Areas Under and Between Curves – OCR FSMQ Additional Maths

1 The Area Under a Curve

CA12: find the area between a curve, two ordinates and the $x$-axis. An ordinate is a vertical line at a given $x$-value — the two ordinates are the left and right edges of the region.

The area formula

$\text{Area} = \displaystyle\int_a^b y \,\mathrm{d}x$ (provided the curve is above the axis)

Worked Example 1 — A straightforward area

Find the area between $y = x^2 + 1$, the $x$-axis, and the ordinates $x = 1$ and $x = 3$.

①The curve is above the axis throughout (since $x^2 + 1 \geqslant 1$), so a single integral works.

②$\displaystyle\int_1^3 \left(x^2+1\right)\mathrm{d}x = \Bigg[\dfrac{x^3}{3} + x\Bigg]_1^3$

③At $x=3$: $9 + 3 = 12$

④At $x=1$: $\dfrac13 + 1 = \dfrac43$

⑤Area $= 12 - \dfrac43 = \dfrac{32}{3} = 10\tfrac23$ square units.

Sense check with a trapezium. The curve runs from $(1,2)$ to $(3,10)$ over a width of $2$, so a straight line would give $\tfrac12(2+10)(2) = 12$. The curve sags below that chord, so an answer a little under $12$ is right ✓

Always state the units as "square units" (or cm², m², as appropriate). An area is not a bare number.

2 When the Curve Goes Below the Axis

The crucial distinction

An integral over a region below the axis comes out negative.
The area is the size of that number, ignoring the sign.

Never integrate straight across a root. If the curve crosses the $x$-axis inside your limits, the positive and negative parts will partly cancel, and the answer will be neither the area nor anything useful.

Find where the curve crosses the $x$-axis by solving $y = 0$.
Split the integral at each crossing that lies inside the limits.
Evaluate each piece separately.
Take the size of each result, then add them.

Worked Example 2 — Splitting at a root

Find the total area enclosed between $y = x^2 - 4$, the $x$-axis, and the ordinates $x = 0$ and $x = 3$.

①Roots: $x^2 - 4 = 0$ gives $x = \pm 2$. Only $x = 2$ is inside $[0, 3]$.

②So split at $x = 2$. Between $0$ and $2$ the curve is below the axis; between $2$ and $3$ it is above.

③$\displaystyle\int_0^2 \left(x^2-4\right)\mathrm{d}x = \Bigg[\dfrac{x^3}{3} - 4x\Bigg]_0^2 = \dfrac83 - 8 = -\dfrac{16}{3}$

④$\displaystyle\int_2^3 \left(x^2-4\right)\mathrm{d}x = \left(9 - 12\right) - \left(\dfrac83 - 8\right) = -3 + \dfrac{16}{3} = \dfrac{7}{3}$

⑤Total area $= \dfrac{16}{3} + \dfrac{7}{3} = \dfrac{23}{3} = 7\tfrac23$ square units.

⑥Compare: integrating straight across from $0$ to $3$ would give $-\tfrac{16}{3} + \tfrac{7}{3} = -3$ — the two parts partly cancelling. That is not the area.

Sketch the curve first, every time. It is the only reliable way to see whether a split is needed.

3 The Area Between Two Curves (CA13)

CA13 — the formula

$\text{Area} = \displaystyle\int_a^b \Big(y_{\text{upper}} - y_{\text{lower}}\Big)\,\mathrm{d}x$
Upper curve minus lower curve, integrated between the crossing points.

Find the limits by solving the two equations simultaneously.
Decide which curve is on top — test an $x$-value between the limits.
Integrate (upper $-$ lower).
The answer will be positive; if it is negative, you had them the wrong way round.

Why "upper minus lower" handles negatives automatically. The vertical gap between the curves is positive wherever the upper one is above, whether or not either curve is below the $x$-axis. No splitting is needed — provided the curves do not cross inside the interval.

Worked Example 3 — A line and a parabola

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

①Crossing points: $x^2 = x + 2$, so $x^2 - x - 2 = 0$.

②$(x-2)(x+1) = 0$, giving $x = -1$ and $x = 2$. These are the limits.

③At $x = 0$: the line gives $2$, the parabola gives $0$. The line is on top.

④$\text{Area} = \displaystyle\int_{-1}^{2}\Big[(x+2) - x^2\Big]\mathrm{d}x$

⑤$= \Bigg[\dfrac{x^2}{2} + 2x - \dfrac{x^3}{3}\Bigg]_{-1}^{2}$

⑥At $x=2$: $2 + 4 - \dfrac83 = 6 - \dfrac83 = \dfrac{10}{3}$

⑦At $x=-1$: $\dfrac12 - 2 + \dfrac13 = -\dfrac{7}{6}$

⑧Area $= \dfrac{10}{3} - \left(-\dfrac76\right) = \dfrac{20}{6} + \dfrac{7}{6} = \dfrac{27}{6} = \dfrac{9}{2} = 4.5$ square units.

Combine before integrating. Simplifying $(x+2) - x^2$ into one expression first is much safer than working out two separate integrals and subtracting.

Worked Example 4 — Two parabolas

Find the area enclosed between $y = 6 - x^2$ and $y = x^2 - 2$.

①Crossings: $6 - x^2 = x^2 - 2$, so $2x^2 = 8$ and $x = \pm 2$.

②At $x=0$: $6 - 0 = 6$ and $0 - 2 = -2$. The first curve is on top.

③Difference: $\left(6-x^2\right) - \left(x^2-2\right) = 8 - 2x^2$

④$\displaystyle\int_{-2}^{2}\left(8 - 2x^2\right)\mathrm{d}x = \Bigg[8x - \dfrac{2x^3}{3}\Bigg]_{-2}^{2}$

⑤At $x=2$: $16 - \dfrac{16}{3} = \dfrac{32}{3}$

⑥At $x=-2$: $-16 + \dfrac{16}{3} = -\dfrac{32}{3}$

⑦Area $= \dfrac{32}{3} + \dfrac{32}{3} = \dfrac{64}{3} = 21\tfrac13$ square units.

Use symmetry when you can. The region is symmetric about the $y$-axis, so the area is $2\displaystyle\int_0^2\left(8-2x^2\right)\mathrm{d}x = 2 \times \tfrac{32}{3}$ — the same answer with half the arithmetic.

4 Quick Reference

Area under

$\displaystyle\int_a^b y\,\mathrm{d}x$, if above the axis.

Ordinate

A vertical boundary line $x = a$.

Below the axis

The integral is negative; the area is its size.

Crossing inside

Split the integral at each root.

Never cancel

Do not integrate across a root in one go.

Between curves

$\displaystyle\int_a^b (\text{upper} - \text{lower})\,\mathrm{d}x$.

Find the limits

Solve the two equations simultaneously.

Which is upper?

Test an $x$-value in between.

Simplify first

Combine into one expression, then integrate.

Units

Square units, always.

5 Practice Questions

Question 1

Find the area under $y = 2x$ between $x = 0$ and $x = 4$.

▶ Show solution

$\Big[x^2\Big]_0^4 = 16$ square units.

Check: the region is a triangle with base $4$ and height $8$, area $\tfrac12(4)(8) = 16$ ✓

Question 2

Find the area under $y = x^2$ between $x = 0$ and $x = 3$.

▶ Show solution

$\Bigg[\dfrac{x^3}{3}\Bigg]_0^3 = 9$ square units.

Question 3

Find the area under $y = x^2 + 2x$ between $x = 1$ and $x = 2$.

▶ Show solution

$\Bigg[\dfrac{x^3}{3} + x^2\Bigg]_1^2$

At $x=2$: $\dfrac83 + 4 = \dfrac{20}{3}$. At $x=1$: $\dfrac13 + 1 = \dfrac43$.

Area $= \dfrac{16}{3} = 5\tfrac13$ square units.

Question 4

Find the area enclosed between $y = 4 - x^2$ and the $x$-axis.

▶ Show solution

Roots at $x = \pm 2$, and the curve is above the axis between them.

$\displaystyle\int_{-2}^{2}\left(4-x^2\right)\mathrm{d}x = \Bigg[4x - \dfrac{x^3}{3}\Bigg]_{-2}^{2}$

At $x=2$: $8 - \dfrac83 = \dfrac{16}{3}$. At $x=-2$: $-\dfrac{16}{3}$.

Area $= \dfrac{32}{3} = 10\tfrac23$ square units.

Question 5

Find the area between $y = x^2 - 1$, the $x$-axis and the ordinates $x = -1$ and $x = 1$.

▶ Show solution

Between $-1$ and $1$ the curve is entirely below the axis (at $x=0$, $y = -1$).

$\displaystyle\int_{-1}^{1}\left(x^2-1\right)\mathrm{d}x = \Bigg[\dfrac{x^3}{3} - x\Bigg]_{-1}^{1}$

At $x=1$: $\dfrac13 - 1 = -\dfrac23$. At $x=-1$: $-\dfrac13 + 1 = \dfrac23$.

Integral $= -\dfrac23 - \dfrac23 = -\dfrac43$.

Area $= \dfrac{4}{3} = 1\tfrac13$ square units (taking the size).

Question 6

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

▶ Show solution

Crossings: $x = x^2$, so $x(x-1) = 0$ and $x = 0$ or $1$.

At $x = 0.5$: the line gives $0.5$, the parabola $0.25$ — the line is on top.

$\displaystyle\int_0^1 \left(x - x^2\right)\mathrm{d}x = \Bigg[\dfrac{x^2}{2} - \dfrac{x^3}{3}\Bigg]_0^1 = \dfrac12 - \dfrac13 = \dfrac16$

Area $= \dfrac{1}{6}$ square units.

Question 7

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

▶ Show solution

Crossings: $x^2 = 2x$, so $x = 0$ or $2$.

At $x=1$: the line gives $2$, the parabola $1$ — the line is on top.

$\displaystyle\int_0^2 \left(2x - x^2\right)\mathrm{d}x = \Bigg[x^2 - \dfrac{x^3}{3}\Bigg]_0^2 = 4 - \dfrac83 = \dfrac43$

Area $= \dfrac{4}{3}$ square units.

Question 8

Find the total area between $y = x^3$, the $x$-axis and the ordinates $x = -1$ and $x = 2$.

▶ Show solution

The curve crosses the axis at $x = 0$, which lies inside the limits — so split there.

$\displaystyle\int_{-1}^{0} x^3\,\mathrm{d}x = \Bigg[\dfrac{x^4}{4}\Bigg]_{-1}^{0} = 0 - \dfrac14 = -\dfrac14$

$\displaystyle\int_0^2 x^3\,\mathrm{d}x = \dfrac{16}{4} = 4$

Total area $= \dfrac14 + 4 = \dfrac{17}{4} = 4.25$ square units.

(Integrating straight through would give $4 - \tfrac14 = 3.75$, which is not the area.)

Question 9

Find the area enclosed between $y = 9 - x^2$ and $y = x^2 + 1$.

▶ Show solution

Crossings: $9 - x^2 = x^2 + 1$, so $2x^2 = 8$ and $x = \pm 2$.

At $x=0$: $9$ versus $1$ — the first curve is on top.

Difference: $\left(9-x^2\right) - \left(x^2+1\right) = 8 - 2x^2$

By symmetry, $2\displaystyle\int_0^2 \left(8-2x^2\right)\mathrm{d}x = 2\Bigg[8x - \dfrac{2x^3}{3}\Bigg]_0^2$

$= 2\left(16 - \dfrac{16}{3}\right) = 2 \times \dfrac{32}{3} = \dfrac{64}{3}$

Area $= \dfrac{64}{3} = 21\tfrac13$ square units.

Question 10

The curve $C$ has equation $y = x^3 - 4x$.

(a) Find where $C$ crosses the $x$-axis. (b) Evaluate $\displaystyle\int_{-2}^{2}\left(x^3-4x\right)\mathrm{d}x$ and explain why the answer does not give the area. (c) Find the total area enclosed between $C$ and the $x$-axis. (d) Find the area enclosed between $C$ and the line $y = 0$ for $0 \leqslant x \leqslant 3$.

▶ Show solution

(a) $x^3 - 4x = x\left(x^2-4\right) = x(x-2)(x+2)$

Crossings at $x = -2$, $0$ and $2$.

(b) $\Bigg[\dfrac{x^4}{4} - 2x^2\Bigg]_{-2}^{2}$

At $x=2$: $4 - 8 = -4$. At $x=-2$: $4 - 8 = -4$.

Integral $= -4 - (-4) = 0$.

The answer is zero because $C$ is a curve with odd symmetry about the origin: the loop above the axis on $[-2,0]$ has exactly the same size as the loop below it on $[0,2]$, and the two signed contributions cancel completely. An area of zero is clearly wrong, which is why the integral must be split at $x = 0$.

(c) On $[-2, 0]$ the curve is above the axis (at $x=-1$, $y = -1+4 = 3 > 0$):

$\displaystyle\int_{-2}^{0} = 0 - (-4) = 4$

On $[0, 2]$ it is below (at $x=1$, $y = 1-4 = -3 < 0$):

$\displaystyle\int_0^2 = -4 - 0 = -4$, so this area is $4$.

Total area $= 4 + 4 = \mathbf{8}$ square units.

(d) On $[0,3]$ the curve crosses at $x = 2$, so split there.

$[0,2]$: area $4$, from part (c).

$[2,3]$: $\Bigg[\dfrac{x^4}{4} - 2x^2\Bigg]_2^3 = \left(\dfrac{81}{4} - 18\right) - (-4) = \dfrac{9}{4} + 4 = \dfrac{25}{4}$

This piece is above the axis, so its area is $\dfrac{25}{4} = 6.25$.

Total $= 4 + 6.25 = \mathbf{10.25}$ square units.

Areas Under and Between Curves (CA12–CA13) · OCR FSMQ Additional Maths · Created with MathJax