Integration as the reverse of differentiation, integrating x^n, finding constants of integration.
You already know how to differentiate — given a curve, you can find its gradient function. Integration is the reverse process. Given the gradient function, you can recover the original curve.
But integration does far more than undo differentiation. It lets you calculate areas, volumes, and solve equations involving rates of change. At A-Level, integration appears in almost every Paper 1, and the skills you build here carry through to Year 13 and beyond.
Think of it this way: differentiation tells you how fast something is changing at each moment. Integration adds up all those tiny changes to find the total. In this lesson we cover the reverse power rule and definite integrals. In the next lesson, we apply these skills to find areas and recover curves from their gradient functions.
If dxd(x4)=4x3, then working backwards:
∫4x3dx=x4+c
The reverse power rule says: raise the power by one, then divide by the new power.
∫xndx=n+1xn+1+c(n=−1)
For a constant multiple:
∫axndx=n+1axn+1+c
Why +c? When you differentiate x4+7, you get 4x3. The same is true for x4+100 or x4−3. The constant vanishes on differentiation, so when you integrate you must include an arbitrary constant c to account for the information that was lost.
Find ∫(6x2+3x−5)dx.
Integrate term by term:
∫6x2dx=36x3=2x3
∫3xdx=23x2
∫−5dx=−5x
Putting it together:
∫(6x2+3x−5)dx=2x3+23x2−5x+c
Common error: Students write 23x2 as 3x2. Always divide the coefficient by the new power.
Find ∫x23dx.
You cannot integrate x23 directly. Rewrite it using a negative index:
x23=3x−2
Now apply the reverse power rule:
∫3x−2dx=−13x−1+c=−3x−1+c=−x3+c
Watch out: ∫x−2dx=ln∣x∣+c. That formula only applies to ∫x−1dx. Here n=−2, so the power rule works normally.
Find ∫10xdx.
Rewrite x=x1/2:
∫10x1/2dx=3/210x3/2+c=10×32×x3/2+c=320x3/2+c
The key step: dividing by 23 is the same as multiplying by 32.
Answer at least 3 of 4 correctly to complete this section.
A definite integral has limits of integration — a lower limit a and an upper limit b:
∫abf(x)dx=[F(x)]ab=F(b)−F(a)
where F(x) is the integral of f(x). Notice that the constant c cancels out, so you do not need it for definite integrals.
Evaluate ∫13(2x+1)dx.
Integrate:
∫(2x+1)dx=x2+x
Apply the limits:
[x2+x]13=(9+3)−(1+1)=12−2=10
Always substitute the upper limit first, then subtract the lower limit. Writing it as F(b)−F(a) avoids sign errors.
Evaluate ∫−12(x2+2)dx.
∫(x2+2)dx=3x3+2x
[3x3+2x]−12=(38+4)−(3−1+(−2))
=38+4+31+2=3+6=9
Be careful with negative numbers. When you substitute x=−1, you get (−1)3=−1, not +1. And subtracting a negative gives a positive.
Given that ∫0k4xdx=18, find the positive value of k.
∫0k4xdx=[2x2]0k=2k2−0=2k2
Set equal to 18:
2k2=18⟹k2=9⟹k=3
(taking the positive value).
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
First principles, differentiating x^n, gradient functions, rates of change.
Evaluating definite integrals, area under a curve, area between curves.
Using substitution to integrate composite functions, reversing the chain rule.
Integration by parts formula, choosing u and dv, repeated application.
Integrating sin, cos, tan, sec² and their compositions. Using trig identities to integrate.
Using partial fraction decomposition to integrate rational functions.
Using differentiation and integration for variable acceleration, displacement-velocity-acceleration relationships.