Mathematics 9709 · Pure Mathematics 1

Reversing differentiation & the constant

Integration is differentiation done backwards. You are given the gradient $\tfrac{dy}{dx}$ . You have to find $y$ .
The whole chapter is built on this one skill. Every area and volume later is the same $\int$ with a different thing put in.

The whole chapter in one line: integration is one method that does four jobs. Find a curve from its gradient, find an area, find the gap between two curves, find a volume. Each one is the same $\int$ with a different thing put in. Learn the method here. After that you just choose what to put in.

Add one to the power. Then divide by the new power. Always write $+c$ . A given point lets you find $c$ .

The power rule, in reverse

When you differentiate $x^n$ , you multiply by the power. Then you drop the power by one.
To undo that, do the opposite in the opposite order. Add one to the power. Then divide by the new power.

$\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+c\qquad(n\neq-1)$

The curly $\int$ sign means integrate. Read it as “the thing whose gradient is…”.

Why add one then divide (check it by differentiating back)

Differentiate $x^{n+1}$ . You bring the $n+1$ to the front and lower the power. You get $(n+1)x^{n}$ . But we wanted just $x^n$ . So we get rid of that extra $n+1$ by dividing by it at the start. That is why you divide by the new power, not the old one. Want to check any integral? Differentiate your answer. It should go back to what you started with.

Why the +c is always there

When you differentiate, any plain number disappears. $x^3+7$ , $x^3-2$ and just $x^3$ all have the same gradient $3x^2$ .
So going backwards, you cannot know which number was there. Many curves have that gradient. They sit one above the other.
Write the unknown number as $+c$ . It is called the constant of integration. If you leave it off, you lose a mark.
A given point on the curve tells you which curve you have. You use the same three steps:

Integrate the gradient term by term, and write $+c$ straight away so you cannot forget it.
Substitute the given point into your integrated expression: put its $x$ and $y$ in.
Solve for $c$ , then write out the full equation of the curve with that value in place.

Worked example

A curve has $\dfrac{dy}{dx}=3x^2-2$ and passes through $(1,\ 4)$ : find $y$

Integrate first. Add one to each power, divide, and add $+c$ :

$y=\int(3x^2-2)\,dx=x^3-2x+c$

Now use the point $(1,4)$ to find $c$ :

$4=(1)^3-2(1)+c=-1+c\;\Rightarrow\;c=5$

Answer

y=x^3-2x+5

A real Paper 1 question uses the same steps but with a root: rewrite as powers first, then do the three steps.

Worked example

A curve has $\dfrac{dy}{dx}=4x-3\sqrt{x}+1$ and passes through $(4,\ 11)$ . Find the equation of the curve.

Write the root as a power so the rule works. Then integrate term by term and add $+c$ :

$y=\int\big(4x-3x^{\frac12}+1\big)\,dx=2x^2-3\cdot\dfrac{x^{\frac32}}{\frac32}+x+c=2x^2-2x^{\frac32}+x+c$

Substitute the point $(4,11)$ . Note $4^{\frac32}=8$ :

$11=2(16)-2(8)+4+c=32-16+4+c=20+c\;\Rightarrow\;c=-9$

Answer

y=2x^2-2x^{\frac32}+x-9

Part (a) of the same paper first asks where the gradient is $\tfrac{11}{2}$ : set $4x-3\sqrt{x}+1=\tfrac{11}{2}$ . This is really a quadratic in $\sqrt{x}$ . It gives $x=\tfrac94$ . (9709/11 Oct/Nov 2024 Q5)

Roots and fractions: rewrite as powers first

Negative and fractional powers work in the same way. The one extra step is to rewrite first so the power rule works.
Turn $\dfrac{1}{x^k}$ into $x^{-k}$ . Turn a root like $\sqrt{x}$ into $x^{1/2}$ .
If a fraction has several terms on top, split it into separate powers before you integrate.

Worked example

Find $\mathrm{f}(x)$ given $\mathrm{f}'(x)=4x^3-\dfrac{2}{x^2}+4$

Write $\tfrac{2}{x^2}$ as $2x^{-2}$ so every term is a power:

$\mathrm{f}(x)=\int\big(4x^3-2x^{-2}+4\big)\,dx$

Add one to each power and divide. The middle term: power $-2$ goes to $-1$ , divide by $-1$ :

$=x^4-\dfrac{2x^{-1}}{-1}+4x+c=x^4+\dfrac{2}{x}+4x+c$

Answer

\mathrm{f}(x)=x^4+\tfrac{2}{x}+4x+c

Where the marks go

Common mistake

Never forget

+c

on an indefinite integral. It is worth a mark on its own, and without it you cannot use a given point to finish the question.

Common mistake

You divide by the new power

n+1

, not the old one. The single power this rule cannot touch is

n=-1

(it would divide by zero), and that case is not in P1.

Now you try

The gradient of a curve is $\tfrac{dy}{dx}=3\sqrt{x}-\dfrac{3}{\sqrt{x}}$ , and the point $(4,7)$ lies on it. Find the equation of the curve. (9709/13 May/June 2020 Q2)

Your turn— tap to reveal the worked answer (9709/13 May/June 2020 Q2)

Rewrite as powers: $\tfrac{dy}{dx}=3x^{\frac12}-3x^{-\frac12}$ .

Integrate term by term and add $+c$ : $y=3\cdot\dfrac{x^{\frac32}}{\frac32}-3\cdot\dfrac{x^{\frac12}}{\frac12}+c=2x^{\frac32}-6x^{\frac12}+c$ .

Put in $(4,7)$ : $4^{\frac32}=8$ and $4^{\frac12}=2$ , so $7=2(8)-6(2)+c=4+c\Rightarrow c=3$ .

Answer

y=2x^{\frac32}-6x^{\frac12}+3

PreviousRates of change NextIntegrating (ax + b)ⁿ

Reversing differentiation & the constant

Integration is differentiation done backwards. You are given the gradient $\tfrac{dy}{dx}$ . You have to find $y$ .
The whole chapter is built on this one skill. Every area and volume later is the same $\int$ with a different thing put in.

Add one to the power. Then divide by the new power. Always write $+c$ . A given point lets you find $c$ .

The power rule, in reverse

When you differentiate $x^n$ , you multiply by the power. Then you drop the power by one.
To undo that, do the opposite in the opposite order. Add one to the power. Then divide by the new power.

$\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+c\qquad(n\neq-1)$

The curly $\int$ sign means integrate. Read it as “the thing whose gradient is…”.

Why add one then divide (check it by differentiating back)

Why the +c is always there

When you differentiate, any plain number disappears. $x^3+7$ , $x^3-2$ and just $x^3$ all have the same gradient $3x^2$ .
So going backwards, you cannot know which number was there. Many curves have that gradient. They sit one above the other.
Write the unknown number as $+c$ . It is called the constant of integration. If you leave it off, you lose a mark.
A given point on the curve tells you which curve you have. You use the same three steps:

Integrate the gradient term by term, and write $+c$ straight away so you cannot forget it.
Substitute the given point into your integrated expression: put its $x$ and $y$ in.
Solve for $c$ , then write out the full equation of the curve with that value in place.

Worked example

A curve has $\dfrac{dy}{dx}=3x^2-2$ and passes through $(1,\ 4)$ : find $y$

Integrate first. Add one to each power, divide, and add $+c$ :

$y=\int(3x^2-2)\,dx=x^3-2x+c$

Now use the point $(1,4)$ to find $c$ :

$4=(1)^3-2(1)+c=-1+c\;\Rightarrow\;c=5$

Answer

y=x^3-2x+5

A real Paper 1 question uses the same steps but with a root: rewrite as powers first, then do the three steps.

Worked example

A curve has $\dfrac{dy}{dx}=4x-3\sqrt{x}+1$ and passes through $(4,\ 11)$ . Find the equation of the curve.

Write the root as a power so the rule works. Then integrate term by term and add $+c$ :

$y=\int\big(4x-3x^{\frac12}+1\big)\,dx=2x^2-3\cdot\dfrac{x^{\frac32}}{\frac32}+x+c=2x^2-2x^{\frac32}+x+c$

Substitute the point $(4,11)$ . Note $4^{\frac32}=8$ :

$11=2(16)-2(8)+4+c=32-16+4+c=20+c\;\Rightarrow\;c=-9$

Answer

y=2x^2-2x^{\frac32}+x-9

Roots and fractions: rewrite as powers first

Negative and fractional powers work in the same way. The one extra step is to rewrite first so the power rule works.
Turn $\dfrac{1}{x^k}$ into $x^{-k}$ . Turn a root like $\sqrt{x}$ into $x^{1/2}$ .
If a fraction has several terms on top, split it into separate powers before you integrate.

Worked example

Find $\mathrm{f}(x)$ given $\mathrm{f}'(x)=4x^3-\dfrac{2}{x^2}+4$

Write $\tfrac{2}{x^2}$ as $2x^{-2}$ so every term is a power:

$\mathrm{f}(x)=\int\big(4x^3-2x^{-2}+4\big)\,dx$

Add one to each power and divide. The middle term: power $-2$ goes to $-1$ , divide by $-1$ :

$=x^4-\dfrac{2x^{-1}}{-1}+4x+c=x^4+\dfrac{2}{x}+4x+c$

Answer

\mathrm{f}(x)=x^4+\tfrac{2}{x}+4x+c

Where the marks go

Common mistake

Never forget

+c

on an indefinite integral. It is worth a mark on its own, and without it you cannot use a given point to finish the question.

Common mistake

You divide by the new power

n+1

, not the old one. The single power this rule cannot touch is

n=-1

(it would divide by zero), and that case is not in P1.

Now you try

The gradient of a curve is $\tfrac{dy}{dx}=3\sqrt{x}-\dfrac{3}{\sqrt{x}}$ , and the point $(4,7)$ lies on it. Find the equation of the curve. (9709/13 May/June 2020 Q2)

Your turn— tap to reveal the worked answer (9709/13 May/June 2020 Q2)

Rewrite as powers: $\tfrac{dy}{dx}=3x^{\frac12}-3x^{-\frac12}$ .

Integrate term by term and add $+c$ : $y=3\cdot\dfrac{x^{\frac32}}{\frac32}-3\cdot\dfrac{x^{\frac12}}{\frac12}+c=2x^{\frac32}-6x^{\frac12}+c$ .

Put in $(4,7)$ : $4^{\frac32}=8$ and $4^{\frac12}=2$ , so $7=2(8)-6(2)+c=4+c\Rightarrow c=3$ .

Answer

y=2x^{\frac32}-6x^{\frac12}+3