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Learn9MA0Paper 1 · Pure Mathematics 1IntegrationAreas under and between curves
Integration · Subtopic 3 of 14

Areas Under and Between Curves

Using definite integrals to find areas: under a curve, regions below the axis, between a curve and a line, between two curves, and against the yy-axis.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 50 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Find the area between a curve and the xx-axis with a definite integral.
  • Deal with regions below the xx-axis, where the integral is negative.
  • Find the area between a curve and a line, or two curves, using "top minus bottom".
  • Find an area between a curve and the yy-axis by integrating with respect to yy.
Pause and think

Split the region under y=x2y=x^2 from x=0x=0 to x=1x=1 into thin vertical strips. What is the approximate area of one strip of width δx\delta x at position xx?

Section 2 of 14

Core idea

Start hereWhen the area comes out as zero

Integrate x3−xx^3-x from −1-1 to 11 and you get 00. Yet the graph clearly encloses some area. What went wrong? Areas below the axis count as negative in an integral, so they cancelled out the areas above. Handle the signs and integration measures any region, even between two curves.

The big idea

An integral counts area above the xx-axis as positive and below as negative. For total area, split at the roots and add the sizes. The area between two curves is ∫ab(top−bottom) dx\displaystyle\int_a^b(\text{top}-\text{bottom})\,dx.

1. Areas below the axis

y = x³ − x: equal areas above and below
−1.5−1−0.50.511.5−0.4−0.20.20.40.60xyy = x³ − x+¼−¼
∫−11(x3−x) dx=0\displaystyle\int_{-1}^1(x^3-x)\,dx=0 because the two pieces cancel, but the total shaded area is 14+14=12\tfrac14+\tfrac14=\tfrac12.
Worked exampleTotal area between y=x3−xy=x^3-x and the xx-axis

Find the total area enclosed between x=−1x=-1 and x=1x=1.

Goal
Positive total area.
Strategy
Roots at −1,0,1-1,0,1; integrate each piece separately; add magnitudes.
  1. ∫−10(x3−x) dx=[x44−x22]−10=0−(14−12)=14\int_{-1}^0(x^3-x)\,dx=\left[\tfrac{x^4}4-\tfrac{x^2}2\right]_{-1}^0=0-\left(\tfrac14-\tfrac12\right)=\tfrac14
    Above the axis.
  2. ∫01(x3−x) dx=14−12=−14\int_0^1(x^3-x)\,dx=\tfrac14-\tfrac12=-\tfrac14
    Below: negative.
  3. 14+∣−14∣=12\tfrac14+\left|-\tfrac14\right|=\tfrac12
    Add sizes.
Answer12\frac12.
✓ CheckThe graph has odd symmetry, so the two pieces are equal ✓; one integral over [−1,1][-1,1] gives 00 ✗.

2. Between two curves

Area between y = x + 2 and y = x²
−2−1123123450xytop: y = x + 2bottom: y = x²(−1, 1)(2, 4)
Area =∫−12((x+2)−x2) dx=[x22+2x−x33]−12=4.5=\displaystyle\int_{-1}^{2}\big((x+2)-x^2\big)\,dx=\left[\tfrac{x^2}{2}+2x-\tfrac{x^3}{3}\right]_{-1}^{2}=4.5.
Why top minus bottom
  1. strip height=ytop−ybottom\text{strip height}=y_{\text{top}}-y_{\text{bottom}}
    A vertical strip between the curves.
  2. strip area≈(ytop−ybottom) δx\text{strip area}\approx(y_{\text{top}}-y_{\text{bottom}})\,\delta x
    Thin rectangle.
  3. Area=∫ab(ytop−ybottom) dx\text{Area}=\int_a^b(y_{\text{top}}-y_{\text{bottom}})\,dx
    Sum and take the limit.
This works even if parts lie below the xx-axis: the difference is always positive.
Worked exampleBetween y=x+2y=x+2 and y=x2y=x^2

Find the enclosed area.

Goal
A positive number.
Strategy
Intersections for limits; check which is on top.
  1. x2=x+2⇒(x−2)(x+1)=0⇒x=−1,2x^2=x+2\Rightarrow(x-2)(x+1)=0\Rightarrow x=-1,2
    Limits.

  2. At x=0x=0: line 22, curve 00: the line is on top.

  3. ∫−12(x+2−x2) dx=[x22+2x−x33]−12\int_{-1}^2(x+2-x^2)\,dx=\left[\tfrac{x^2}2+2x-\tfrac{x^3}3\right]_{-1}^2
    Top − bottom.
  4. =(2+4−83)−(12−2+13)=103+76=92=\left(2+4-\tfrac83\right)-\left(\tfrac12-2+\tfrac13\right)=\tfrac{10}3+\tfrac76=\tfrac92
    Evaluate carefully.
Answer92\frac92.
✓ CheckFor a parabola and chord, area =∣a∣(β−α)36=276=92=\frac{|a|(\beta-\alpha)^3}6=\frac{27}6=\frac92 ✓.

3. Against the y-axis

Worked exampleRegion between y=xy=\sqrt x, the yy-axis, y=1y=1 and y=3y=3

Find the area.

Goal
Area.
Strategy
Horizontal strips: ∫x dy\int x\,dy with xx in terms of yy.
  1. x=y2x=y^2
    Rearrange.
  2. ∫13y2 dy=[y33]13=9−13=263\int_1^3y^2\,dy=\left[\tfrac{y^3}3\right]_1^3=9-\tfrac13=\tfrac{26}3
    Integrate.
Answer263\frac{26}3.
✓ CheckPositive, since x≥0x\ge0 ✓.
Common misconceptionIgnoring signs Always sketch first; split at roots when the curve crosses the axis.
Pause and think

Area enclosed by y=4−x2y=4-x^2 and the xx-axis?

Summary

  • Below the axis: negative; split and add sizes.
  • Between curves: top − bottom.
  • Horizontal strips: ∫x dy\int x\,dy.
Key words
Enclosed region
Area bounded on all sides.
HistoryArchimedes and the parabola. Around 250 BC Archimedes proved that a segment of a parabola has exactly 43\tfrac43 the area of a certain triangle inside it. He did it by filling the region with ever smaller triangles and summing a geometric series, nearly two thousand years before calculus.
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