Mini-Lesson

Integration

4 missions. Guess, learn, then check. Your place is saved.

Mission 1A speedometer shows speed, not distance. How far did the car go?
Mission 2Why does $\int \frac{1}{x} d x$ ∫ 1/x dx give a logarithm, not a power?
Mission 3The product rule runs backwards. When does that help you integrate?
Mission 4Rumours spread in proportion to those yet to hear. How do they grow?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

A speedometer shows speed, not distance. How far did the car go?

Make a guess: A speedometer shows speed, not distance. How far did the car go?

Integration · powers and areas

Integration reverses differentiation

If $\frac{d}{d x} (F (x)) = f (x)$ d/dx(F(x)) = f(x), then $\int f (x) d x = F (x) + c$ ∫ f(x) dx = F(x) + c. This is the Fundamental Theorem of Calculus.
Power rule: add 1 to the power, then divide by the new power:

\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c, n \neq = - 1

$c$ c is the constant of integration. Every indefinite integral needs it: $x^{4} + 7$ x⁴ + 7 and $x^{4} - 2$ x⁴ − 2 both differentiate to $4 x^{3}$ 4x³.
Integrate a sum term by term; constant multiples stay in front.

Worked example Integration · powers and areas

Rewrite as powers, then integrate

Question

Find

\int (6 x^{2} + 3 x - \frac{4}{x ^{3}}) d x

.

1
Write every term as a power of $x$ x: $6 x^{2} + 3 x^{\frac{1}{2}} - 4 x^{- 3}$ 6x² + 3x^(1/2) − 4x⁻³
2
Integrate $6 x^{2}$ 6x²: $\frac{6 x ^{3}}{3} = 2 x^{3}$ 6x³/3 = 2x³
3
Integrate $3 x^{\frac{1}{2}}$ 3x^(1/2): $\frac{3 x ^{\frac{3}{2}}}{\frac{3}{2}} = 2 x^{\frac{3}{2}}$ (3x^(3/2))/(3/2) = 2x^(3/2)
4
Integrate $- 4 x^{- 3}$ −4x⁻³: the power goes up to $- 2$ −2, so $\frac{- 4 x ^{- 2}}{- 2} = 2 x^{- 2}$ (−4x⁻²)/(−2) = 2x⁻²

Answer

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

⚠️ Watch out

with negative powers, adding 1 moves towards zero (

- 3 \to - 2

), not away from it. Check by differentiating your answer.

Quick check

?Find

\int \frac{3}{x} d x

.

Integration · powers and areas

Finding the curve from its gradient

Integrating $\frac{d y}{d x}$ dy/dx gives a family of curves, one for each value of $c$ c, all vertical translations of each other.
One known point picks out the curve: substitute its $x$ x and $y$ y and solve for $c$ c.
Then write the full equation $y = \dots$ y = … with the number in place of $c$ c.

Calculate

?A curve has gradient

\frac{d y}{d x} = 4 x^{3} - 6 x + 1

and passes through the point

(2, 9)

. Find the value of

y

on the curve when

x = 3

. Give your answer as an exact integer.

Integration · powers and areas

Definite integrals and area

A definite integral has limits and gives a number (no $c$ c):

\int_{a b} f (x) d x = [F (x)]_{a b} = F (b) - F (a)

Where the curve is above the $x$ x-axis, this is the area between the curve, the axis and the lines $x = a$ x = a, $x = b$ x = b.
Below the axis the integral is negative. Split at the roots and add the sizes of the parts.
On a speed–time graph the area under the curve is the distance travelled: a rectangle at 20 m/s for 30 s gives 600 m.

Worked example Integration · powers and areas

A region partly below the axis

Question

Find the total area between

y = x^{2} - 2 x

, the

x

-axis and the lines

x = 0

and

x = 3

.

1
Find where the curve crosses the axis: $x (x - 2) = 0$ x(x − 2) = 0, so $x = 0$ x = 0 or $x = 2$ x = 2. Below the axis for $0 < x < 2$ 0 < x < 2.
2
Integrate: $F (x) = \frac{x ^{3}}{3} - x^{2}$ F(x) = x³/3 − x²
3
Part below the axis: $F (2) - F (0) = \frac{8}{3} - 4 = - \frac{4}{3}$ F(2) − F(0) = 8/3 − 4 = −4/3, an area of $\frac{4}{3}$ 4/3
4
Part above the axis: $F (3) - F (2) = 0 - (- \frac{4}{3}) = \frac{4}{3}$ F(3) − F(2) = 0 − (−4/3) = 4/3
5
Add the sizes: $\frac{4}{3} + \frac{4}{3}$ 4/3 + 4/3

Answer

total area

= \frac{8}{3}

square units

⚠️ Watch out

\int_{03} (x^{2} - 2 x) d x = 0

, because the two parts cancel. Always sketch first and split at the roots.

Calculate

?Evaluate

\int_{14} \frac{x ^{2} + 2}{x} d x

. Give your answer as an exact decimal.

Calculate

?CCEA also asks for areas against the $y$ y-axis:

area = \int_{c d} x d y

, with

x

written in terms of

y

. Find the area of the region bounded by

y = x^{2}

(

x \geq 0

), the

y

-axis and the lines

y = 1

and

y = 4

. Give your answer to 3 significant figures.

Mission check

?A car's speed is

v = 3 t - 0.1 t^{2}

m/s from

t = 0

to

t = 30

s, when it stops. How far does it travel?

Mission 2

Why does $\int \frac{1}{x} d x$ ∫ 1/x dx give a logarithm, not a power?

Make a guess: Why does $\int \frac{1}{x} d x$ ∫ 1/x dx give a logarithm, not a power?

Integration · standard integrals and areas

The one power the power rule misses (A2 only)

The power rule gives $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + c. For $n = - 1$ n = −1 it would need $\frac{x ^{0}}{0}$ x⁰/0, which is undefined.
But you already know a function with derivative $\frac{1}{x}$ 1/x: $\frac{d}{d x} (ln x) = \frac{1}{x}$ d/dx(ln x) = 1/x.
Reverse it:

\int \frac{1}{x} d x = ln ∣ x ∣ + c

The modulus is there because $\frac{1}{x}$ 1/x also exists for negative $x$ x, where $ln (- x)$ ln (−x) has derivative $\frac{1}{x}$ 1/x too.

Integration · standard integrals and areas

Standard integrals

$f (x)$ f(x)	$\int f (x) d x$ ∫ f(x) dx
$e^{k x}$ e^kx	$\frac{1}{k} e^{k x} + c$ 1/ke^kx + c
$sin k x$ sin kx	$- \frac{1}{k} cos k x + c$ −1/kcos kx + c
$cos k x$ cos kx	$\frac{1}{k} sin k x + c$ 1/ksin kx + c
$sec^{2} k x$ sec² kx	$\frac{1}{k} tan k x + c$ 1/ktan kx + c

Reverse chain rule: if the top is the derivative of the bottom, $\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + c$ ∫ (f'(x))/(f(x)) dx = ln |f(x)| + c. For example $tan x = \frac{s i n x}{c o s x}$ tan x = (sin x)/(cos x) integrates to $- ln ∣ cos x ∣ + c$ −ln |cos x| + c.
For $cos^{2} x$ cos²x or $sin^{2} x$ sin²x, first use $cos 2 x = 2 cos^{2} x - 1 = 1 - 2 sin^{2} x$ cos 2x = 2cos²x − 1 = 1 − 2sin²x. Angles are in radians.

Worked example Integration · standard integrals and areas

Three standard integrals at once

Question

Find

\int (e^{2 x} + \frac{4}{x} - 6 sin 3 x) d x

.

1
Exponential, $k = 2$ k = 2: $\int e^{2 x} d x = \frac{1}{2} e^{2 x}$ ∫ e^2x dx = 1/2e^2x
2
Reciprocal: $\int \frac{4}{x} d x = 4 ln ∣ x ∣$ ∫ 4/x dx = 4ln |x|
3
Sine, $k = 3$ k = 3: $\int - 6 sin 3 x d x = - 6 \times (- \frac{1}{3} cos 3 x) = 2 cos 3 x$ ∫ −6sin 3x dx = −6 × (−1/3cos 3x) = 2cos 3x
4
Add the parts and one constant

Answer

\frac{1}{2} e^{2 x} + 4 ln ∣ x ∣ + 2 cos 3 x + c

⚠️ Watch out

sine integrates to minus cosine. Writing

- 2 cos 3 x

gives a function whose derivative is

+ 6 sin 3 x

. Differentiate your answer to check.

Quick check

?Find

\int \frac{6 x}{x ^{2} + 5} d x

.

Calculate

?Use the identity

cos 2 x = 2 cos^{2} x - 1

to evaluate

\int_{0 \frac{π}{2}} cos^{2} x d x

. Give your answer to 3 significant figures.

Integration · standard integrals and areas

Area as the limit of a sum, and areas between curves

Split the area into thin strips of width $δ x$ δ x. Their total is close to the area, and the error vanishes as the strips get thinner:

\int_{a b} f (x) d x = δ x \to 0 lim x = a \sum b f (x) δ x

Between two curves, each strip has height top − bottom. So the area is $\int_{a b} (top - bottom) d x$ ∫_a^b (top − bottom) dx.
The limits $a$ a, $b$ b are usually where the curves meet: solve their equations simultaneously.
Top − bottom works even where both curves are below the $x$ x-axis.

Worked example Integration · standard integrals and areas

A curve and a line

Question

Find the area of the finite region bounded by

y = 6 x - x^{2}

and

y = 2 x

.

1
Find where they meet: $6 x - x^{2} = 2 x$ 6x − x² = 2x, so $x (4 - x) = 0$ x(4 − x) = 0, giving $x = 0$ x = 0 and $x = 4$ x = 4
2
Decide which is on top: at $x = 1$ x = 1 the curve gives 5 and the line gives 2, so the curve is on top
3
Top − bottom: $(6 x - x^{2}) - 2 x = 4 x - x^{2}$ (6x − x²) − 2x = 4x − x²
4
Integrate: $[2 x^{2} - \frac{x ^{3}}{3}]_{04} = 32 - \frac{64}{3}$ [2x² − x³/3]₀⁴ = 32 − 64/3

Answer

area

= \frac{32}{3}

square units

⚠️ Watch out

subtract in the right order. Bottom − top gives

- \frac{32}{3}

; an area is never negative.

Calculate

?The curves

y = x^{2} - 2 x

and

y = 4 - x^{2}

meet at two points. Find the area of the finite region enclosed between them. Give your answer as an exact integer.

Worked example Integration · standard integrals and areas

Volume of revolution

Question

The region under

y = x^{2} + 1

between

x = 0

and

x = 1

is rotated through

2 π

about the

x

-axis. Find the exact volume of the solid formed.

1
Each strip turns into a thin disc of volume $π y^{2} δ x$ π y² δ x, so $V = π \int_{a b} y^{2} d x$ V = π∫_a^b y² dx
2
Square $y$ y first: $(x^{2} + 1)^{2} = x^{4} + 2 x^{2} + 1$ (x² + 1)² = x⁴ + 2x² + 1
3
Integrate: $π [\frac{x ^{5}}{5} + \frac{2 x ^{3}}{3} + x]_{01} = π (\frac{1}{5} + \frac{2}{3} + 1)$ π[x⁵/5 + 2x³/3 + x]₀¹ = π(1/5 + 2/3 + 1)

Answer

V = \frac{28 π}{15} \approx 5.86

cubic units

⚠️ Watch out

(x^{2} + 1)^{2}

is not

x^{4} + 1

. Dropping the middle term gives

\frac{6 π}{5}

.

Mission check

?Why does

\int \frac{1}{x} d x

give a logarithm when other powers of

x

give powers?

Mission 3

The product rule runs backwards. When does that help you integrate?

Make a guess: The product rule runs backwards. When does that help you integrate?

Integration · substitution and parts

Substitution reverses the chain rule (A2 only)

Choose a new variable $u$ u, usually the "inside" of a bracket or root, e.g. $u = 2 x - 1$ u = 2x − 1.
Differentiate it: $\frac{d u}{d x} = 2$ du/dx = 2, so replace $d x$ dx by $\frac{1}{2} d u$ 1/2 du.
Rewrite every $x$ x in terms of $u$ u, then integrate with respect to $u$ u.
For a definite integral, change the limits to $u$ u-values; then you never need to go back to $x$ x.

Worked example Integration · substitution and parts

Substitution with limits

Question

Use the substitution

u = 2 x - 1

to find the exact value of

\int_{15} \frac{x}{2 x - 1} d x

.

1
Replace $x$ x and $d x$ dx: $x = \frac{u + 1}{2}$ x = (u + 1)/2 and $d x = \frac{1}{2} d u$ dx = 1/2 du
2
Change the limits: $x = 1 \Rightarrow u = 1$ x = 1 ⇒ u = 1 and $x = 5 \Rightarrow u = 9$ x = 5 ⇒ u = 9
3
Rewrite and simplify: $\int_{19} \frac{u + 1}{2 u} \times \frac{1}{2} d u = \frac{1}{4} \int_{19} (u^{\frac{1}{2}} + u^{- \frac{1}{2}}) d u$ ∫₁⁹ (u + 1)/(2√u) × 1/2 du = 1/4∫₁⁹ (u^(1/2) + u^(−1/2))du
4
Integrate: $\frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}}]_{19} = \frac{1}{4} (24 - \frac{8}{3})$ 1/4[2/3u^(3/2) + 2u^(1/2)]₁⁹ = 1/4(24 − 8/3)

Answer

\frac{16}{3}

⚠️ Watch out

keeping the old limits 1 and 5 with the new variable

u

gives a wrong value. Change the limits as soon as you change the variable.

Calculate

?Use the substitution

u = 1 + x^{2}

to evaluate

\int_{01} x^{3} 1 + x^{2} d x

. Give your answer to 3 significant figures.

Integration · substitution and parts

Integration by parts reverses the product rule

Integrate the product rule and rearrange:

\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x

It helps when the new integral $\int v \frac{d u}{d x} d x$ ∫ vdu/dx dx is easier than the old one.
So choose $u$ u to be the factor that gets simpler when differentiated (a power of $x$ x), and $\frac{d v}{d x}$ dv/dx one you can integrate ( $e^{k x}$ e^kx, $sin k x$ sin kx, $cos k x$ cos kx).
Exception: with $ln x$ ln x, make $u = ln x$ u = ln x, since you cannot easily integrate it. Write $ln x$ ln x as $1 \times ln x$ 1 × ln x if it is alone.
With $x^{2}$ x², apply parts twice.

Worked example Integration · substitution and parts

Parts with a logarithm

Question

Find the exact value of

\int_{1 e} x ln x d x

.

1
Choose: $u = ln x$ u = ln x, $\frac{d v}{d x} = x$ dv/dx = x, so $\frac{d u}{d x} = \frac{1}{x}$ du/dx = 1/x and $v = \frac{x ^{2}}{2}$ v = x²/2
2
Apply the formula: $\frac{x ^{2}}{2} ln x - \int \frac{x ^{2}}{2} \cdot \frac{1}{x} d x = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4}$ x²/2ln x − ∫ x²/2 · 1/x dx = x²/2ln x − x²/4
3
Upper limit, using $ln e = 1$ ln e = 1: $\frac{e ^{2}}{2} - \frac{e ^{2}}{4} = \frac{e ^{2}}{4}$ e²/2 − e²/4 = e²/4
4
Lower limit, using $ln 1 = 0$ ln 1 = 0: $0 - \frac{1}{4} = - \frac{1}{4}$ 0 − 1/4 = −1/4

Answer

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

⚠️ Watch out

choosing

u = x

here needs

\int ln x d x

straight away, which is harder. When

ln x

appears, it is almost always

u

.

Calculate

?Evaluate

\int_{0 π} x^{2} sin x d x

. Give your answer to 3 significant figures.

Integration · substitution and parts

Choosing a method; partial fractions

You see	Try
a function and (a multiple of) its inner derivative, e.g. $x (x^{2} + 4)^{5}$ x(x² + 4)⁵	substitution (or by inspection)
a product where one factor simplifies when differentiated, e.g. $x e^{3 x}$ xe^3x, or $ln x$ ln x	by parts
a fraction with a factorised polynomial denominator	partial fractions

Partial fractions give terms you can integrate at once: $\int \frac{A}{a x + b} d x = \frac{A}{a} ln ∣ a x + b ∣ + c$ ∫ A/(ax + b) dx = A/aln |ax + b| + c.
A repeated factor gives $\int \frac{B}{( x + 1 ) ^{2}} d x = - \frac{B}{x + 1} + c$ ∫ B/((x + 1)²) dx = −B/(x + 1) + c, not a logarithm.

Sort it

?Choose the best first method for each integral.

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?Express

\frac{3 x + 1}{( x - 1 ) ( x + 1 )}

in partial fractions and hence evaluate

\int_{23} \frac{3 x + 1}{( x - 1 ) ( x + 1 )} d x

. Give your answer to 3 significant figures.

Mission check

?The product rule runs backwards as integration by parts. When does that help?

Mission 4

Rumours spread in proportion to those yet to hear. How do they grow?

Make a guess: Rumours spread in proportion to those yet to hear. How do they grow?

Integration · differential equations

Turning words into a differential equation (A2 only)

"Rate of change of $N$ N" means $\frac{d N}{d t}$ dN/dt.
"Proportional to $X$ X" means $= k X$ = kX, with $k$ k a positive constant. "Inversely proportional to $X$ X" means $= \frac{k}{X}$ = k/X.
If the quantity is decreasing, put in a minus sign so that $k$ k stays positive.
Example: a liquid cools at a rate proportional to how far it is above room temperature $20^{\circ} C$ 20 °C: $\frac{d θ}{d t} = - k (θ - 20)$ dθ/dt = −k(θ − 20).

Quick check

?Water leaks from a tank. The volume

V

decreases at a rate proportional to the square root of

V

. Which differential equation models this, with

k > 0

?

Integration · differential equations

Separating the variables

If $\frac{d y}{d x} = f (x) g (y)$ dy/dx = f(x) g(y), move all the $y$ y terms to one side and all the $x$ x terms to the other:

\int \frac{1}{g ( y )} d y = \int f (x) d x

You may need to factorise first: $2 x y + 2 x = 2 x (y + 1)$ 2xy + 2x = 2x(y + 1).
Integrate both sides, adding one constant $c$ c. This is the general solution: a family of curves.
A given condition (e.g. $y = 0$ y = 0 when $x = 0$ x = 0) fixes $c$ c and gives the particular solution.

Worked example Integration · differential equations

Factorise, separate, solve

Question

Solve

\frac{d y}{d x} = 2 x y + 2 x

, given that

y = 0

when

x = 0

. Give

y

in terms of

x

.

1
Factorise the right-hand side: $\frac{d y}{d x} = 2 x (y + 1)$ dy/dx = 2x(y + 1)
2
Separate: $\int \frac{1}{y + 1} d y = \int 2 x d x$ ∫ 1/(y + 1) dy = ∫ 2x dx
3
Integrate: $ln ∣ y + 1∣ = x^{2} + c$ ln |y + 1| = x² + c
4
Use $y = 0$ y = 0 when $x = 0$ x = 0: $ln 1 = 0 + c$ ln 1 = 0 + c, so $c = 0$ c = 0
5
Undo the logarithm: $y + 1 = e^{x^{2}}$ y + 1 = e^x²

Answer

y = e^{x^{2}} - 1

⚠️ Watch out

e^{x^{2} + c}

is

A e^{x^{2}}

with

A = e^{c}

, not

e^{x^{2}} + c

. Find

c

before you exponentiate, or work with

A

.

Calculate

?Solve

\frac{d y}{d x} = \frac{x ^{2}}{y}

, given that

y = 2

when

x = 0

, and

y > 0

. Find the value of

y

when

x = 3

, to 3 significant figures.

Integration · differential equations

Reading a solution in context

Look at what happens as $t$ t grows. $e^{- k t} \to 0$ e^(−kt) → 0, so $θ = 20 + A e^{- k t}$ θ = 20 + Ae^(−kt) settles at 20: the room temperature.
The general solution is a family of curves, one for each value of the constant. Sketch a few to see the pattern.
Limitations: check whether the model still makes sense for large $t$ t or small values. Exponential growth $P = P_{0} e^{k t}$ P = P₀e^kt grows without limit, which no real population can.
In kinematics, a solution such as $v = 20 - 20 e^{- t}$ v = 20 − 20e^(−t) describes a speed rising towards 20 m/s.

Worked example Integration · differential equations

A cooling model

Question

A metal block is taken from a furnace at

200^{\circ} C

into a room at

20^{\circ} C

. Its temperature

θ

after

t

minutes satisfies

\frac{d θ}{d t} = - k (θ - 20)

. After 5 minutes it is at

120^{\circ} C

. Find

θ

in terms of

t

and find

k

.

1
Separate and integrate: $\int \frac{1}{θ - 20} d θ = \int - k d t$ ∫ 1/(θ − 20) dθ = ∫ −k dt, so $ln (θ - 20) = - k t + c$ ln (θ − 20) = −kt + c
2
Exponentiate: $θ - 20 = A e^{- k t}$ θ − 20 = Ae^(−kt)
3
Use $θ = 200$ θ = 200 at $t = 0$ t = 0: $A = 180$ A = 180, so $θ = 20 + 180 e^{- k t}$ θ = 20 + 180e^(−kt)
4
Use $θ = 120$ θ = 120 at $t = 5$ t = 5: $100 = 180 e^{- 5 k}$ 100 = 180e^(−5k), so $e^{- 5 k} = \frac{5}{9}$ e^(−5k) = 5/9
5
Take logs: $k = \frac{1}{5} ln \frac{9}{5}$ k = 1/5ln 9/5

Answer

θ = 20 + 180 e^{- k t}

with

k = \frac{1}{5} ln 1.8 \approx 0.118

⚠️ Watch out

A

is the starting difference from room temperature (180), not the starting temperature (200).

Calculate

?For the block in screen 7,

θ = 20 + 180 e^{- k t}

with

k = \frac{1}{5} ln 1.8

. How long after leaving the furnace does the block reach

50^{\circ} C

? Give your answer to 3 significant figures.

minutes

Quick check

?A biologist models a bacteria colony with

\frac{d P}{d t} = 0.2 P

,

P = 500

when

t = 0

(

t

in hours). Which is the strongest criticism of the model?

Mission check

?A rumour spreads at a rate proportional to the number

P - N

who haven't heard it. How does

N

grow?

Recap

Big ideas

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c$ ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + c ( $n \neq = - 1$ n ≠ −1): rewrite roots and fractions as powers first, and never forget $+ c$ + c.
A point on the curve fixes $c$ c; a definite integral $F (b) - F (a)$ F(b) − F(a) needs no $c$ c.
Area under a curve = definite integral, but parts below the $x$ x-axis come out negative: split at the roots.
Against the $y$ y-axis, use $\int x d y$ ∫ x dy with $x$ x in terms of $y$ y.
$\int \frac{1}{x} d x = ln ∣ x ∣ + c$ ∫ 1/x dx = ln |x| + c; also $\int e^{k x} d x = \frac{1}{k} e^{k x} + c$ ∫ e^kx dx = 1/ke^kx + c, $\int sin k x d x = - \frac{1}{k} cos k x + c$ ∫ sin kx dx = −1/kcos kx + c, $\int cos k x d x = \frac{1}{k} sin k x + c$ ∫ cos kx dx = 1/ksin kx + c.
$\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + c$ ∫ (f'(x))/(f(x)) dx = ln |f(x)| + c; use double-angle identities for $sin^{2} x$ sin²x and $cos^{2} x$ cos²x.
An integral is the limit of a sum of thin strips; between two curves, integrate top − bottom between their meeting points.
Volume of revolution about the $x$ x-axis: $V = π \int_{a b} y^{2} d x$ V = π∫_a^b y² dx.
Substitution reverses the chain rule: replace $x$ x, $d x$ dx and the limits.
By parts reverses the product rule: $\int u \frac{d v}{d x} d x = uv - \int v \frac{d u}{d x} d x$ ∫ udv/dx dx = uv − ∫ vdu/dx dx, with $u$ u the factor that simplifies (or $u = ln x$ u = ln x).
Rational functions with linear factors: split into partial fractions, then each term gives a logarithm.
Translate words into a differential equation: rate → derivative, proportional → $k \times$ k×, decreasing → minus sign.
Separate the variables, integrate both sides with one constant (general solution), then use a condition to fix it (particular solution).
Read the solution in context: what happens for large $t$ t, and where the model stops being realistic.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Integration for CCEA GCE Mathematics (2210). 🎉

Mission 1A speedometer shows speed, not distance. How far did the car go?
Mission 2Why does $\int \frac{1}{x} d x$ ∫ 1/x dx give a logarithm, not a power?
Mission 3The product rule runs backwards. When does that help you integrate?
Mission 4Rumours spread in proportion to those yet to hear. How do they grow?

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Specification points and sources for this lesson are listed at the bottom of the page.

Integration

A speedometer shows speed, not distance. How far did the car go?

Integration reverses differentiation

Rewrite as powers, then integrate

Finding the curve from its gradient

Definite integrals and area

A region partly below the axis

Why does ∫x1​dx∫ 1/x dx give a logarithm, not a power?

The one power the power rule misses (A2 only)

Standard integrals

Three standard integrals at once

Area as the limit of a sum, and areas between curves

A curve and a line

Volume of revolution

The product rule runs backwards. When does that help you integrate?

Substitution reverses the chain rule (A2 only)

Substitution with limits

Integration by parts reverses the product rule

Parts with a logarithm

Choosing a method; partial fractions

Rumours spread in proportion to those yet to hear. How do they grow?

Turning words into a differential equation (A2 only)

Separating the variables

Factorise, separate, solve

Reading a solution in context

A cooling model

Big ideas

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Why does $\int \frac{1}{x} d x$ ∫ 1/x dx give a logarithm, not a power?