^a = -\int_a^b

, and the interval can be split,

\displaystyle\int_a^c = \int_a^b + \int_b^c

. These, together with the area interpretation, let you evaluate areas that would be awkward to compute in one piece.

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

The fundamental theorem of calculus

Evaluate any antiderivative at the two limits and subtract; the constant cancels.

\text{Area} = \int_a^b \big(f(x) - g(x)\big)\,dx

Area between two curves

Integrate upper minus lower between the x-coordinates of the intersections.

Area between a line and a curve

Fig. 3Between

y = x + 2

and

y = x^2

(meeting at

x = -1

and

x = 2

), the enclosed area is

\tfrac{9}{2}

Worked example

Area between a line and a curve

Find the area enclosed between $y = x + 2$ and $y = x^2$ .

Find the intersections
$x + 2 = x^2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow (x - 2)(x + 1) = 0$ , so $x = -1$ and $x = 2$ .
Integrate top minus bottom
$\displaystyle\int_{-1}^{2}\big((x + 2) - x^2\big)\,dx = \left[\tfrac{x^2}{2} + 2x - \tfrac{x^3}{3}\right]_{-1}^{2}$ .
$\int_{-1}^{2}(x + 2 - x^2)\,dx$
Evaluate
At $x = 2$ : $2 + 4 - \tfrac{8}{3} = \tfrac{10}{3}$ . At $x = -1$ : $\tfrac{1}{2} - 2 + \tfrac{1}{3} = -\tfrac{7}{6}$ . Difference $= \tfrac{10}{3} + \tfrac{7}{6} = \tfrac{9}{2}$ .

Result: The enclosed area is $\dfrac{9}{2} = 4.5$ .

Typical mistakes

Reporting a negative definite integral as an area without accounting for the sign.
Using the wrong order (bottom minus top) when finding the area between two curves.

Active revision

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

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Techniques: substitution and by parts#

AdvancedDfE GCE Mathematics H4-H5AQA 7357 H

Integration by substitution reverses the chain rule by changing to a new variable

u

. You choose

u = g(x)

, compute

\dfrac{du}{dx}

so that

du = g'(x)\,dx

, and rewrite the whole integral — integrand and

dx

— in terms of

u

. The substitution is well chosen when the integral in

u

is a standard one. For a definite integral you must also change the limits to

u

-values, or convert back to

x

before substituting the original limits.

A frequent pattern is an integral containing a function and (a multiple of) its derivative, such as

\displaystyle\int 2x\,e^{x^2}\,dx

: putting

u = x^2

gives

du = 2x\,dx

and the integral becomes

\displaystyle\int e^{u}\,du = e^{u} + c = e^{x^2} + c

. Recognising this 'function-and-its-derivative' structure is the main skill, and it also gives the standard result

\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c

Integration by parts reverses the product rule and is used for integrals of products such as

x e^{x}

x\cos x

. The formula is

\displaystyle\int u\,\dfrac{dv}{dx}\,dx = uv - \int v\,\dfrac{du}{dx}\,dx

; you choose

u

to be the part that simplifies when differentiated (often the polynomial) and

\dfrac{dv}{dx}

to be the part you can integrate. A helpful priority for choosing

u

is 'logs, then powers, then exponentials/trigonometric'.

Some integrals need parts applied twice, or a clever trick such as integrating

\ln x

by taking

u = \ln x

and

\dfrac{dv}{dx} = 1

. Choosing the parts wisely is what turns a hard integral into a routine one; a poor choice makes the remaining integral worse, which is the signal to swap the roles of

u

and

\dfrac{dv}{dx}

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx

Integration by parts

Choose $u$ to simplify on differentiating; $\tfrac{dv}{dx}$ must be integrable.

\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c

A standard substitution result

When the numerator is the derivative of the denominator, the integral is a logarithm.

Worked example

Integration by parts

Find $\displaystyle\int x e^{x}\,dx$ .

Choose the parts
Let $u = x$ (simplifies on differentiating) and $\dfrac{dv}{dx} = e^{x}$ , so $\dfrac{du}{dx} = 1$ and $v = e^{x}$ .
Apply the formula
$\displaystyle\int x e^{x}\,dx = x e^{x} - \int e^{x}\cdot 1\,dx$ .
$\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx$
Finish
$= x e^{x} - e^{x} + c = e^{x}(x - 1) + c$ .

Result: $\displaystyle\int x e^{x}\,dx = e^{x}(x - 1) + c$ .

Typical mistakes

Forgetting to change the limits when substituting in a definite integral.
Choosing $u$ and $\dfrac{dv}{dx}$ so that the remaining integral is harder than the original.

Active revision

Find (a) $\displaystyle\int x\sqrt{x^2 + 1}\,dx$ by substitution and (b) $\displaystyle\int x\sin x\,dx$ by parts.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Differential equations#

AdvancedDfE GCE Mathematics H6AQA 7357 H

Solution of a growth equation

Fig. 4

\dfrac{dP}{dt} = 0.15P

with

P(0) = 100

solves to

P = 100e^{0.15t}

A differential equation relates a quantity to its rate of change; the A-Level treats first-order equations that separate. Many models arise directly from a modelling statement: 'the rate of growth is proportional to the population' becomes

\dfrac{dP}{dt} = kP

; 'the rate of cooling is proportional to the temperature difference' becomes

\dfrac{d\theta}{dt} = -k(\theta - \theta_0)

. Translating the words into an equation is the first, examined, step.

The method of separation of variables solves any equation of the form

\dfrac{dy}{dx} = f(x)g(y)

. You gather all the

y

-terms on one side and all the

x

-terms on the other,

\dfrac{1}{g(y)}\,dy = f(x)\,dx

, and integrate both sides. This yields the general solution, containing one arbitrary constant, which represents a whole family of solution curves.

The general solution becomes a particular solution once a boundary or initial condition fixes the constant. For

\dfrac{dP}{dt} = kP

, separating gives

\ln P = kt + c

, so

P = Ae^{kt}

; the condition

P = P_0

t = 0

fixes

A = P_0

, recovering the exponential growth model met earlier. This is why exponential models are exactly the solutions of proportional-rate equations.

Interpreting the solution in context and checking its behaviour completes the modelling cycle. You should be able to state what the solution predicts, identify any long-term (steady-state) value, and comment on the model's limitations — for example that unbounded exponential growth cannot continue indefinitely, motivating a refined model.

\frac{dy}{dx} = f(x)g(y) \ \Rightarrow\ \int \frac{1}{g(y)}\,dy = \int f(x)\,dx

Separation of variables

Separate the variables to opposite sides and integrate both.

\frac{dP}{dt} = kP \ \Rightarrow\ P = P_0 e^{kt}

The proportional-growth solution

The constant is fixed by the initial value $P_0$ ; growth for $k>0$ , decay for $k<0$ .

Worked example

Separating the variables

Solve $\dfrac{dy}{dx} = xy$ given that $y = 2$ when $x = 0$ .

Separate
$\dfrac{1}{y}\,dy = x\,dx$ (valid for $y \neq 0$ ).
Integrate both sides
$\ln|y| = \tfrac{1}{2}x^2 + c$ , so $y = Ae^{x^2/2}$ where $A = e^{c}$ .
$\ln|y| = \tfrac{1}{2}x^2 + c$
Apply the condition
At $x = 0$ , $y = 2$ : $2 = Ae^{0} = A$ , so $A = 2$ .

Result: $y = 2e^{x^2/2}$ .

Typical mistakes

Integrating without separating the variables first.
Leaving the answer as a general solution when a boundary condition was given (so a particular solution was required).

Active revision

A tank drains so that $\dfrac{dV}{dt} = -0.4\sqrt{V}$ . Given $V = 100$ when $t = 0$ , find $V$ as a function of $t$ and the time at which the tank is empty.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Integration

Key points

Integrating a sum of powers

Key points

Area between a line and a curve

Key points

Integration by parts

Key points

Separating the variables

Sources