SYSTEM ONLINE · INTEGRATION
Edexcel International A Level Mathematics

Integration
Without the Confusion.

A complete Integration learning system: understand what an integral means, master standard techniques, solve definite-integral and area problems, handle parameters and multi-step questions, recognise traps, and finish with a 100-question progressive exam.

Theory + intuition
Worked examples
Exam methods
Parameter problems
Area & volume
100-question exam
01 · Foundation

The Big Idea

Integration is not a collection of disconnected formulas. It is the reverse process of differentiation and, in definite form, a method for measuring accumulated change.

Core concept

Integration as reverse differentiation

If differentiating \(F(x)\) gives \(f(x)\), then integrating \(f(x)\) gives \(F(x)\), up to a constant.

\[ F'(x)=f(x) \quad\Longrightarrow\quad \int f(x)\,dx=F(x)+C \]

The constant \(C\) is necessary because differentiation removes constants:

\[ \frac{d}{dx}(x^3+7)=3x^2, \qquad \frac{d}{dx}(x^3-100)=3x^2. \]

Therefore the integral of \(3x^2\) cannot identify one unique constant.

Accumulation

What a definite integral measures

A definite integral adds up infinitely many tiny contributions. Geometrically it gives signed area.

\[ \int_a^b f(x)\,dx = \lim_{n\to\infty} \sum f(x_i)\Delta x \]

You normally do not calculate this limit directly in IAL questions. Instead, the Fundamental Theorem of Calculus connects the accumulation to an antiderivative.

Indefinite vs definite

  • Indefinite integral: produces a family of functions and includes \(+C\).
  • Definite integral: produces a number and does not need \(+C\).
  • Definite integration evaluates an antiderivative at two endpoints.

First major trap

Never write \[ \int x^{-1}\,dx=\frac{x^0}{0}. \] The power rule specifically excludes \(n=-1\).

\[ \boxed{\int\frac1x\,dx=\ln|x|+C} \]
02 · Core Rules

The Power Rule

The most important basic integration rule is obtained directly by reversing the differentiation power rule.

\[ \boxed{ \int x^n\,dx= \frac{x^{n+1}}{n+1}+C \qquad n\ne-1 } \]

Why does it work?

Differentiate the proposed answer:

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

Example 1 · Basic

Find:

\[ \int(6x^4-5x^2+3)\,dx \]
Step 1: integrate each term.
\[ 6\cdot\frac{x^5}{5} -5\cdot\frac{x^3}{3} +3x+C \]
\[ \boxed{\frac65x^5-\frac53x^3+3x+C} \]

Example 2 · Negative powers

Find:

\[ \int\left(4x^{-3}-7x^{-2}\right)dx \]
\[ 4\frac{x^{-2}}{-2} -7\frac{x^{-1}}{-1}+C \]
\[ \boxed{-2x^{-2}+7x^{-1}+C} \]

Example 3 · Fractional powers

Find:

\[ \int(3\sqrt{x}+2x^{3/2})dx \]
Rewrite: \[ 3x^{1/2}+2x^{3/2}. \]
\[ 3\frac{x^{3/2}}{3/2} + 2\frac{x^{5/2}}{5/2} +C \]
\[ \boxed{2x^{3/2}+\frac45x^{5/2}+C} \]

Example 4 · Rewrite first

Find:

\[ \int\frac{3x^2+2x-1}{x}\,dx \]
Divide each term by \(x\): \[ \frac{3x^2}{x}+\frac{2x}{x}-\frac1x = 3x+2-\frac1x. \]
Therefore: \[ \int(3x+2-x^{-1})dx = \frac32x^2+2x-\ln|x|+C. \]
\[ \boxed{\frac32x^2+2x-\ln|x|+C} \]
03 · Special Functions

\(\frac1x\), Exponentials & Trigonometry

These functions appear constantly. Memorise the standard forms, but also understand how to recognise scaled or composite versions.

Logarithmic integral

\[ \boxed{\int\frac1x\,dx=\ln|x|+C} \]

The absolute value is important because \[ \frac{d}{dx}\ln|x|=\frac1x \] on every interval where \(x\ne0\).

Scaled form

\[ \int\frac{5}{2x}\,dx = \frac52\ln|x|+C. \]

Exponentials

\[ \boxed{\int e^x\,dx=e^x+C} \] \[ \boxed{\int e^{ax}\,dx=\frac1a e^{ax}+C} \] \[ \boxed{\int a^x\,dx=\frac{a^x}{\ln a}+C} \qquad(a>0,\ a\ne1) \]

Standard trigonometric integrals

\[ \int\sin x\,dx=-\cos x+C \] \[ \int\cos x\,dx=\sin x+C \] \[ \int\sec^2x\,dx=\tan x+C \] \[ \int\cosec^2x\,dx=-\cot x+C \]

More standard forms

\[ \int\sec x\tan x\,dx=\sec x+C \] \[ \int\cosec x\cot x\,dx=-\cosec x+C \]

Example 5 · Exponential with coefficient

\[ \int7e^{3x}\,dx \]
Since the derivative of \(3x\) is \(3\), divide by \(3\): \[ \frac73e^{3x}+C. \]
\[ \boxed{\frac73e^{3x}+C} \]

Example 6 · Trigonometric coefficient

\[ \int4\cos(5x)\,dx \]
The derivative of \(5x\) is \(5\): \[ 4\cdot\frac15\sin(5x)+C. \]
\[ \boxed{\frac45\sin(5x)+C} \]
04 · Technique

Substitution

Substitution reverses the chain rule. The central skill is recognising when part of the integrand is the derivative of another part.

The recognition pattern

\[ \boxed{ \int f(g(x))g'(x)\,dx = \int f(u)\,du } \]

Set \(u=g(x)\), then calculate \(du=g'(x)\,dx\).

Example 7 · Basic substitution

\[ \int2x(x^2+4)^5dx \]
Let \[ u=x^2+4. \]
Then \[ du=2x\,dx. \]
So \[ \int u^5du=\frac{u^6}{6}+C. \]
\[ \boxed{\frac{(x^2+4)^6}{6}+C} \]

Example 8 · Coefficient adjustment

\[ \int x(3x^2+1)^4dx \]
Let \[ u=3x^2+1,\qquad du=6x\,dx. \]
Therefore \[ x\,dx=\frac16du. \]
\[ \frac16\int u^4du = \frac{u^5}{30}+C. \]
\[ \boxed{\frac{(3x^2+1)^5}{30}+C} \]

Example 9 · Exponential recognition

\[ \int(2x+3)e^{x^2+3x}dx \]
Let \[ u=x^2+3x. \]
Then \[ du=(2x+3)dx. \]
Hence \[ \int e^u\,du=e^u+C. \]
\[ \boxed{e^{x^2+3x}+C} \]

Example 10 · Important pattern

\[ \boxed{ \int(2ax+b)e^{ax^2+bx+c}dx = e^{ax^2+bx+c}+C } \]

This works because the derivative of \(ax^2+bx+c\) is exactly \(2ax+b\).

\[ \int(6x-4)e^{3x^2-4x+7}dx = \boxed{e^{3x^2-4x+7}+C}. \]

Example 11 · Logarithmic recognition

\[ \int\frac{2x+1}{x^2+x+7}dx \]
Let \[ u=x^2+x+7. \]
Since \[ du=(2x+1)dx, \] the integral becomes \[ \int\frac1u\,du=\ln|u|+C. \]
\[ \boxed{\ln(x^2+x+7)+C} \] because \(x^2+x+7>0\) for all real \(x\).

Example 12 · When the derivative is only a multiple

\[ \int\frac{5x}{x^2+9}dx \]
Let \(u=x^2+9\), so \(du=2x\,dx\).
Therefore \[ \frac52\int\frac1u\,du. \]
\[ \boxed{\frac52\ln(x^2+9)+C} \]

Substitution decision rule

  1. Look for a bracket/composite expression.
  2. Differentiate that expression.
  3. Check whether its derivative appears elsewhere in the integrand.
  4. If it appears up to a constant multiplier, substitution is usually appropriate.
  5. If no useful derivative appears, do not force substitution.
05 · Definite Integration

Limits, FTC & Properties

Definite integration turns an antiderivative into a numerical result.

\[ \boxed{ \int_a^b f(x)\,dx=F(b)-F(a) } \] where \(F'(x)=f(x)\).

Fundamental Theorem of Calculus

If \[ A(x)=\int_a^x f(t)\,dt, \] then, under the standard continuity conditions, \[ \boxed{A'(x)=f(x)}. \]

Example 13 · Direct definite integral

\[ \int_1^3(2x+1)dx \]
An antiderivative is \[ x^2+x. \]
\[ [x^2+x]_1^3 =(9+3)-(1+1)=10. \]
\[ \boxed{10} \]

Example 14 · Definite substitution

\[ \int_0^1 2x(x^2+1)^3dx \]
Let \(u=x^2+1\). Then \(du=2x\,dx\).
Change the limits: \[ x=0\Rightarrow u=1, \qquad x=1\Rightarrow u=2. \]
\[ \int_1^2u^3du = \left[\frac{u^4}{4}\right]_1^2 = \frac{16-1}{4}. \]
\[ \boxed{\frac{15}{4}} \]

Example 15 · Reverse the limits

\[ \boxed{ \int_b^a f(x)dx = -\int_a^b f(x)dx } \]

Therefore:

\[ \int_5^2 3x\,dx = -\int_2^5 3x\,dx. \]

Example 16 · Splitting an interval

\[ \boxed{ \int_a^c f(x)dx = \int_a^b f(x)dx+ \int_b^c f(x)dx } \]

This is especially useful when a curve changes sign or when a complicated region must be divided into simpler regions.

Linearity

\[ \int(af+bg)dx = a\int fdx+b\int gdx. \]

Zero integral

\[ \int_a^a f(x)dx=0. \]

Constant function

\[ \int_a^b k\,dx=k(b-a). \]
06 · Technique

Integration by Parts

When an integrand is a product and substitution does not simplify it, integration by parts can reverse the product rule.

\[ \boxed{ \int u\,dv=uv-\int v\,du } \]

Choosing \(u\)

A useful practical priority is often: logarithmic → inverse trigonometric → algebraic → trigonometric → exponential. The exact choice depends on which choice makes the remaining integral simpler.

Example 17 · \(\int xe^x dx\)

Choose \[ u=x,\qquad dv=e^xdx. \]
Therefore \[ du=dx,\qquad v=e^x. \]
\[ \int xe^xdx = xe^x-\int e^xdx. \]
\[ \boxed{e^x(x-1)+C} \]

Example 18 · \(\int x\cos xdx\)

Let \[ u=x,\qquad dv=\cos xdx. \]
Then \[ du=dx,\qquad v=\sin x. \]
\[ x\sin x-\int\sin xdx. \]
\[ \boxed{x\sin x+\cos x+C} \]

Example 19 · Repeated by parts

\[ \int x^2e^xdx \]

First application: \[ x^2e^x-2\int xe^xdx. \]

Using Example 17: \[ \int xe^xdx=e^x(x-1). \]

\[ \boxed{e^x(x^2-2x+2)+C} \]

Example 20 · Definite by parts

\[ \int_0^1 xe^xdx \]
From Example 17: \[ \int xe^xdx=e^x(x-1). \]
Therefore: \[ [e^x(x-1)]_0^1 = 0-(-1)=1. \]
\[ \boxed{1} \]
07 · Rational Functions

Partial Fractions & Rational Integration

For rational expressions, first simplify or decompose the expression before choosing an integration technique.

Distinct linear factors

\[ \frac{1}{(x+1)(x+2)} = \frac{A}{x+1}+\frac{B}{x+2}. \]

Multiply through by \((x+1)(x+2)\): \[ 1=A(x+2)+B(x+1). \]

Setting \(x=-1\): \[ A=1. \]

Setting \(x=-2\): \[ B=-1. \]

\[ \boxed{ \frac1{x+1}-\frac1{x+2} } \]

Integrate after decomposition

\[ \int\frac{1}{(x+1)(x+2)}dx \]
\[ = \int\left( \frac1{x+1}-\frac1{x+2} \right)dx. \]
\[ \boxed{ \ln|x+1|-\ln|x+2|+C } \]

Repeated linear factor

\[ \frac{1}{x(x+1)^2} = \frac Ax+\frac B{x+1}+\frac C{(x+1)^2}. \]

Every power of a repeated factor requires its own partial-fraction term.

Quadratic denominator

If a quadratic cannot be factored over the reals, completing the square may reveal a standard arctangent form.

\[ x^2+4x+13 = (x+2)^2+9. \]

Hence \[ \int\frac{dx}{(x+2)^2+3^2} = \boxed{\frac13 \tan^{-1}\left(\frac{x+2}{3}\right)+C}. \]

08 · Geometry

Area Under & Between Curves

Integration gives signed area. Exam questions often require you to convert that signed result into geometric area.

Area under a curve above the x-axis

\[ \boxed{ A=\int_a^b f(x)dx } \]

This is directly the geometric area only when \(f(x)\ge0\) throughout the interval.

Below the x-axis

If \(f(x)<0\), the definite integral is negative but geometric area is positive.

\[ \boxed{ \text{Geometric area} = \int_a^b|f(x)|dx } \]

Area between two curves

\[ \boxed{ A=\int_a^b (\text{upper curve}-\text{lower curve})dx } \]

Never assume which curve is upper. Test a point in the interval or compare the functions algebraically.

Finding the limits

If two curves form the boundaries, solve:

\[ f(x)=g(x). \]

Those intersection values normally become the limits.

Example 21 · Area under a parabola

\[ A=\int_0^2(4-x^2)dx. \]
\[ \left[4x-\frac{x^3}{3}\right]_0^2 = 8-\frac83 = \frac{16}{3}. \]
\[ \boxed{\frac{16}{3}} \]

Example 22 · Area below the axis

\[ y=x-2,\qquad 0\le x\le2. \]

The line is below the x-axis. Its geometric area is therefore \[ -\int_0^2(x-2)dx. \]

\[ \boxed{2} \]

Example 23 · Between two curves

\[ y=4-x^2,\qquad y=0. \]

Intersections with the x-axis: \[ 4-x^2=0 \Rightarrow x=\pm2. \]

\[ A=\int_{-2}^{2}(4-x^2)dx. \]
\[ \boxed{\frac{32}{3}} \]

Example 24 · Two curves

\[ y=x+2,\qquad y=x^2. \]

Intersections: \[ x+2=x^2 \Rightarrow x^2-x-2=0 \] so \[ x=-1,\ 2. \]

At \(x=0\), \(x+2=2\) while \(x^2=0\), so the line is above.

\[ A=\int_{-1}^{2}(x+2-x^2)dx. \]
\[ \boxed{\frac92} \]

When curves cross inside the region

If \(f(x)-g(x)\) changes sign inside the interval, a single \(\int(f-g)dx\) can produce cancellation.

For geometric area, split the integral at every crossing:

\[ A= \int_a^c|f-g|dx+ \int_c^b|f-g|dx. \]
09 · Applications

Volumes of Revolution

Where required by the specification/course route, integration can be used to calculate volumes generated by rotating regions.

Rotation about the x-axis

\[ \boxed{ V=\pi\int_a^b y^2dx } \]

Each thin slice generates a disc with radius \(y\), so its area is \(\pi y^2\).

Example 25

\[ y=x,\qquad0\le x\le2. \]
\[ V=\pi\int_0^2x^2dx = \pi\left[\frac{x^3}{3}\right]_0^2 = \frac{8\pi}{3}. \]
\[ \boxed{\frac{8\pi}{3}} \]

Rotation of a region

If the region is bounded by two curves, determine the correct radius first. Depending on the geometry, the volume may involve the difference of squared radii.

Units

If \(x\) and \(y\) are measured in metres, an integral for volume must end in cubic metres: \[ \boxed{\text{m}^3}. \]

10 · Advanced Reasoning

Parameters & Unknown Constants

Integration questions become significantly harder when the constant or parameter is hidden inside a condition.

Example 26 · Find \(C\)

\[ \frac{dy}{dx}=6x^2-4x,\qquad y=5\text{ when }x=1. \]

Integrate: \[ y=2x^3-2x^2+C. \]

Use \(x=1,y=5\): \[ 5=2-2+C. \]

\[ \boxed{C=5} \]

Example 27 · Two unknown constants

\[ y''=12x+6. \]

First integration: \[ y'=6x^2+6x+A. \]

Second integration: \[ y=2x^3+3x^2+Ax+B. \]

There must be two independent constants because the equation was integrated twice.

Example 28 · Parameter from an area

\[ y=x^2+k,\qquad0\le x\le2. \]

Suppose the area is \(12\): \[ \int_0^2(x^2+k)dx=12. \]

Therefore \[ \frac83+2k=12. \]

\[ \boxed{k=\frac{14}{3}} \]

Example 29 · Parameter + intersection

If a curve \[ y=x^2-kx \] passes through \((2,4)\), substitute both coordinates:

\[ 4=4-2k \Rightarrow \boxed{k=0}. \]

Always apply the given condition exactly as stated.

Parameter checklist

  1. Write the general mathematical expression.
  2. Apply every given condition.
  3. Solve simultaneously if necessary.
  4. Check whether the parameter gives real intersections.
  5. Check whether it satisfies the stated geometric region.
  6. Reject values that contradict the question.
11 · Applications

Accumulation, Motion & Reconstruction

Integration becomes especially powerful when the integral represents a physical quantity or reconstructs a function.

Velocity → displacement

\[ \boxed{ s(b)-s(a)=\int_a^b v(t)dt } \]

This gives displacement, not automatically total distance.

Speed vs velocity

If velocity changes sign, displacement can involve cancellation. Total distance requires the magnitude of velocity:

\[ \boxed{ \text{distance}=\int_a^b|v(t)|dt } \]

Example 30 · Displacement

\[ v(t)=3t^2-4t+2. \]

From \(t=1\) to \(t=4\):

\[ \int_1^4(3t^2-4t+2)dt = [t^3-2t^2+2t]_1^4. \]

At \(t=4\): \[ 64-32+8=40. \] At \(t=1\): \[ 1-2+2=1. \]

\[ \boxed{39\text{ m}} \]

Example 31 · Accumulation function

\[ A(x)=\int_2^x(3t^2+1)dt. \]

By the Fundamental Theorem: \[ \boxed{A'(x)=3x^2+1}. \]

Also: \[ A(2)=0. \]

Example 32 · Solving an accumulation condition

\[ \int_0^x2t\,dt=18. \]

Integrate: \[ [t^2]_0^x=x^2. \]

Therefore: \[ x^2=18. \]

\[ \boxed{x=\pm3\sqrt2} \] and if the context requires \(x\ge0\), then \[ \boxed{x=3\sqrt2}. \]

Example 33 · Variable upper limit

\[ F(x)=\int_1^x(t^3-2t)dt. \]

Directly: \[ \boxed{F'(x)=x^3-2x}. \]

Notice that the dummy variable \(t\) disappears after differentiation.

Example 34 · Find the original function

\[ f'(x)=4x^3-6x,\qquad f(2)=10. \]

Integrate: \[ f(x)=x^4-3x^2+C. \]

Apply \(f(2)=10\): \[ 16-12+C=10. \]

\[ \boxed{C=6} \] so \[ \boxed{f(x)=x^4-3x^2+6}. \]

Example 35 · Integrating a gradient

\[ \frac{dy}{dx}=2x+\frac3x. \]
\[ y=x^2+3\ln|x|+C. \]

The \(\ln|x|\) term is essential because the derivative of \(\ln|x|\) is \(1/x\).

12 · Exam Intelligence

Choosing the Right Technique

A strong integration student does not merely know methods; they know when to use them.

Look at the integrand First method to consider Reason
Simple powers Power rule Direct reversal of differentiation
\(1/x\) Logarithm Power rule does not apply
Composite function × derivative of inside Substitution Reverse chain rule
Product such as \(xe^x\) Integration by parts Reverse product rule
Rational function with factorable denominator Partial fractions Split into simpler fractions
Quadratic denominator Complete square / substitution Expose standard form
Area problem Sketch + intersections + integral Geometry determines limits/sign

Trap 1 · Forgetting \(+C\)

Add \(+C\) to indefinite integrals. Do not add it to a final numerical definite integral.

Trap 2 · Wrong sign

\[ \int\sin xdx=-\cos x+C, \] not \(+\cos x+C\).

Trap 3 · Missing inner derivative

\[ \int e^{3x}dx\ne e^{3x}+C. \] The correct answer is \[ \frac13e^{3x}+C. \]

Trap 4 · Area ≠ signed integral

If the graph lies below the axis, the integral is negative but the geometric area is positive.

Trap 5 · Wrong upper/lower curve

Always determine which curve is above before writing \[ \int(\text{upper}-\text{lower})dx. \]

Trap 6 · Not splitting at crossings

If two curves cross within the region, one unsplit signed integral can cancel part of the geometric area.

Mark-scheme-style workflow

  1. Identify the required quantity.
  2. Sketch if geometry is involved.
  3. Find intersections before choosing limits.
  4. Select the method deliberately.
  5. Show substitutions rather than jumping to the result.
  6. Apply limits carefully.
  7. Check sign and units.
  8. Differentiate your indefinite answer whenever practical.
13 · Formula Atlas

Complete Integration Formula Sheet

Use this as a final revision reference, not as a replacement for understanding the methods.

Core algebra

\[ \int x^n dx= \frac{x^{n+1}}{n+1}+C,\quad n\ne-1 \] \[ \int\frac1x dx=\ln|x|+C \] \[ \int(af+bg)dx = a\int fdx+b\int gdx \]

Exponentials

\[ \int e^x dx=e^x+C \] \[ \int e^{ax}dx=\frac1a e^{ax}+C \] \[ \int a^x dx=\frac{a^x}{\ln a}+C \]

Trigonometry

\[ \int\sin xdx=-\cos x+C \] \[ \int\cos xdx=\sin x+C \] \[ \int\sec^2xdx=\tan x+C \] \[ \int\cosec^2xdx=-\cot x+C \] \[ \int\sec x\tan xdx=\sec x+C \] \[ \int\cosec x\cot xdx=-\cosec x+C \]

Substitution

\[ u=g(x),\qquad du=g'(x)dx \] \[ \int f(g(x))g'(x)dx = \int f(u)du \]

Definite integration

\[ \int_a^b f(x)dx=F(b)-F(a) \] \[ \int_a^a f(x)dx=0 \] \[ \int_a^b fdx=-\int_b^a fdx \] \[ \int_a^c fdx= \int_a^b fdx+\int_b^c fdx \]

Integration by parts

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

Area

\[ A=\int_a^bf(x)dx \] \[ A=\int_a^b|f(x)|dx \] \[ A=\int_a^b(\text{upper}-\text{lower})dx \]

Volume

\[ V=\pi\int_a^by^2dx \] for rotation about the x-axis.

Final verification principle

If you claim \[ \int f(x)dx=F(x)+C, \] differentiate the result:

\[ \boxed{F'(x)\stackrel{?}{=}f(x)} \]

If the derivative does not return the original integrand, the answer is wrong.

14 · Final Assessment

100-Question Integration Exam

One question at a time. No skipping. Questions progress from foundation to multi-step reasoning. Your score is calculated at the end.

INTEGRATION // 100 QUESTIONS
Foundation → Standard → Application → Hard → Challenge
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