Integration: Undoing Differentiation

Integrate sums of powers of x (and 1/x for 0606) as the reverse of differentiation, add + c, and use a point to find the curve's equation.

In the exam for Edexcel Further Pure Maths 4PM1 · 5 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Rewind differentiation

Differentiation takes a curve and gives you its gradient. Integration goes the other way: you know the gradient, and you find the curve. Think of it as the rewind button.

Example

You know that $\frac{d}{d x} (x^{3}) = 3 x^{2}$ . So if you rewind $3 x^{2}$ , you get back to $x^{3}$ .

But $x^{3} + 5$ and $x^{3} - 2$ also differentiate to $3 x^{2}$ . A number on its own disappears when you differentiate, so when you rewind you cannot know what it was. You write $+ c$ to stand for it.

Formula

\int 3 x^{2} d x = x^{3} + c

$\int$ — the integral sign. It means “integrate this”.
$d x$ — tells you the letter you are working with is x.
$c$ — the constant of integration: an unknown number.

Worked example

1
Check the answer by differentiating it.
2
$\frac{d}{d x} (x^{3} + c) = 3 x^{2} + 0$ — The c is a plain number, so its gradient is 0.
3
$= 3 x^{2}$ — We are back to the expression we integrated, so the answer is right.

Rule

An integral with no limits is called an indefinite integral. Always end it with + c. Mark schemes expect it, and leaving it out can cost the final mark.

Key idea

The integral sign is a long, stretched letter S. Gottfried Leibniz chose it in 1675 for “summa”, the Latin word for sum.

The power rule backwards

To differentiate, you multiply by the power, then take 1 off the power. To integrate, you undo both, in reverse order: add 1 to the power, then divide by the new power.

Formula

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

$n$ — the power of x. Any number except −1, because then you would divide by 0.
$c$ — the constant of integration.

A number in front just waits, then multiplies the answer:

Formula

\int a x^{n} d x = \frac{a x ^{n + 1}}{n + 1} + c

$a$ — the number in front of the power of x.

Rule

Learn this rule. 0606 and 4037 give no calculus formulas. 4PM1 lists it as a formula to learn.

Worked example

1
Find $\int (6 x^{2} - 4 x + 5) d x$ .
2
$6 x^{2} \to \frac{6 x ^{3}}{3} = 2 x^{3}$ — The power 2 becomes 3. Divide by the new power, 3.
3
$- 4 x \to \frac{- 4 x ^{2}}{2} = - 2 x^{2}$ — x is x to the power 1. The power becomes 2, so divide by 2.
4
$5 \to 5 x$ — A number on its own is 5 times x to the power 0. The power becomes 1, so divide by 1.
5
$\int (6 x^{2} - 4 x + 5) d x$ $= 2 x^{3} - 2 x^{2} + 5 x + c$ — Put the terms together and add + c.
6
$\frac{d}{d x} (2 x^{3} - 2 x^{2} + 5 x + c)$ $= 6 x^{2} - 4 x + 5$ — Check: differentiating takes you back to the start.

Common mistake

Multiplying by the power, out of differentiation habit. $\int x^{3} d x$ is $\frac{x ^{4}}{4} + c$ , not $3 x^{4}$ .

Common mistake

Leaving a plain number alone. When you integrate, a number gains an x: 5 becomes 5x.

Key idea

Up one, divide by the new power, + c. Then differentiate your answer to check — it takes ten seconds.

Roots and powers on the bottom

The power rule only works on terms that look like a number times x to a power. Roots, and x on the bottom of a fraction, must be rewritten first.

Formula

m x^{k} = x^{\frac{k}{m}}

$m$ — the root: 2 for a square root, 3 for a cube root.
$k$ — the power under the root. So $x = x^{\frac{1}{2}}$ .

Formula

\frac{1}{x ^{n}} = x^{- n}

$n$ — the power on the bottom. Moving it to the top makes it negative.

Worked example

1
Find $\int (x + \frac{3}{x ^{2}}) d x$ .
2
$\int (x^{\frac{1}{2}} + 3 x^{- 2}) d x$ — Rewrite first: the root is the power one half; x squared on the bottom is the power −2.
3
$x^{\frac{1}{2}} \to \frac{x ^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} x^{\frac{3}{2}}$ — One half plus 1 is three halves. Dividing by 3/2 is the same as multiplying by 2/3.
4
$3 x^{- 2} \to \frac{3 x ^{- 1}}{- 1} = - 3 x^{- 1}$ — −2 plus 1 is −1. Divide by −1.
5
$= \frac{2}{3} x^{\frac{3}{2}} - \frac{3}{x} + c$ — Tidy up: x to the power −1 is 1 over x.

Common mistake

Getting −3 from −2 + 1. Adding 1 to a negative power moves it towards zero: −2 + 1 = −1.

Common mistake

Dividing by a fraction the wrong way. Dividing by 3/2 means multiplying by 2/3, not by 3/2.

Key idea

Rewrite as powers first. Integrate second. Tidy last.

Brackets and fractions

There is no product rule for integration. The integral of a product is NOT the product of the integrals. So before you integrate, expand brackets and split fractions into single terms.

Worked example

1
Find $\int (x - 3) (2 x + 1) d x$ .
2
$(x - 3) (2 x + 1) = 2 x^{2} - 5 x - 3$ — Expand the brackets first.
3
$\int (2 x^{2} - 5 x - 3) d x$ — Now each term is a single power of x.
4
$= \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} - 3 x + c$ — Power rule on each term, then + c.

Worked example

1
Find $\int \frac{x ^{4} - 6 x}{x ^{2}} d x$ .
2
$\frac{x ^{4}}{x ^{2}} - \frac{6 x ^{\frac{1}{2}}}{x ^{2}}$ — Split: each term on top goes over the bottom on its own.
3
$= x^{2} - 6 x^{- \frac{3}{2}}$ — Dividing powers means subtracting: 4 − 2 = 2, and one half − 2 = −3/2.
4
$x^{2} \to \frac{x ^{3}}{3}$ — Power rule.
5
$- 6 x^{- \frac{3}{2}} \to \frac{- 6 x ^{- \frac{1}{2}}}{- \frac{1}{2}} = 12 x^{- \frac{1}{2}}$ — −3/2 plus 1 is −1/2. Dividing by −1/2 is multiplying by −2.
6
$\int \frac{x ^{4} - 6 x}{x ^{2}} d x = \frac{x ^{3}}{3} + \frac{12}{x} + c$ — Put together, tidy, and add + c.

Common mistake

Integrating the top and the bottom of a fraction separately. Split the fraction into single terms first.

Common mistake

Integrating each bracket on its own and multiplying the results. Always expand first.

Key idea

A bracket or a fraction is a signal: turn it into a list of single powers of x before you integrate.

When the power is −1

Rule

This step is for 0606 and 4037 only. 4PM1 does not ask you to integrate $\frac{1}{x}$ .

Try the power rule on $x^{- 1}$ : the power becomes 0, and you would divide by 0. That is impossible, so the power rule fails here.

But you already know a function whose gradient is $\frac{1}{x}$ : $\frac{d}{d x} (ln x) = \frac{1}{x}$ . Rewind it:

Formula

\int \frac{1}{x} d x = ln x + c

$x$ — must be positive here, because ln only works on positive numbers.
$ln$ — the natural logarithm, log to base e.

Rule

Learn it: 0606 and 4037 do not give it. On the non-calculator paper, leave answers in exact ln form.

Worked example

1
Find $\int \frac{x ^{2} + 5}{x} d x$ .
2
$\frac{x ^{2} + 5}{x} = x + \frac{5}{x}$ — Split the fraction into single terms.
3
$x \to \frac{x ^{2}}{2}$ — Power rule.
4
$\frac{5}{x} = 5 \times \frac{1}{x} \to 5 ln x$ — The 5 waits in front of the ln.
5
$\int \frac{x ^{2} + 5}{x} d x = \frac{x ^{2}}{2} + 5 ln x + c$ — Put together and add + c.

Common mistake

Writing $\frac{1}{x ^{2}}$ as $ln x^{2}$ . Only the power −1 gives ln. $\int x^{- 2} d x = - \frac{1}{x} + c$ by the power rule.

Key idea

Every power of x uses the power rule — except −1, which gives ln x.

Find the curve: work out c

Many exam questions give you the gradient, $\frac{d y}{d x}$ , and one point on the curve. Integrating gives the curve with an unknown c. The point tells you what c is.

Steps

1
1. Integrate $\frac{d y}{d x}$ to get y. Include + c.
2. Put the x and y of the point into your equation for y.
3. Solve for c.
4. Write the full equation of the curve with the number for c.

Worked example

1
A curve has gradient $\frac{d y}{d x} = 3 x - 2$ and passes through $(4, 10)$ . Find the equation of the curve.
2
$y = \int (3 x^{\frac{1}{2}} - 2) d x$ — Rewrite the root as a power, then integrate.
3
$y = 2 x^{\frac{3}{2}} - 2 x + c$ — 3 divided by 3/2 is 2. Do not forget + c.
4
$10 = 2 (4)^{\frac{3}{2}} - 2 (4) + c$ — Put x = 4 and y = 10 into the equation for y.
5
$4^{\frac{3}{2}} = (4)^{3} = 8$ — Square root first, then cube. No calculator needed.
6
$10 = 16 - 8 + c, c = 2$ — Solve for c.
7
$y = 2 x^{\frac{3}{2}} - 2 x + 2$ — The full equation of the curve.

The same idea works with function notation: if you are given $f^{'} (x)$ and a point, integrate to find $f (x)$ .

Marks usually go: one for integrating, one for the correct terms, one for using the point, one for the final equation. So show the substitution line.

Common mistake

Putting the point into $\frac{d y}{d x}$ . The point is on the curve, so it goes into the equation for y.

Key idea

No + c, no way to use the point. The c is exactly what the point fixes.

“Hence” questions

Some questions ask you to differentiate first, then say “hence find” an integral. Hence means: use the answer you just found. Integration undoes differentiation.

Formula

\int k g (x) d x = k F (x) + c

$F (x)$ — the function you differentiated in the first part.
$g (x)$ — its derivative, so $F^{'} (x) = g (x)$ .
$k$ — a number. Find it by comparing the integral with your derivative.

Worked example

1
(a) Given $y = \frac{x ^{3} + 2}{x}$ , find $\frac{d y}{d x}$ . (b) Hence find $\int (5 x^{\frac{3}{2}} - 2 x^{- \frac{3}{2}}) d x$ .
2
$y = x^{\frac{5}{2}} + 2 x^{- \frac{1}{2}}$ — Split the fraction: 3 − one half = 5/2, and 2 over root x is 2x to the power −1/2.
3
$\frac{d y}{d x} = \frac{5}{2} x^{\frac{3}{2}} - x^{- \frac{3}{2}}$ — Differentiate each term.
4
$5 x^{\frac{3}{2}} - 2 x^{- \frac{3}{2}} = 2 (\frac{5}{2} x^{\frac{3}{2}} - x^{- \frac{3}{2}})$ — Compare: the expression to integrate is exactly 2 times dy/dx, so k = 2.
5
$\int (5 x^{\frac{3}{2}} - 2 x^{- \frac{3}{2}}) d x = \frac{2 ( x ^{3} + 2 )}{x} + c$ — Integrating 2 times dy/dx gives 2y. Then add + c.

In real exams the derivative in part (a) is often a hard one. The move in part (b) is always the same: find k, write k times the original function, add + c.

Common mistake

Forgetting the number k. If the integral is 2 times your derivative, the answer is 2y + c, not y + c.

Key idea

“Hence” means the hard work is already done. Spot the k, then write k times what you differentiated, + c.

Which method? Read the question

Read the question first. What you see tells you what to do.

You see	You do
“Find $\int \dots d x$ ” with powers of x	Add 1 to the power, divide by the new power, + c
A root, or x on the bottom	Rewrite as $x^{n}$ first
Brackets or a fraction	Expand or split into single terms first
$\frac{1}{x}$ (0606 / 4037)	$ln x + c$
The gradient and a point	Integrate with + c, put the point into y, find c
“Hence” after a derivative	k times what you differentiated, + c

Common mistake

Stopping without + c on an indefinite integral. It is the easiest mark to lose.

Key idea

Integration is differentiation run backwards. When unsure, differentiate your answer: you should get back the question.

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Real past-paper questions on integration: undoing differentiation

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