Integration

What you'll learn

How integration reverses differentiation.
How to integrate polynomials, brackets, roots and reciprocal powers.
How to find the constant of integration using a point on a curve.
How definite integrals give areas, including areas between curves.

Integration as reverse differentiation

You already know that differentiation finds a gradient function. Integration goes the other way: it starts with a gradient function and works back to the original function.

A flow diagram showing differentiation and integration as reverse processes between a function and its gradient function.

For example, if

ddx(x3)=3x2\frac{d}{dx}(x^3)=3x^2

then integrating $3x^2$ gives $x^3$ , plus possibly a constant.

Definition

Integral and antiderivative

If $F^{'} (x) = f (x)$ , then $F (x)$ is called an antiderivative of $f (x)$ . The symbol $dx\int f(x)\,dx$ means “integrate $f (x)$ with respect to $x$ ”.

The little $d x$ tells you the variable you are integrating with respect to. In AS Pure, this will usually be $x$ .

The power rule for integration

Key Idea

The reverse power rule

To integrate a power of $x$ , increase the power by 1, then divide by the new power.

A visual summary of the reverse power rule for integrating a power of x.

For any constant $a$ ,

dx=an+1xn+1+C,n≠−1\int ax^n\,dx=\frac{a}{n+1}x^{n+1}+C,\qquad n\neq -1

The $C$ is called the constant of integration. It is needed because differentiating any constant gives zero.

Example

Integrating a polynomial

Find $dx\int \left(4x^3-6x^2+5x-9\right)\,dx$ .

Integrate each term separately.
$dx=x4\int 4x^3\,dx=x^4$
Continue term by term, remembering that a constant integrates to a multiple of $x$ .
$dx=−9x\int -6x^2\,dx=-2x^3,\qquad \int 5x\,dx=\frac{5}{2}x^2,\qquad \int -9\,dx=-9x$
Add the constant of integration.
$dx=x4−2x3+52x2−9x+C\int \left(4x^3-6x^2+5x-9\right)\,dx=x^4-2x^3+\frac{5}{2}x^2-9x+C$

Tip

Quick check

Differentiate your answer. If you get back the original expression, your integration is correct.

Expanding before integrating

Sometimes you are asked to integrate an expression in brackets. The power rule works best when the expression is written as a sum of powers of $x$ , so expand first.

Example

Integrating a product of brackets

Find $dx\int (x+5)(x-2)\,dx$ .

Expand the brackets first.
$x+5)(x-2)=x^2+3x-10$
Integrate the expanded expression term by term.
$dx=13x3+32x2−10x+C\int \left(x^2+3x-10\right)\,dx=\frac{1}{3}x^3+\frac{3}{2}x^2-10x+C$

Common Mistake

Do not integrate each bracket separately

There is no rule that lets you integrate $(x + 5) (x - 2)$ by integrating each bracket on its own. Expand first, then integrate.

Roots and fractions as powers

Before integrating roots or fractions involving $x$ , rewrite them using powers.

Useful rewrites include:

$x=x12\sqrt{x}=x^{\frac{1}{2}}$
$1x2=x−2\frac{1}{x^2}=x^{-2}$
$1x=x−12\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$

Common Mistake

The power rule exception

The reverse power rule above does not work for $x^{-1}$ . At AS level, questions in this topic usually avoid needing to integrate $1x\frac{1}{x}$ .

Example

Integrating roots and reciprocal powers

Find $dx\int \left(6\sqrt{x}+\frac{4}{x^2}+3\right)\,dx$ .

Rewrite the expression using powers of $x$ .
$6x+4x2+3=6x12+4x−2+36\sqrt{x}+\frac{4}{x^2}+3=6x^{\frac{1}{2}}+4x^{-2}+3$
Apply the reverse power rule to each term.
$dx=3x\int 6x^{\frac{1}{2}}\,dx=4x^{\frac{3}{2}},\qquad \int 4x^{-2}\,dx=-4x^{-1},\qquad \int 3\,dx=3x$
Write the final answer in a neat form.
$dx=4x32−4x+3x+C\int \left(6\sqrt{x}+\frac{4}{x^2}+3\right)\,dx=4x^{\frac{3}{2}}-\frac{4}{x}+3x+C$

Finding the constant using a point

An indefinite integral gives a family of curves because of the unknown $C$ . If you are told a point lies on the curve, substitute its coordinates to find $C$ .

Several vertically shifted antiderivative curves show how different constants of integration give a family of curves, with one curve selected by a given point.

Example

Finding a function from its derivative

Given that $dydx=6x2−4x+7\frac{dy}{dx}=6x^2-4x+7$ and the curve passes through $(1, 10)$ , find $y$ in terms of $x$ .

Integrate the derivative.
$dxy=\int \left(6x^2-4x+7\right)\,dx$
Write the general solution.
$y=2x^3-2x^2+7x+C$
Substitute the point $(1, 10)$ , so $x = 1$ and $y = 10$ .
$10=2(1)^3-2(1)^2+7(1)+C$
Solve for $C$ .
$10=7+C,C=310=7+C,\qquad C=3$
Give the final expression for the curve.
$y=2x^3-2x^2+7x+3$

Definite integrals

A definite integral has a lower limit and an upper limit. It gives a number, not a function.

If a curve is above the $x$ -axis, the definite integral represents the area under the curve between the two limits.

The definite integral is represented by the area under a curve between the vertical lines x = a and x = b.

Diagram showing area under a curve between x = a and x = b

To evaluate a definite integral:

Find an antiderivative.
Substitute the upper limit.
Subtract the value at the lower limit.

Example

Evaluating a definite integral

Find $dx\int_1^4 \left(6+\frac{2}{\sqrt{x}}\right)\,dx$ .

Rewrite the integrand using powers.
$6+2x=6+2x−126+\frac{2}{\sqrt{x}}=6+2x^{-\frac{1}{2}}$
Integrate to get an antiderivative.
$dx=6x+4x12\int \left(6+2x^{-\frac{1}{2}}\right)\,dx=6x+4x^{\frac{1}{2}}$
Substitute the limits using upper minus lower.
$[6x+4x12]14=(24+8)−(6+4)\left[6x+4x^{\frac{1}{2}}\right]_1^4=(24+8)-(6+4)$
Calculate the final value.
$dx=22\int_1^4 \left(6+\frac{2}{\sqrt{x}}\right)\,dx=22$

Common Mistake

Forgetting upper minus lower

The notation $[F(x)]ab\left[F(x)\right]_a^b$ means $F (b) - F (a)$ , not $F (a) - F (b)$ .

The bracket notation for a definite integral means substitute the upper limit first, then subtract the lower limit.

Areas between curves

For the area between two curves, find the intersection points first. These give the limits of integration.

The intersection points of two curves determine the limits for the area between them.

If one curve is above the other throughout the interval, use:

dx\text{Area}=\int_a^b \left(\text{top curve}-\text{bottom curve}\right)\,dx

Diagram showing area between a parabola and a straight line

Example

Area between a curve and a line

The curve $y=6x-x^2$ and the line $y = 2 x$ enclose a finite region. Find the coordinates of their intersections and the area of the region.

The parabola y = 6x − x² lies above the line y = 2x between their intersections at (0, 0) and (4, 8).

Set the two expressions for $y$ equal to find the intersections.
$6x-x^2=2x$
Rearrange and factorise.
$−x2+4x=0,x(4−x)=0-x^2+4x=0,\qquad x(4-x)=0$
Find the coordinates using $y = 2 x$ .
$\Rightarrow y=0,\qquad x=4 \Rightarrow y=8$
Decide which curve is on top between the limits. At $x = 2$ , the curve gives 8 and the line gives 4, so the curve is above the line.
$dx\text{Area}=\int_0^4 \left((6x-x^2)-2x\right)\,dx$
Simplify and integrate.
$dx=[2x2−13x3]04\int_0^4 \left(4x-x^2\right)\,dx=\left[2x^2-\frac{1}{3}x^3\right]_0^4$
Evaluate to get an area of $323\frac{32}{3}$ square units.
$(32−643)−0=323\left(32-\frac{64}{3}\right)-0=\frac{32}{3}$

Tip

When the curve crosses the x-axis

If you are finding total area between a curve and the $x$ -axis, split the calculation at any roots where the curve crosses the axis. Areas below the axis need to be made positive.

When a curve crosses the x-axis, total area is found by splitting at the roots and treating below-axis regions as positive areas.

Exam technique

In the exam

Rewrite roots and fractions as powers before integrating.
Always include $+ C$ for indefinite integrals, unless you are evaluating between limits.
For area questions, find intersections first and use top curve minus bottom curve.

Self review

Check yourself

Can you explain why $dx=5x+C\int 5\,dx=5x+C$ ?
When should you include the constant of integration?
How do you decide which expression goes first in an area-between-curves integral?

What you'll learn

How integration reverses differentiation.
How to integrate polynomials, brackets, roots and reciprocal powers.
How to find the constant of integration using a point on a curve.
How definite integrals give areas, including areas between curves.

Integration as reverse differentiation

You already know that differentiation finds a gradient function. Integration goes the other way: it starts with a gradient function and works back to the original function.

A flow diagram showing differentiation and integration as reverse processes between a function and its gradient function.

For example, if

ddx(x3)=3x2\frac{d}{dx}(x^3)=3x^2

then integrating $3x^2$ gives $x^3$ , plus possibly a constant.

Definition

Integral and antiderivative

If $F^{'} (x) = f (x)$ , then $F (x)$ is called an antiderivative of $f (x)$ . The symbol $dx\int f(x)\,dx$ means “integrate $f (x)$ with respect to $x$ ”.

The little $d x$ tells you the variable you are integrating with respect to. In AS Pure, this will usually be $x$ .

The power rule for integration

Key Idea

The reverse power rule

To integrate a power of $x$ , increase the power by 1, then divide by the new power.

A visual summary of the reverse power rule for integrating a power of x.

For any constant $a$ ,

dx=an+1xn+1+C,n≠−1\int ax^n\,dx=\frac{a}{n+1}x^{n+1}+C,\qquad n\neq -1

The $C$ is called the constant of integration. It is needed because differentiating any constant gives zero.

Example

Integrating a polynomial

Find $dx\int \left(4x^3-6x^2+5x-9\right)\,dx$ .

Integrate each term separately.
$dx=x4\int 4x^3\,dx=x^4$
Continue term by term, remembering that a constant integrates to a multiple of $x$ .
$dx=−9x\int -6x^2\,dx=-2x^3,\qquad \int 5x\,dx=\frac{5}{2}x^2,\qquad \int -9\,dx=-9x$
Add the constant of integration.
$dx=x4−2x3+52x2−9x+C\int \left(4x^3-6x^2+5x-9\right)\,dx=x^4-2x^3+\frac{5}{2}x^2-9x+C$

Tip

Quick check

Differentiate your answer. If you get back the original expression, your integration is correct.

Expanding before integrating

Sometimes you are asked to integrate an expression in brackets. The power rule works best when the expression is written as a sum of powers of $x$ , so expand first.

Example

Integrating a product of brackets

Find $dx\int (x+5)(x-2)\,dx$ .

Expand the brackets first.
$x+5)(x-2)=x^2+3x-10$
Integrate the expanded expression term by term.
$dx=13x3+32x2−10x+C\int \left(x^2+3x-10\right)\,dx=\frac{1}{3}x^3+\frac{3}{2}x^2-10x+C$

Common Mistake

Do not integrate each bracket separately

There is no rule that lets you integrate $(x + 5) (x - 2)$ by integrating each bracket on its own. Expand first, then integrate.

Roots and fractions as powers

Before integrating roots or fractions involving $x$ , rewrite them using powers.

Useful rewrites include:

$x=x12\sqrt{x}=x^{\frac{1}{2}}$
$1x2=x−2\frac{1}{x^2}=x^{-2}$
$1x=x−12\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$

Common Mistake

The power rule exception

The reverse power rule above does not work for $x^{-1}$ . At AS level, questions in this topic usually avoid needing to integrate $1x\frac{1}{x}$ .

Example

Integrating roots and reciprocal powers

Find $dx\int \left(6\sqrt{x}+\frac{4}{x^2}+3\right)\,dx$ .

Rewrite the expression using powers of $x$ .
$6x+4x2+3=6x12+4x−2+36\sqrt{x}+\frac{4}{x^2}+3=6x^{\frac{1}{2}}+4x^{-2}+3$
Apply the reverse power rule to each term.
$dx=3x\int 6x^{\frac{1}{2}}\,dx=4x^{\frac{3}{2}},\qquad \int 4x^{-2}\,dx=-4x^{-1},\qquad \int 3\,dx=3x$
Write the final answer in a neat form.
$dx=4x32−4x+3x+C\int \left(6\sqrt{x}+\frac{4}{x^2}+3\right)\,dx=4x^{\frac{3}{2}}-\frac{4}{x}+3x+C$

Finding the constant using a point

An indefinite integral gives a family of curves because of the unknown $C$ . If you are told a point lies on the curve, substitute its coordinates to find $C$ .

Several vertically shifted antiderivative curves show how different constants of integration give a family of curves, with one curve selected by a given point.

Example

Finding a function from its derivative

Given that $dydx=6x2−4x+7\frac{dy}{dx}=6x^2-4x+7$ and the curve passes through $(1, 10)$ , find $y$ in terms of $x$ .

Integrate the derivative.
$dxy=\int \left(6x^2-4x+7\right)\,dx$
Write the general solution.
$y=2x^3-2x^2+7x+C$
Substitute the point $(1, 10)$ , so $x = 1$ and $y = 10$ .
$10=2(1)^3-2(1)^2+7(1)+C$
Solve for $C$ .
$10=7+C,C=310=7+C,\qquad C=3$
Give the final expression for the curve.
$y=2x^3-2x^2+7x+3$

Definite integrals

A definite integral has a lower limit and an upper limit. It gives a number, not a function.

If a curve is above the $x$ -axis, the definite integral represents the area under the curve between the two limits.

The definite integral is represented by the area under a curve between the vertical lines x = a and x = b.

Diagram showing area under a curve between x = a and x = b

To evaluate a definite integral:

Find an antiderivative.
Substitute the upper limit.
Subtract the value at the lower limit.

Example

Evaluating a definite integral

Find $dx\int_1^4 \left(6+\frac{2}{\sqrt{x}}\right)\,dx$ .

Rewrite the integrand using powers.
$6+2x=6+2x−126+\frac{2}{\sqrt{x}}=6+2x^{-\frac{1}{2}}$
Integrate to get an antiderivative.
$dx=6x+4x12\int \left(6+2x^{-\frac{1}{2}}\right)\,dx=6x+4x^{\frac{1}{2}}$
Substitute the limits using upper minus lower.
$[6x+4x12]14=(24+8)−(6+4)\left[6x+4x^{\frac{1}{2}}\right]_1^4=(24+8)-(6+4)$
Calculate the final value.
$dx=22\int_1^4 \left(6+\frac{2}{\sqrt{x}}\right)\,dx=22$

Common Mistake

Forgetting upper minus lower

The notation $[F(x)]ab\left[F(x)\right]_a^b$ means $F (b) - F (a)$ , not $F (a) - F (b)$ .

The bracket notation for a definite integral means substitute the upper limit first, then subtract the lower limit.

Areas between curves

For the area between two curves, find the intersection points first. These give the limits of integration.

The intersection points of two curves determine the limits for the area between them.

If one curve is above the other throughout the interval, use:

dx\text{Area}=\int_a^b \left(\text{top curve}-\text{bottom curve}\right)\,dx

Diagram showing area between a parabola and a straight line

Example

Area between a curve and a line

The curve $y=6x-x^2$ and the line $y = 2 x$ enclose a finite region. Find the coordinates of their intersections and the area of the region.

The parabola y = 6x − x² lies above the line y = 2x between their intersections at (0, 0) and (4, 8).

Set the two expressions for $y$ equal to find the intersections.
$6x-x^2=2x$
Rearrange and factorise.
$−x2+4x=0,x(4−x)=0-x^2+4x=0,\qquad x(4-x)=0$
Find the coordinates using $y = 2 x$ .
$\Rightarrow y=0,\qquad x=4 \Rightarrow y=8$
Decide which curve is on top between the limits. At $x = 2$ , the curve gives 8 and the line gives 4, so the curve is above the line.
$dx\text{Area}=\int_0^4 \left((6x-x^2)-2x\right)\,dx$
Simplify and integrate.
$dx=[2x2−13x3]04\int_0^4 \left(4x-x^2\right)\,dx=\left[2x^2-\frac{1}{3}x^3\right]_0^4$
Evaluate to get an area of $323\frac{32}{3}$ square units.
$(32−643)−0=323\left(32-\frac{64}{3}\right)-0=\frac{32}{3}$

Tip

When the curve crosses the x-axis

If you are finding total area between a curve and the $x$ -axis, split the calculation at any roots where the curve crosses the axis. Areas below the axis need to be made positive.

When a curve crosses the x-axis, total area is found by splitting at the roots and treating below-axis regions as positive areas.

Exam technique

In the exam

Rewrite roots and fractions as powers before integrating.
Always include $+ C$ for indefinite integrals, unless you are evaluating between limits.
For area questions, find intersections first and use top curve minus bottom curve.

Self review

Check yourself

Can you explain why $dx=5x+C\int 5\,dx=5x+C$ ?
When should you include the constant of integration?
How do you decide which expression goes first in an area-between-curves integral?

What you'll learn

Integration as reverse differentiation

The power rule for integration

Expanding before integrating

Roots and fractions as powers

Finding the constant using a point

Definite integrals

Areas between curves

Recap questions

Review Integration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Integration

What you'll learn

Integration as reverse differentiation

The power rule for integration

Expanding before integrating

Roots and fractions as powers

Finding the constant using a point

Definite integrals

Areas between curves

Recap questions

Review Integration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Integration

What you'll learn

Integration as reverse differentiation

The power rule for integration

Expanding before integrating

Roots and fractions as powers

Finding the constant using a point

Definite integrals

Areas between curves

Recap questions

Review Integration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Integration Revision Guide

Integration

What you'll learn

Integration as reverse differentiation

The power rule for integration

Expanding before integrating

Roots and fractions as powers

Finding the constant using a point

Definite integrals

Areas between curves

Recap questions

Review Integration by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Integration Revision Guide