Launch price £9.99 until 31 Jan, then £14.99

TMUA Integration and Area

Integration reverses differentiation, and a definite integral gives signed area. The TMUA asks about area, so watch for regions below the axis.

Paper 1 and Paper 2 topic. Specification codes: MM7.1, MM7.2, MM7.3, MM7.4, MM7.6. Last checked 12 November 2026.

Rules

∫xn dx=xn+1n+1+C\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C for n≠−1n\ne-1, and you integrate sums term by term. A definite integral ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx=F(b)-F(a) gives the signed area.

Area below the axis

Area below the xx-axis counts as negative in an integral. To find a total area, split at each root and add the positive sizes.

Area between curves

Find where the curves meet, then integrate (upper curve minus lower curve) between those xx-values.

Checks and shortcuts

Differentiate your answer to check an integral. For areas, sketch first and mark where the curve crosses the axis and which curve is on top.

Over a range symmetric about zero, an odd function such as x3x^3 integrates to zero. The trapezium rule overestimates the area under a curve that bends upwards and underestimates it for a curve that bends downwards.

Rewrite before integrating: x+1x=x1/2+x−1/2\dfrac{x+1}{\sqrt x}=x^{1/2}+x^{-1/2}. For dy/dx=f(x)dy/dx=f(x) with a given point on the curve, integrate and then use the point to find the constant CC.

Worked examples

Example 1

Evaluate ∫143x dx\int_1^4 3\sqrt x\,dx.

Show the worked solution

∫3x1/2 dx=2x3/2\int3x^{1/2}\,dx=2x^{3/2}. So the value is 2(8)−2(1)=142(8)-2(1)=14.

Answer: 14

Example 2

Find the area enclosed between y=x2y=x^2 and y=x+2y=x+2.

Show the worked solution

They meet where x2=x+2x^2=x+2, i.e. x=−1x=-1 and x=2x=2. The area is ∫−12(x+2−x2) dx=[x22+2x−x33]−12=103+76=92\int_{-1}^{2}(x+2-x^2)\,dx=\left[\tfrac{x^2}2+2x-\tfrac{x^3}3\right]_{-1}^{2}=\tfrac{10}3+\tfrac76=\tfrac92.

Answer: 92\tfrac92

Example 3

Find the total area between y=x2−4x+3y=x^2-4x+3 and the xx-axis for 0≤x≤30\le x\le3.

Show the worked solution

The roots are 1 and 3. ∫01=43\int_0^1=\tfrac43 (above the axis) and ∫13=−43\int_1^3=-\tfrac43 (below). The total area is 43+43=83\tfrac43+\tfrac43=\tfrac83, though the integral over [0,3][0,3] is 00.

Answer: 83\tfrac83

Common mistakes

  • Giving the integral when the question asks for area, and so letting positive and negative parts cancel.
  • Forgetting to divide by the new power when integrating.
  • Subtracting the curves in the wrong order and getting a negative area.
  • Differentiating by mistake.

Test yourself on this topic with our free practice questions.

Quick answers

How do you find the area between two curves?

Find their intersection points, then integrate the upper curve minus the lower curve between those xx-values.

Why can a definite integral be zero when there is area?

Parts of the curve above and below the axis give positive and negative contributions that can cancel.

What is the integral of x to the n?

xn+1n+1+C\dfrac{x^{n+1}}{n+1}+C, provided n≠−1n\ne-1.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

Related topics

Written by Proven Papers, an independent practice site, from the public UAT-UK content specification. Not affiliated with UAT-UK, Pearson VUE or any university.