Integration: A-level Maths OCR revision notes

1.08

Indefinite and standard integrals

The indefinite integral and the constant of integration 1.08

Definitions

Indefinite integral: $\int f (x) d x$ denotes the family of all functions whose derivative is $f (x)$ .
Constant of integration: the arbitrary constant $c$ appended to every indefinite integral, because any constant differentiates to zero and so cannot be recovered by reversing the process.

Key results

$\displaystyle\int f'(x)\,dx = f(x)+c$ , which is the whole content of integration as an operation.
Geometrically, the members of the family $y = f (x) + c$ are vertical translations of one another, all with the same gradient function.
A single point on the required curve fixes $c$ and so selects one member of that family.

Notes

Every indefinite integral can be checked instantly by differentiating the answer; doing so should be automatic, because it costs seconds and catches almost every error.
Omitting $+ c$ is the most commonly penalised slip in the whole topic.
For a definite integral the $+ c$ is never needed at all (it cancels between the limits), so only add it back for an indefinite integral — mixing the two conventions up is a common source of confusion.

Standard integrals 1.08

Key results

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ , for $n \neq = - 1$ — raise the power by one and divide by the new power.
$\int \frac{1}{x} d x = ln ∣ x ∣ + c$ , the missing case $n = - 1$ .
$\int e^{k x} d x = \frac{1}{k} e^{k x} + c$ .
$\int cos x d x = sin x + c$ and $\int sin x d x = - cos x + c$ , with $x$ in radians.
Integration is linear: constants come outside, and sums integrate term by term.

Notes

Each of these is the direct reverse of a derivative from Differentiation, so the surest way to recall an integral is to ask which function differentiates to the integrand.
The modulus in $ln ∣ x ∣$ matters, because $\frac{1}{x}$ is defined for negative $x$ as well; it is dropped only when the limits guarantee $x > 0$ .
As with differentiation, rewrite every term in index form first: $\frac{2}{x ^{3}}$ becomes $2 x^{- 3}$ and $3 x$ becomes $x^{\frac{1}{3}}$ before the power rule is applied.
The signs on the trigonometric integrals are the opposite way round to the derivatives — integrating sine introduces the minus sign, differentiating cosine introduces it.
The $n \neq = - 1$ restriction on the power rule for integration is exactly why $\frac{1}{x}$ needs its own separate rule — dividing by $n + 1 = 0$ would be meaningless.

Reverse chain rule 1.08

Key results

$\int (a x + b)^{n} d x = \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + c$ , for $n \neq = - 1$ .
$\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$ .
$\int cos (a x + b) d x = \frac{1}{a} sin (a x + b) + c$ and $\int sin (a x + b) d x = - \frac{1}{a} cos (a x + b) + c$ .
$\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln\left|f(x)\right|+c$ , whenever the numerator is exactly the derivative of the denominator.

Notes

The extra factor of $\frac{1}{a}$ compensates for the $a$ that the chain rule would produce on differentiating; omitting it is the standard error here.
These shortcuts work only when the inner function is linear, or when the derivative of the inner function is already present as a factor; anything else needs a full substitution.
Spotting the $\dfrac{f'(x)}{f(x)}$ pattern is easiest by checking the numerator against the derivative of the denominator directly — a numerator merely 'similar to' the denominator's derivative (off by a constant factor) still counts, just with that factor adjusted outside.

Worked example

Find:

\int (6 x^{2} - \frac{4}{x ^{2}} + x) d x

Show worked solution

Rewrite in index form:

6 x^{2} - 4 x^{- 2} + x^{\frac{1}{2}}

Integrating term by term,

\int 6 x^{2} d x = 2 x^{3}

\int - 4 x^{- 2} d x = - 4 \times \frac{x ^{- 1}}{- 1} = \frac{4}{x}

and:

\int x^{\frac{1}{2}} d x = \frac{x ^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} x^{\frac{3}{2}}

So the integral is:

2 x^{3} + \frac{4}{x} + \frac{2}{3} x^{\frac{3}{2}} + c

Check by differentiating:

6 x^{2} - 4 x^{- 2} + \frac{2}{3} \times \frac{3}{2} x^{\frac{1}{2}} = 6 x^{2} - \frac{4}{x ^{2}} + x

as required.

1.08

Definite integrals

Definite integrals 1.08

Definitions

Definite integral: $\int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} = F (b) - F (a)$ , where $F$ is any antiderivative of $f$ .

Key results

No constant of integration is needed, because $c$ appears in both $F (b)$ and $F (a)$ and cancels on subtraction.
Reversing the limits reverses the sign: $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ .
Adjacent intervals add: $\int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x = \int_{a}^{c} f (x) d x$ .

Notes

The Fundamental Theorem of Calculus is what licenses this: the accumulated area up to $x$ is a function whose derivative is $f (x)$ , so areas can be evaluated by antidifferentiation rather than by summing strips.
Keep exact values such as $ln 2$ , $e^{2}$ and $π$ in the working and round only at the very end, since rounding early can corrupt the third significant figure of the answer.
Write the substituted limits clearly as $F (b) - F (a)$ before subtracting — combining the two substitutions mentally is a common place to drop a sign or a term.

Worked example

Evaluate:

\int_{1}^{4} (3 x - \frac{2}{x}) d x

giving your answer in exact form and to three significant figures.

Show worked solution

In index form the integrand is $3 x^{\frac{1}{2}} - 2 x^{- 1}$ .

Its antiderivative is:

3 \times \frac{x ^{\frac{3}{2}}}{\frac{3}{2}} - 2 ln x = 2 x^{\frac{3}{2}} - 2 ln x

valid since $x > 0$ throughout.

Evaluating:

[2 x^{\frac{3}{2}} - 2 ln x]_{1}^{4} = (2 \times 8 - 2 ln 4) - (2 \times 1 - 0) = 16 - 2 ln 4 - 2 = 14 - 4 ln 2

using $2 ln 4 = 4 ln 2$ .

Numerically:

4 ln 2 = 2.7726

so the value is $11.2$ (3 s.f.).

1.08

Integration by substitution and by parts

Integration by substitution 1.08

Key results

Substitution reverses the chain rule, turning a composite expression into a standard integral.

Method

Choose $u$ to be the inner function of the composite expression.
Differentiate to find $\frac{d u}{d x}$ , and hence write $d x$ in terms of $d u$ .
Rewrite the integral entirely in terms of $u$ , with no $x$ remaining anywhere.
Integrate with respect to $u$ , then substitute back for an indefinite integral, or evaluate between converted limits for a definite one.

Notes

The substitution to use is usually signalled by a function and something proportional to its derivative appearing together in the same integral.
For a definite integral, either convert the limits to the new variable at the substitution step, or find the antiderivative in terms of the original variable first and use the original limits — both are valid, provided the two approaches are not mixed part-way through.
Converted limits are $u$ -values, not $x$ -values, so the bracket $[]_{u = 1}^{u = 5}$ is worth labelling explicitly to avoid substituting the wrong numbers.
Check that literally no $x$ remains after substitution before integrating — a leftover $x$ term signals either the wrong choice of $u$ or an algebra slip in expressing $d x$ .

Integration by parts 1.08

Key results

$\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x$ , the reverse of the product rule.
$\int ln x d x = x ln x - x + c$ , obtained by taking $u = ln x$ and $\frac{d v}{d x} = 1$ .

Method

Choose $u$ to be the factor that simplifies when differentiated, typically a power of $x$ or $ln x$ .
Let $\frac{d v}{d x}$ be the remaining factor, which must be one you can integrate, such as $e^{x}$ , $sin x$ or a power of $x$ .
Write down all four of $u$ , $\frac{d u}{d x}$ , $\frac{d v}{d x}$ and $v$ before substituting into the formula.
Substitute, then integrate the new integral — repeating the whole process if it is still a product.

Notes

Choosing which factor is $u$ is the whole difficulty: the wrong choice produces an integral harder than the one you started with, which is itself a clear signal to start again with the choices swapped.
An integrand with $x^{2}$ against $e^{x}$ or $sin x$ needs the method applied twice, the power dropping by one each time.
Some integrals need a preliminary rearrangement before any standard form appears; persistent simplification matters more than any single clever trick.
A useful mnemonic for choosing $u$ (LATE: Logarithm, Algebraic/power, Trig, Exponential, in that priority) picks $u$ from whichever type appears earliest in that list — it differentiates to something simpler almost every time.

Worked examples

Worked example

Use the substitution $u = x^{2} + 1$ to evaluate:

\int_{0}^{2} x (x^{2} + 1)^{3} d x

Show worked solution

With $u = x^{2} + 1$ ,

\frac{d u}{d x} = 2 x

so:

x d x = \frac{d u}{2}

The limits convert as $x = 0 \Rightarrow u = 1$ and $x = 2 \Rightarrow u = 5$ .

The integral becomes:

\int_{u = 1}^{u = 5} \frac{u ^{3}}{2} d u = \frac{1}{2} [\frac{u ^{4}}{4}]_{1}^{5} = \frac{1}{8} (625 - 1) = \frac{624}{8} = 78

Check by expanding instead:

x (x^{2} + 1)^{3} = x^{7} + 3 x^{5} + 3 x^{3} + x

whose integral:

\frac{x ^{8}}{8} + \frac{x ^{6}}{2} + \frac{3 x ^{4}}{4} + \frac{x ^{2}}{2}

evaluated at $x = 2$ gives:

32 + 32 + 12 + 2 = 78

and at $x = 0$ gives $0$ .

Worked example

Find:

\int x e^{x} d x

using integration by parts.

Show worked solution

Let $u = x$ (so:

\frac{d u}{d x} = 1

) and:

\frac{d v}{d x} = e^{x}

(so $v = e^{x}$ ).

By integration by parts:

\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + c = e^{x} (x - 1) + c

1.08

Area under and between curves

Area under a curve 1.08

Key results

The area between a curve and the $x$ -axis over $[a, b]$ is $\int_{a}^{b} y d x$ , provided the curve does not cross the axis in that interval.
A region lying entirely below the axis returns a negative value from the integral, even though its area is positive — take the modulus of the result.

Method

Find where the curve crosses the $x$ -axis by solving $y = 0$ .
Split the interval at every crossing point that lies strictly inside it.
Integrate over each piece separately.
Take the modulus of each piece before adding, so that regions below the axis contribute positively.

Notes

Integrating straight through a crossing point without splitting gives the signed total, in which area above the axis silently cancels area below it — a correct calculation of the wrong quantity.
Areas measured with respect to the $y$ -axis use $\int_{c}^{d} x d y$ instead, with the curve rearranged to give $x$ in terms of $y$ .
Sketching the curve first, even roughly, immediately shows how many times (if any) it crosses the axis in the given interval — far more reliable than assuming it doesn't.

Area between two curves 1.08

Key results

The area between $y = f (x)$ and $y = g (x)$ over an interval where $f (x) \geq g (x)$ is $\int_{a}^{b} (f (x) - g (x)) d x$ .
The limits $a$ and $b$ are the $x$ -coordinates of the points of intersection, found by solving $f (x) = g (x)$ .

Method

Solve $f (x) = g (x)$ to locate the intersections and hence the limits.
Decide which curve is the upper one on that interval, by testing a single convenient value of $x$ between the limits.
Integrate upper minus lower in one go, rather than integrating each curve separately.
Evaluate and state the answer in square units.

Notes

Subtracting upper minus lower works even where part of the region lies below the $x$ -axis, because the negative contributions cancel in the difference; this is why the single combined integral is safer than two separate ones.
A negative answer means the curves were taken in the wrong order — swap them rather than simply dropping the sign, so the working stays honest.
If the curves cross within the interval, the 'upper' curve switches partway through — split the integral at each crossing exactly as for a single curve crossing the $x$ -axis.

Worked example

Find the area of the region enclosed between the curve $y = 4 x - x^{2}$ and the line $y = x$ .

Show worked solution

Find the intersections by solving $4 x - x^{2} = x$ : $3 x - x^{2} = 0$ , so $x (3 - x) = 0$ , giving $x = 0$ or $x = 3$ .

The area is:

\int_{0}^{3} [(4 x - x^{2}) - x] d x = \int_{0}^{3} (3 x - x^{2}) d x = [\frac{3 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{3} = (\frac{27}{2} - 9) - 0 = 13.5 - 9 = 4.5

square units.

1.08

Differential equations

Separable differential equations 1.08

Definitions

Separable equation: a first-order differential equation of the form $\frac{d y}{d x} = f (x) g (y)$ , in which the right-hand side factorises into a function of $x$ alone and a function of $y$ alone.
General solution: the solution containing an arbitrary constant, representing a whole family of curves that satisfy the equation.
Particular solution: the single member of that family fixed by a boundary condition, such as a known value of $y$ at a specific $x$ .

Method

Rearrange to put every $y$ term with $d y$ on one side and every $x$ term with $d x$ on the other.
Integrate both sides independently, writing a single arbitrary constant on one side only.
Rearrange to make $y$ the subject where the question asks for it, combining constants as you go.
Substitute the boundary condition to evaluate the constant, and state the particular solution.

Notes

When integration produces $ln$ on both sides, exponentiate at once and replace $e^{c}$ by a new constant $A$ ; carrying an unexponentiated $c$ through later working is where most marks are lost.
Modelling questions usually supply the differential equation in words — 'the rate of decrease is proportional to the amount present' means $\frac{d m}{d t} = - km$ with $k > 0$ — and the constant of proportionality is then found from the data.
Always check the boundary condition is applied to the GENERAL solution, after integrating and combining constants — applying it too early, before both integrals are done, gives a wrong constant.

Worked example

A hot liquid cools in a room at $2 0^{\circ}$ C so that:

\frac{d θ}{d t} = - k (θ - 20)

where $\theta\,^\circ$ C is its temperature after $t$ minutes.

Initially $θ = 80$ , and after $10$ minutes $θ = 50$ .

Find $θ$ when $t = 25$ .

Show worked solution

Separate the variables:

\int \frac{d θ}{θ - 20} = \int - k d t

so:

ln ∣ θ - 20∣ = - k t + c

Exponentiating gives:

θ - 20 = A e^{- k t}

with $A = e^{c}$ .

At $t = 0$ , $θ = 80$ , so $A = 60$ .

At $t = 10$ , $θ = 50$ :

30 = 60 e^{- 10 k}

so:

e^{- 10 k} = \frac{1}{2}

and:

k = \frac{ln 2}{10}

Then at $t = 25$ :

θ = 20 + 60 e^{- 2.5 l n 2} = 20 + 60 \times 2^{- 2.5} = 20 + \frac{60}{4 2} = 20 + \frac{15 2}{2} = 30. 6^{\circ}

(3 s.f.).

The answer sits sensibly between the $5 0^{\circ}$ C at $10$ minutes and the limiting room temperature of $2 0^{\circ}$ C.