Integration as the reverse of differentiation

AS · P1 · 12 min

Differentiation turns a function into its gradient function. Integration runs the process backwards: given the gradient function, it recovers the original function. Because a constant disappears when you differentiate, integration can only recover the function up to an unknown constant, and a known point on the curve is needed to fix it. This note covers integrating powers of $x$ and the classic exam question "find the equation of the curve", which appears on almost every Paper 1. Integrating brackets like $(3x - 2)^5$ is in integrating $(ax + b)^n$ , and areas are in definite integrals.

Undoing a derivative

If $\dfrac{dy}{dx} = 2x$ , what was $y$ ? You know that $x^2$ differentiates to $2x$ , so $y = x^2$ is one answer. But $x^2 + 5$ , $x^2 - 3$ and $x^2 + 100$ all differentiate to $2x$ as well, because the constant differentiates to zero. So the full answer is $y = x^2 + c$ , where $c$ is an arbitrary constant.

Definition

An integral (or antiderivative) of $f(x)$ is a function $F(x)$ with $F'(x) = f(x)$ . The indefinite integral of $f(x)$ is

\int f(x)\, dx = F(x) + c

where $c$ is the constant of integration. Finding it is called integrating $f(x)$ with respect to $x$ .

The symbol $\int \ldots\, dx$ means "integrate with respect to $x$ ". The $dx$ tells you the variable, and it closes the expression, so $\int (3x + 1)\, dx$ means integrate the whole bracket.

Geometrically, all the curves $y = x^2 + c$ have exactly the same gradient at each value of $x$ : they are vertical translations of one another. Integration gives you the whole family; one point picks out a single member.

y = x^2 + x - 4 y = x^2 + x - 2 y = x^2 + x y = x^2 + x + 2 (1, -2)

These are curves $y = x^2 + x + c$ for four values of $c$ . All have gradient $2x + 1$ . The one through $(1, -2)$ is $y = x^2 + x - 4$ , the example from the syllabus worked in Example 4 below.

The power rule in reverse

To differentiate $x^n$ you multiply by the power and reduce the power by one. To integrate, do the opposite in the opposite order: increase the power by one, then divide by the new power.

\frac{d}{dx}\left(\frac{x^{n+1}}{n + 1}\right) = \frac{(n + 1)x^n}{n + 1} = x^n

so $\dfrac{x^{n+1}}{n + 1}$ is an integral of $x^n$ . The rule fails only for $n = -1$ , where it would mean dividing by zero. The integral of $x^{-1} = \dfrac{1}{x}$ is a logarithm, which is in Paper 3, not Paper 1.

Key result

\int x^n\, dx = \frac{x^{n + 1}}{n + 1} + c \qquad \text{for any rational } n \neq -1

Constant multiples stay: $\displaystyle\int kx^n\, dx = \frac{kx^{n+1}}{n + 1} + c$ .
Sums and differences integrate term by term, with a single $+\,c$ at the end.
A constant integrates to a multiple of $x$ : $\displaystyle\int k\, dx = kx + c$ .

Check every integral by differentiating it in your head. If you do not get back to the integrand, something is wrong.

Rewrite as powers first

Exactly as for differentiation, roots and reciprocals must become powers of $x$ , products must be expanded and fractions split before you integrate. There is no product or quotient rule for integration.

Integrand	Rewrite as	Integral (without $+\,c$ )
$\sqrt{x}$	$x^{\frac{1}{2}}$	$\tfrac{2}{3}x^{\frac{3}{2}}$
$\dfrac{1}{x^2}$	$x^{-2}$	$-x^{-1} = -\dfrac{1}{x}$
$\dfrac{1}{\sqrt{x}}$	$x^{-\frac{1}{2}}$	$2x^{\frac{1}{2}} = 2\sqrt{x}$
$\dfrac{6}{x^4}$	$6x^{-4}$	$-2x^{-3}$
$x\sqrt{x}$	$x^{\frac{3}{2}}$	$\tfrac{2}{5}x^{\frac{5}{2}}$
$(x + 1)(x - 3)$	$x^2 - 2x - 3$	$\tfrac{1}{3}x^3 - x^2 - 3x$
$\dfrac{x^3 + 2}{x^2}$	$x + 2x^{-2}$	$\tfrac{1}{2}x^2 - 2x^{-1}$

Dividing by a fraction is multiplying by its reciprocal: $\dfrac{x^{3/2}}{3/2} = \tfrac{2}{3}x^{\frac{3}{2}}$ . Negative powers need care: $-2 + 1 = -1$ , and $-\tfrac{1}{2} + 1 = +\tfrac{1}{2}$ .

Finding the equation of a curve

The constant of integration is fixed by one extra piece of information, usually a point on the curve. This is the syllabus's own example: "find the equation of the curve through $(1, -2)$ for which $\dfrac{dy}{dx} = 2x + 1$ ."

Equation of a curve from its gradient

Rewrite $\dfrac{dy}{dx}$ as a sum of powers of $x$ .
Integrate term by term and include $+\,c$ .
Substitute the coordinates of the given point into $y = \ldots + c$ and solve for $c$ .
Write the final equation $y = \ldots$ with the value of $c$ in place.

Information in other forms

The extra information is not always a plain point.

A tangent or normal: "the normal at $(2, 7)$ has gradient $-\tfrac{1}{4}$ " tells you the curve's gradient at $x = 2$ is $4$ . That can fix an unknown constant inside $\dfrac{dy}{dx}$ .
A stationary point: " $\dfrac{dy}{dx} = 0$ at $x = 1$ " fixes a constant in the derivative.
A second derivative: if $\dfrac{d^2y}{dx^2}$ is given, integrate twice. Each integration brings its own constant, so you need two pieces of information.

Worked examples

A polynomial

Find $\displaystyle\int \left(6x^2 - 4x + 5\right) dx$ .

Solution

Integrate term by term: raise each power by one and divide by the new power.

\int \left(6x^2 - 4x + 5\right) dx = \frac{6x^3}{3} - \frac{4x^2}{2} + 5x + c = 2x^3 - 2x^2 + 5x + c

Check: $\dfrac{d}{dx}\left(2x^3 - 2x^2 + 5x\right) = 6x^2 - 4x + 5$ .

Roots and reciprocals

Find $\displaystyle\int \left(3\sqrt{x} - \frac{4}{x^3} + \frac{2}{\sqrt{x}}\right) dx$ .

Solution

Rewrite:

\int \left(3x^{\frac{1}{2}} - 4x^{-3} + 2x^{-\frac{1}{2}}\right) dx

Integrate each term:

= \frac{3x^{\frac{3}{2}}}{\tfrac{3}{2}} - \frac{4x^{-2}}{-2} + \frac{2x^{\frac{1}{2}}}{\tfrac{1}{2}} + c = 2x^{\frac{3}{2}} + 2x^{-2} + 4x^{\frac{1}{2}} + c

or $2x^{\frac{3}{2}} + \dfrac{2}{x^2} + 4\sqrt{x} + c$ .

Expanding a fraction first

Find $\displaystyle\int \frac{(x^2 - 3)^2}{x^2}\, dx$ .

Solution

Expand the numerator and divide each term by $x^2$ :

\frac{x^4 - 6x^2 + 9}{x^2} = x^2 - 6 + 9x^{-2}

Then

\int \left(x^2 - 6 + 9x^{-2}\right) dx = \frac{x^3}{3} - 6x - 9x^{-1} + c = \frac{x^3}{3} - 6x - \frac{9}{x} + c

The curve through a point

Find the equation of the curve through $(1, -2)$ for which $\dfrac{dy}{dx} = 2x + 1$ .

Solution

y = \int (2x + 1)\, dx = x^2 + x + c

At $(1, -2)$ : $-2 = 1 + 1 + c$ , so $c = -4$ .

The curve is $y = x^2 + x - 4$ .

Fractional powers and a point

A curve is such that $\dfrac{dy}{dx} = 3\sqrt{x} - \dfrac{6}{x^2}$ , and the point $(4, 9)$ lies on the curve. Find the equation of the curve.

Solution

y = \int \left(3x^{\frac{1}{2}} - 6x^{-2}\right) dx = 2x^{\frac{3}{2}} + 6x^{-1} + c

At $x = 4$ : $4^{\frac{3}{2}} = 8$ , so

9 = 2(8) + \frac{6}{4} + c = 16 + 1.5 + c \quad\Rightarrow\quad c = -8.5

The curve is $y = 2x^{\frac{3}{2}} + \dfrac{6}{x} - \dfrac{17}{2}$ .

A gradient condition from a normal

A curve has $\dfrac{dy}{dx} = kx^2 - 4x$ , where $k$ is a constant. The normal to the curve at the point $(2, 7)$ has gradient $-\tfrac{1}{4}$ . Find $k$ and the equation of the curve.

Solution

The normal has gradient $-\tfrac{1}{4}$ , so the tangent (and hence the curve) has gradient $4$ at $x = 2$ :

4k - 8 = 4 \quad\Rightarrow\quad k = 3

Then $\dfrac{dy}{dx} = 3x^2 - 4x$ and

y = x^3 - 2x^2 + c

At $(2, 7)$ : $7 = 8 - 8 + c$ , so $c = 7$ . The curve is $y = x^3 - 2x^2 + 7$ .

Integrating twice

A curve has $\dfrac{d^2y}{dx^2} = 6x - 4$ . It has a stationary point at $(1, 5)$ .

(a) Find the equation of the curve.

(b) Find the coordinates of the other stationary point and determine the nature of both.

Solution

(a) Integrate once:

\frac{dy}{dx} = 3x^2 - 4x + c_1

The gradient is $0$ at $x = 1$ : $3 - 4 + c_1 = 0$ , so $c_1 = 1$ . Integrate again:

y = x^3 - 2x^2 + x + c_2

At $(1, 5)$ : $5 = 1 - 2 + 1 + c_2$ , so $c_2 = 5$ . The curve is $y = x^3 - 2x^2 + x + 5$ .

(b) $\dfrac{dy}{dx} = 3x^2 - 4x + 1 = (3x - 1)(x - 1) = 0$ gives $x = \tfrac{1}{3}$ (and $x = 1$ ).

y = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} + 5 = \frac{1 - 6 + 9 + 135}{27} = \frac{139}{27}

Using $\dfrac{d^2y}{dx^2} = 6x - 4$ : at $x = \tfrac{1}{3}$ it is $-2 < 0$ , so $\left(\tfrac{1}{3}, \tfrac{139}{27}\right)$ is a maximum; at $x = 1$ it is $2 > 0$ , so $(1, 5)$ is a minimum. See stationary points for the test.

Watch out

Forgetting $+\,c$ . An indefinite integral without the constant loses a mark, and in a "find the curve" question the constant is the whole point.

Integrating a product term by term. $\displaystyle\int (x + 1)(x - 3)\, dx$ is not $\tfrac{1}{2}x^2 + x$ times $\tfrac{1}{2}x^2 - 3x$ . Expand first.

Multiplying instead of dividing by the new power. $\displaystyle\int x^{\frac{1}{2}}\, dx = \dfrac{x^{3/2}}{3/2} = \tfrac{2}{3}x^{\frac{3}{2}}$ , not $\tfrac{3}{2}x^{\frac{3}{2}}$ .

Adding one to a negative power wrongly. $x^{-3}$ becomes $\dfrac{x^{-2}}{-2}$ , not $\dfrac{x^{-4}}{-4}$ (that is differentiating).

Using the gradient of the normal as the gradient of the curve. The curve's gradient is the negative reciprocal of the normal's.

One constant for two integrations. Integrating $\dfrac{d^2y}{dx^2}$ twice gives two independent constants; find the first before integrating again.

Exam tip

Show the rewritten form before integrating, for example $\int \left(3x^{\frac{1}{2}} - 6x^{-2}\right)dx$ . It makes your method clear and earns credit even if one term slips.
Typical marking for "find the equation of the curve": one mark for raising a power correctly, one for the coefficients, one for using the point to find $c$ , one for the final equation. Do not leave $c$ unevaluated.
Give the final answer as an equation " $y = \ldots$ ", not just an expression.
Keep fractions exact. $c = -\tfrac{17}{2}$ is better than $-8.5$ only if the rest is in fractions; either is accepted, but avoid rounded decimals.
Check by differentiating. It takes ten seconds and catches most errors.

Summary

Integration reverses differentiation: $\displaystyle\int f(x)\, dx = F(x) + c$ where $F'(x) = f(x)$ .
$\displaystyle\int x^n\, dx = \frac{x^{n+1}}{n + 1} + c$ for rational $n \neq -1$ : add one to the power, divide by the new power.
Integrate sums term by term; constants multiply through; one $+\,c$ at the end.
Rewrite roots, reciprocals, products and fractions as sums of powers first.
A point on the curve fixes $c$ ; a gradient condition (tangent, normal, stationary point) fixes a constant in $\dfrac{dy}{dx}$ .
From $\dfrac{d^2y}{dx^2}$ , integrate twice, with a new constant each time.
Check by differentiating your answer.

Practice questions

Question

Find $\displaystyle\int \left(4x^3 - 6x + 7\right) dx$ .
Find $\displaystyle\int \left(\frac{5}{x^2} + \sqrt{x}\right) dx$ .
Find $\displaystyle\int (2x + 3)(x - 1)\, dx$ .
Find $\displaystyle\int \frac{x + 4}{\sqrt{x}}\, dx$ .
A curve has $\dfrac{dy}{dx} = 6x^2 - 2x$ and passes through $(1, 4)$ . Find its equation.
A curve has $\dfrac{dy}{dx} = \dfrac{4}{\sqrt{x}} - 1$ and passes through $(9, 10)$ . Find its equation.
A function $f$ has $f''(x) = 12x - 6$ . The curve $y = f(x)$ passes through $(0, 3)$ and has a stationary point where $x = 2$ . Find $f(x)$ , and find the coordinates of the other stationary point.
A curve has $\dfrac{dy}{dx} = 3x^2 + kx$ , where $k$ is a constant, and passes through the points $(1, 0)$ and $(3, 10)$ . Find $k$ and the equation of the curve.
A curve passes through $(1, 1)$ and has $\dfrac{dy}{dx} = 3\sqrt{x} - k$ , where $k$ is a constant. The tangent to the curve at the point where $x = 4$ is parallel to the line $y = 2x$ . (a) Find $k$ and the equation of the curve. (b) Find the coordinates of the stationary point of the curve.

Answers

$x^4 - 3x^2 + 7x + c$ .
$\displaystyle\int \left(5x^{-2} + x^{\frac{1}{2}}\right) dx = -5x^{-1} + \tfrac{2}{3}x^{\frac{3}{2}} + c = -\dfrac{5}{x} + \tfrac{2}{3}x^{\frac{3}{2}} + c$ .
Expand: $2x^2 + x - 3$ . Integral: $\tfrac{2}{3}x^3 + \tfrac{1}{2}x^2 - 3x + c$ .
$\dfrac{x + 4}{\sqrt{x}} = x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}$ . Integral: $\tfrac{2}{3}x^{\frac{3}{2}} + 8x^{\frac{1}{2}} + c$ .
$y = 2x^3 - x^2 + c$ . At $(1, 4)$ : $4 = 2 - 1 + c$ , $c = 3$ . So $y = 2x^3 - x^2 + 3$ .
$y = \displaystyle\int \left(4x^{-\frac{1}{2}} - 1\right) dx = 8x^{\frac{1}{2}} - x + c$ . At $(9, 10)$ : $10 = 24 - 9 + c$ , $c = -5$ . So $y = 8\sqrt{x} - x - 5$ .
$f'(x) = 6x^2 - 6x + a$ . $f'(2) = 0$ : $24 - 12 + a = 0$ , $a = -12$ . $f(x) = 2x^3 - 3x^2 - 12x + b$ ; $f(0) = 3$ gives $b = 3$ . So $f(x) = 2x^3 - 3x^2 - 12x + 3$ . $f'(x) = 6(x - 2)(x + 1)$ , so the other stationary point is at $x = -1$ : $f(-1) = -2 - 3 + 12 + 3 = 10$ . The point is $(-1, 10)$ .
$y = x^3 + \tfrac{k}{2}x^2 + c$ . At $(1, 0)$ : $0 = 1 + \tfrac{k}{2} + c$ . At $(3, 10)$ : $10 = 27 + \tfrac{9k}{2} + c$ . Subtracting: $10 = 26 + 4k$ , so $k = -4$ , and then $c = 1$ . The curve is $y = x^3 - 2x^2 + 1$ .
(a) Parallel to $y = 2x$ means gradient $2$ at $x = 4$ : $3(2) - k = 2$ , so $k = 4$ . Then $y = \displaystyle\int \left(3x^{\frac{1}{2}} - 4\right) dx = 2x^{\frac{3}{2}} - 4x + c$ . At $(1, 1)$ : $1 = 2 - 4 + c$ , $c = 3$ . So $y = 2x^{\frac{3}{2}} - 4x + 3$ . (b) $3\sqrt{x} - 4 = 0$ gives $\sqrt{x} = \tfrac{4}{3}$ , $x = \tfrac{16}{9}$ . Then $x^{\frac{3}{2}} = \tfrac{64}{27}$ , so $y = \tfrac{128}{27} - \tfrac{64}{9} + 3 = \tfrac{128 - 192 + 81}{27} = \tfrac{17}{27}$ . The stationary point is $\left(\tfrac{16}{9}, \tfrac{17}{27}\right)$ .

Undoing a derivative

The power rule in reverse

Rewrite as powers first

Finding the equation of a curve

Information in other forms

Worked examples

Practice questions

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