8 Integration — notes on one page

8.1 · Know and use the Fundamental Theorem of Calculus.

Integration is the reverse process of differentiation: if $\frac{\mathrm dy}{\mathrm dx}=f(x)$ , then $y=\int f(x)\,\mathrm dx$ recovers the family of curves with that gradient function.
Every indefinite integral needs a constant of integration: $\int f(x)\,\mathrm dx=F(x)+c$ .
One known point on the curve fixes the value of $c$ .
A definite integral is evaluated from any antiderivative: $\int_a^b f(x)\,\mathrm dx=F(b)-F(a)$ ; the constant of integration cancels in the subtraction.
A common error is to omit $+c$ from an indefinite integral, or to substitute the known point into $f(x)$ instead of the integrated function when finding $c$ .

Worked example

The function $f$ satisfies $f'(x)=3\sqrt{x}-\dfrac{2}{x^2}$ for $x>0$ , and $f(4)=10$ . Find $f(x)$ .

1.Rewrite as $f'(x)=3x^{1/2}-2x^{-2}$ and integrate: $f(x)=2x^{3/2}+2x^{-1}+c$ .
2.Substituting $x=4$ : $2(8)+\frac{2}{4}+c=10$ , so $16.5+c=10$ and $c=-\frac{13}{2}$ .
3.Hence $f(x)=2x^{3/2}+\frac{2}{x}-\frac{13}{2}$ .

Answer: $f(x)=2x^{3/2}+\dfrac{2}{x}-\dfrac{13}{2}$

Common mistakes

Don't use the supplied boundary value itself as the integration constant without substituting into the antiderivative.
Don't integrate the derivative but omit the arbitrary constant before using the given function value.

Exam tip. When recovering a function from its derivative, include the integration constant and determine it from the boundary condition.

8.2 · Integrate xⁿ (excluding n = −1) and related sums, differences and constant multiples; integrate e^(kx), 1/x, sin kx, cos kx and related sums, differences and constant multiples.

For $n\ne-1$ , $\int xⁿ\,\mathrm dx=\frac{x^{n+1}}{n+1}+C$; rewrite roots and reciprocals as powers before applying the rule. Standard forms include $\int e^{kx}\,\mathrm dx=\frac1k e^{kx}+C$ and $\int\frac1x\,\mathrm dx=\ln|x|+C$ on intervals not crossing $0$ .
Divide by the inner coefficient in trigonometric integrals.
Also use $\int\sec^2(kx)\,\mathrm dx=\frac1k\tan(kx)+C$ , $\int\tan x\,\mathrm dx=-\ln|\cos x|+C$ , and identities for powers such as $\sin^2x=\frac12(1-\cos2x)$ .
A common error is to forget the constant of integration or to multiply by $k$ rather than divide by it when integrating a function of $kx$ .
Integrate term by term: the first three terms give $2e^{3x}+4\ln|x|+\frac52\cos(2x)$ .

Worked example

Find $\int\left(6e^{3x}+\frac4x-5\sin(2x)+3\tan x\right)\,\mathrm dx$ .

1.Integrate term by term: the first three terms give $2e^{3x}+4\ln|x|+\frac52\cos(2x)$ .
2.Since $\int\tan x\,\mathrm dx=-\ln|\cos x|$ , the final term gives $-3\ln|\cos x|$ .
3.Add $C$ .

Answer: $2e^{3x}+4\ln|x|+\frac52\cos(2x)-3\ln|\cos x|+C$

Common mistakes

Don't integrate $e^{kx}$ as $ke^{kx}$ instead of $e^{kx}/k$ .
Don't treat the integral of 1/x as a power-rule case or miss reciprocal chain factors.

Exam tip. Match each term to its standard antiderivative and include the integration constant after combining the results.

8.3 · Evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

Definition. Specification 8.3: evaluate definite integrals; use a definite integral to find the area under a curve and the area between two curves.

$\int_a^b f(x)\,dx=F(b)-F(a)$ where $F$ is an antiderivative; write the antiderivative in square brackets with the limits, substitute both, and subtract in the right order, using $e^0=1$ and $\ln1=0$ where they appear.
The integral is a signed area, so a region below the $x$ -axis integrates to a negative number and its area is the modulus; split at the roots when a region crosses the axis.
For the area between a curve and a line or a second curve, find the intersections first, then integrate $y_{\text{upper}}-y_{\text{lower}}$ between them, or subtract the area of a triangle or trapezium under a straight line from the area under the curve.
Exact answers keep surds, $\pi$ , $e$ and logarithms; “show all stages of working” means calculator-only answers score nothing.
A parametric curve has area $\int y\,\dfrac{dx}{dt}\,dt$ with the limits in $t$ .

For a parametric curve, vertical strips give area

\int y(t)\,\dfrac{dx}{dt}\,dt

.

Worked example

1.

The curve

C

has equation

y=4x-x^2

and the line

l

has equation

y=x

. The finite region

R

is bounded by

C

and

l

. Use algebraic integration to find the exact area of

R

.

(5)

(Total for Question 1 is 5 marks)

Working

1.Intersections: $4x-x^2=x$ , so $x(3-x)=0$ and $x=0$ or $x=3$ ; between them the curve is above the line.
2.Area $=\displaystyle\int_0^3\left[(4x-x^2)-x\right]dx=\int_0^3(3x-x^2)\,dx$ .
3. $=\left[\dfrac{3x^2}{2}-\dfrac{x^3}{3}\right]_0^3=\left(\dfrac{27}{2}-9\right)-0=\dfrac92$ .

Answer: $\dfrac{9}{2}$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Sets $4x-x^2=x$ and solves to find the intersections	M1
	$x=0$ and $x=3$	A1
	Integrates the difference of the two functions, $x^n\to x^{n+1}$	M1
	$\dfrac{3x^2}{2}-\dfrac{x^3}{3}$ or $2x^2-\dfrac{x^3}{3}$ and $\dfrac92$	A1
	Applies the limits $0$ and $3$ correctly to obtain $\dfrac92$	A1
Notes
M1: or area under the curve minus the triangle $\frac12\times3\times3$
A1: cso

(5 marks)

Mark-scheme wording to use

“y = 8x, 12x + y = 48 ⇒ y = 19.2 (or x = 2.4) M1 Triangle area is (1/2) × 4 × "19.2" (= 38.4 or 192/5) or ∫_0^"2.4" 8x dx + ∫_"2.4"^4 "(48 − 12x)" dx dM1 … M1: Attempts to find either the x or y coordinate of the intersection of line l_1 and line l_2 … dM1: Correct method for the area of the triangle.” — Pearson 9MA0, June 2024 Paper 1, Q10(c)
“M1: Substitutes the limits 1 and 0 into an expression of the form ±αx²e^(−3x) ± βxe^(−3x) ± γe^(−3x), α, β, γ ≠ 0 and subtracts the right way round. … Must see evidence of the use of both limits and subtraction and use of e⁰ = 1. … A1: Correct answer of 16/27 − (136/27)e⁻³ but allow equivalent exact fractions and condone 16/27 − 136/(27e³).” — Pearson 9MA0, June 2024 Paper 2, Q11
“M1: Substitute both limits, subtract and set the result equal to ln(7). A1: Reach a=ln(7/2) and state k=7/2 or an exact equivalent” — Pearson 9MA0, Sample Paper 1, Q4

What the examiner wants. The simplest correct strategy, kept exact. The June 2024 Paper 1 report says most found the area of the triangle and subtracted the area under the curve, but a significant proportion used Pythagoras and trigonometry for the triangle, which led to errors or premature rounding, and the Paper 2 report notes sign errors when integration by parts was applied twice.

Common mistakes

Don't use one pair of limits after the upper and lower curves swap, causing part of an enclosed area to cancel.
Don't integrate one curve from the axis instead of subtracting lower curve from upper curve over the intersection limits.

Exam tip. For an enclosed area, solve for every intersection, identify upper minus lower, and check that the final area is positive.

8.4 · Understand and use integration as the limit of a sum.

Partition an interval into strips of width $\Delta x$ and form $\sum f(x_r)\Delta x$ ; its limit as the maximum strip width tends to zero is the definite integral.
To recognise a limit, identify the factor playing the role of $\Delta x$ and rewrite the sampled expression in terms of an endpoint such as $x_r=a+r\Delta x$ .
A sum with factor $1/n$ usually samples an interval of length $1$ ; a different interval width introduces a corresponding scale factor.
A common error is to identify the integrand but omit the width factor, which changes the value of the limiting integral.

Worked example

Evaluate $\lim_{n\to\infty}\frac1n\sum_{r=1}^n\left(1+\frac{3r}{n}\right)^2$ by expressing it as a definite integral.

1.With $x=r/n$ , the limit is $\int_0^1(1+3x)^2\,\mathrm dx$ .
2.Equivalently, with $u=1+3x$ , it is $\frac13\int_1^4u^2\,\mathrm du=\frac19(4^3-1^3)=7$ .

Answer: $7$

Common mistakes

Don't read $i/n$ as the strip width rather than as the sample-point position within the interval.
Don't match a Riemann sum to the wrong interval by overlooking the index increment and endpoint expression.

Exam tip. Rewrite the summand as a function of the sample point and identify both the strip width and integration limits.

8.5 · Carry out simple cases of integration by substitution and integration by parts; understand these methods as the inverse processes of the chain and product rules respectively.

Substitution reverses the chain rule: choose $u=g(x)$ so the remaining factor supplies $\mathrm du=g'(x)\,\mathrm dx$ . In particular, recognise $\int\frac{f'(x)}{f(x)}\,\mathrm dx=\ln|f(x)|+C$ .
Integration by parts reverses the product rule: $\int u\,\mathrm dv=uv-\int v\,\mathrm du$ ; choose $u$ so that differentiating it simplifies the integral.
For a definite integral, either change the limits to the new variable or return fully to the original variable before applying the old limits.
A common error is to omit the minus sign in integration by parts or to mix $x$ and $u$ in the same transformed integral.
A required standard result is $\int\ln x\,dx=x\ln x-x+c$ , obtained by integration by parts with $u=\ln x$ and $dv=dx$ .

Worked example

Find $\int xe^{2x}\,\mathrm dx$ .

1.Use integration by parts with $u=x$ and $\mathrm dv=e^{2x}\,\mathrm dx$ .
2.Then $\mathrm du=\mathrm dx$ and $v=\frac12e^{2x}$ .
3.Hence $\int xe^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac12\int e^{2x}\,\mathrm dx=\frac x2e^{2x}-\frac14e^{2x}+C$ .

Answer: $e^{2x}\left(\frac x2-\frac14\right)+C$

Common mistakes

Don't change variable in a definite integral but keep the original $x$ -limits.
Don't choose integration by parts but reverse the selected u and dv, creating a harder integral.

Exam tip. For a product of an algebraic and exponential term, choose the algebraic factor as u and retain the boundary-free constant.

8.6 · Integrate using partial fractions that are linear in the denominator.

Factor the denominator fully into linear factors, then write one partial-fraction term for each factor before solving for its constants. For $\frac{A}{x-a}$ , the integral is $A\ln|x-a|+C$ ; retain absolute values unless the domain makes the sign known.
Determine constants by equating coefficients or substituting convenient values that make all but one term vanish.
A common error is to integrate the unfactorised denominator as though $\int f'(x)/f(x)\,\mathrm dx$ applied when the numerator is not its derivative.
Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$.
Then $5x+1=A(x+2)+B(x-1)$ .

Worked example

Express $\frac{5x+1}{(x-1)(x+2)}$ in partial fractions and hence integrate it.

1.Set $\frac{5x+1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2}$ .
2.Then $5x+1=A(x+2)+B(x-1)$ .
3.Substituting $x=1$ gives $A=2$ , and $x=-2$ gives $B=3$ .
4.Integrating the decomposition gives $2\ln|x-1|+3\ln|x+2|+C$ .

Answer: $\frac{5x+1}{(x-1)(x+2)}=\frac2{x-1}+\frac3{x+2}$ ; $2\ln|x-1|+3\ln|x+2|+C$

Common mistakes

Don't drop the absolute-value signs after integrating a linear denominator to a logarithm.
Don't split the fraction without solving the coefficient identity, so the logarithmic coefficients are incorrect.

Exam tip. Find partial-fraction constants before integrating and preserve absolute-value signs in logarithmic antiderivatives.

8.7 · Evaluate the analytical solution of simple first order differential equations with separable variables, including finding particular solutions.

For $\frac{\mathrm dy}{\mathrm dx}=g(x)h(y)$, rearrange to place all $y$ -terms with $\mathrm dy$ and all $x$ -terms with $\mathrm dx$ , then integrate both sides.
Include a constant after integration and use the initial condition to determine it.
A family of solutions can be sketched from its equilibrium curves, initial values, gradients and long-term behaviour.
When logarithms arise, exponentiate carefully and use the stated domain or initial condition to choose any required sign or branch.
A common error is to divide by a factor involving $y$ without checking whether doing so loses a constant equilibrium solution.

Worked example

Solve $\frac{\mathrm dy}{\mathrm dx}=\frac{x+1}{y}$ , given that $y=2$ when $x=0$ and $y>0$ .

1.Separate: $y\,\mathrm dy=(x+1)\,\mathrm dx$ .
2.Integration gives $\frac12y^2=\frac12x^2+x+C$ , so $y^2=x^2+2x+C_1$ .
3.The condition $y(0)=2$ gives $C_1=4$ .
4.Since $y>0$ , $y=\sqrt{x^2+2x+4}$ .

Answer: $y=\sqrt{x^2+2x+4}$

Common mistakes

Don't rearrange $\mathrm dy/\mathrm dx=f(x)/g(y)$ as $f(x)\,\mathrm dy=g(y)\,\mathrm dx$ .
Don't separate variables but lose the integration constant or select a branch inconsistent with the initial condition.

Exam tip. After separation and integration, use the initial condition before choosing the branch required by the stated domain.

8.8 · Interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution; includes links to kinematics.

Interpret constants and initial values with their units, and examine limiting behaviour to identify equilibrium values or long-term predictions.
In kinematics, $v=\frac{\mathrm ds}{\mathrm dt}$ and $a=\frac{\mathrm dv}{\mathrm dt}$ ; the sign of velocity gives direction, while a change of sign marks a reversal of motion.
Check whether the mathematical solution remains meaningful on the time interval and within the physical range assumed by the model.
A common error is to state only that a model is 'unrealistic'; name a specific assumption, such as constant environmental conditions or neglect of resistance, and explain its effect.

Worked example

The velocity of a particle is modelled by $v=20(1-e^{-0.5t})\,\text{m s}^{-1}$ for $t\geq0$ . Find the limiting velocity and the time when $v=15\,\text{m s}^{-1}$ . State one limitation of the model.

1.As $t\to\infty$ , $e^{-0.5t}\to0$ , so the limiting velocity is $20\,\text{m s}^{-1}$ .
2.For $v=15$ , $15=20(1-e^{-0.5t})$ , so $e^{-0.5t}=1/4$ and $t=2\ln4\approx2.77\,\text{s}$ .
3.A real resistance law or driving force may change with conditions, so the same constants need not remain valid indefinitely.

Answer: $20\,\text{m s}^{-1}$ ; $t=2\ln4\approx2.77\,\text{s}$ ; One valid limitation, such as a constant resistance law being assumed

Common mistakes

Don't extend a population solution beyond the time at which the model predicts an impossible negative value.
Don't report a limiting value algebraically and fail to interpret its units or challenge the long-term model.

Exam tip. For a contextual differential model, give numerical conclusions with units and state a limitation tied to its assumptions.