Areas and Integration as a Limit – A Level Maths Year 2

1 Integration as the Limit of a Sum

In Year 1 a definite integral was defined as "the area under the curve" (see Definite Integrals and Areas). Year 2 explains why the integral gives the area.

Split the region under $y = \mathrm{f}(x)$ from $x = a$ to $x = b$ into thin vertical strips of width $\delta x$. Each strip is roughly a rectangle of area $y\,\delta x$. Adding them gives an approximation to the area, and the approximation gets better as the strips get thinner. In the limit, the sum becomes the integral.

The definite integral as a limit

$\displaystyle\int_a^b y\,dx = \lim_{\delta x \to 0}\sum_{x=a}^{b} y\,\delta x$

The notation is deliberately similar: the $\displaystyle\int$ sign is an elongated S for "sum", and $dx$ is what $\delta x$ becomes in the limit.

Five strips under $y = x^2$ for $0 \leqslant x \leqslant 1$. Dark rectangles (left heights) underestimate the area; the light rectangles (right heights) overestimate it.

Worked Example 1 — Upper and lower sums for $y = x^2$

Use five strips to find lower and upper bounds for $\displaystyle\int_0^1 x^2\,dx$, and explain what happens as the number of strips increases.

①Width $\delta x = 0.2$. Left heights: $0, 0.04, 0.16, 0.36, 0.64$; right heights: $0.04, 0.16, 0.36, 0.64, 1$.

②Lower sum $= 0.2(0 + 0.04 + 0.16 + 0.36 + 0.64) = 0.24$.

③Upper sum $= 0.2(0.04 + 0.16 + 0.36 + 0.64 + 1) = 0.44$.

④With $10$ strips the bounds are $0.285$ and $0.385$; with $100$ strips, $0.32835$ and $0.33835$.

⑤Both sums are squeezed towards $\displaystyle\int_0^1 x^2\,dx = \tfrac{1}{3}$ as $\delta x \to 0$.

With $n$ strips the two sums differ by exactly $\tfrac{1}{n}$ here (the last right-hand strip, height $1$, width $\tfrac{1}{n}$), so the gap closes as $n \to \infty$.

Worked Example 2 — Recognising a limit as an integral

Write $\displaystyle\lim_{\delta x \to 0}\sum_{x=1}^{2} 3x^2\,\delta x$ as an integral and evaluate it.

①The sum runs from $x = 1$ to $x = 2$ with $y = 3x^2$.

②$= \displaystyle\int_1^2 3x^2\,dx = \Big[x^3\Big]_1^2 = 8 - 1 = 7$

Link to numerical methods: if you replace each rectangle by a trapezium you get the trapezium rule, which is used when the integral cannot be found exactly.

2 Areas Between Two Curves

Area between curves

$\displaystyle\text{Area} = \int_a^b\left(y_{\text{top}} - y_{\text{bottom}}\right)dx$, where $a$ and $b$ are usually the intersections

Each thin strip now has height $y_{\text{top}} - y_{\text{bottom}}$. This difference is positive whichever side of the $x$-axis the curves are on, so no splitting is needed as long as the same curve stays on top throughout.

Worked Example 3 — Two parabolas

Find the area of the region $R$ enclosed by $y = x^2$ and $y = 4x - x^2$.

①Intersections: $x^2 = 4x - x^2 \Rightarrow 2x^2 - 4x = 0 \Rightarrow x = 0$ or $x = 2$.

②At $x = 1$: $4x - x^2 = 3 > 1 = x^2$, so $y = 4x - x^2$ is on top.

③$\displaystyle\int_0^2\left[(4x - x^2) - x^2\right]dx = \int_0^2(4x - 2x^2)\,dx$

④$= \Big[2x^2 - \tfrac{2}{3}x^3\Big]_0^2 = 8 - \tfrac{16}{3} = \tfrac{8}{3}$

Worked Example 4 — A curve and a line, with logs

Find the exact area enclosed by $y = \dfrac{1}{x}$ and the line $x + y = \tfrac{5}{2}$.

①$x + \dfrac{1}{x} = \tfrac{5}{2} \Rightarrow 2x^2 - 5x + 2 = 0 \Rightarrow (2x - 1)(x - 2) = 0$, so $x = \tfrac{1}{2}$ or $2$.

②At $x = 1$ the line gives $1.5$ and the curve $1$, so the line is on top.

③$\displaystyle\int_{1/2}^2\left(\tfrac{5}{2} - x - \frac{1}{x}\right)dx = \Big[\tfrac{5}{2}x - \tfrac{1}{2}x^2 - \ln x\Big]_{1/2}^2$

④$= (5 - 2 - \ln 2) - \left(\tfrac{5}{4} - \tfrac{1}{8} + \ln 2\right) = \tfrac{15}{8} - 2\ln 2$ ($\approx 0.489$)

$-\ln\tfrac{1}{2} = +\ln 2$. The sign of the log at the lower limit is a common slip.

Worked Example 5 — Trig curves

Find the area enclosed between $y = \sin x$ and $y = \cos x$ for $\tfrac{\pi}{4} \leqslant x \leqslant \tfrac{5\pi}{4}$.

①The curves meet where $\tan x = 1$, i.e. at $x = \tfrac{\pi}{4}$ and $\tfrac{5\pi}{4}$. Between them $\sin x \geqslant \cos x$ (check $x = \tfrac{\pi}{2}$: $1 > 0$).

②$\displaystyle\int_{\pi/4}^{5\pi/4}(\sin x - \cos x)\,dx = \Big[-\cos x - \sin x\Big]_{\pi/4}^{5\pi/4}$

③$= \left(\tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2}\right) - \left(-\tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2}\right) = 2\sqrt{2}$

Part of this region is below the $x$-axis, but "top minus bottom" handles it automatically.

Worked Example 6 — A curve crossing the axis

Find the total area between $y = x^3 - 4x$ and the $x$-axis.

①$x^3 - 4x = x(x-2)(x+2)$: roots $-2, 0, 2$. Above the axis for $-2 < x < 0$, below for $0 < x < 2$.

②$\displaystyle\int_{-2}^0(x^3 - 4x)\,dx = \Big[\tfrac{1}{4}x^4 - 2x^2\Big]_{-2}^0 = 0 - (4 - 8) = 4$

③By symmetry $\displaystyle\int_0^2(x^3-4x)\,dx = -4$, an area of $4$ below the axis.

④Total area $= 4 + 4 = 8$ (not $\int_{-2}^2 = 0$).

3 Areas Under Parametric Curves

For a curve given by $x = \mathrm{f}(t)$, $y = \mathrm{g}(t)$ (see Parametric and Cartesian Forms), change the variable in $\int y\,dx$ from $x$ to $t$, exactly as in a substitution.

Parametric area

$\displaystyle\int_{x=a}^{x=b} y\,dx = \int_{t_1}^{t_2} y\,\frac{dx}{dt}\,dt$, where $t_1, t_2$ are the $t$-values at $x = a, b$

Convert the limits. The limits must be values of $t$. If $x$ decreases as $t$ increases, the integral comes out negative — take the modulus, or swap the limits so they run left to right in $x$.

Worked Example 7 — Checking against the Cartesian answer

A curve has $x = t^2$, $y = 2t$, $t \geqslant 0$. Find the area under the curve between $x = 0$ and $x = 4$.

①$x = 0 \Rightarrow t = 0$; $x = 4 \Rightarrow t = 2$. $\dfrac{dx}{dt} = 2t$.

②$\displaystyle\int_0^2 2t \cdot 2t\,dt = \int_0^2 4t^2\,dt = \Big[\tfrac{4}{3}t^3\Big]_0^2 = \tfrac{32}{3}$

③Check: $y = 2\sqrt{x}$, and $\displaystyle\int_0^4 2x^{1/2}\,dx = \Big[\tfrac{4}{3}x^{3/2}\Big]_0^4 = \tfrac{32}{3}$ ✓

Worked Example 8 — The area of an ellipse

The ellipse $x = 3\cos t$, $y = 2\sin t$, $0 \leqslant t < 2\pi$. Find the area it encloses.

①By symmetry, find the quarter in the first quadrant and multiply by $4$. There $x$ runs from $0$ to $3$, i.e. $t$ runs from $\tfrac{\pi}{2}$ to $0$.

②$\dfrac{dx}{dt} = -3\sin t$, so $\displaystyle\int_{\pi/2}^{0}2\sin t\,(-3\sin t)\,dt = \int_0^{\pi/2}6\sin^2 t\,dt$

③$= 3\displaystyle\int_0^{\pi/2}(1 - \cos 2t)\,dt = 3\Big[t - \tfrac{1}{2}\sin 2t\Big]_0^{\pi/2} = \tfrac{3\pi}{2}$

④Total area $= 4 \times \tfrac{3\pi}{2} = 6\pi$.

In general an ellipse with semi-axes $a$ and $b$ has area $\pi ab$ — here $\pi \times 3 \times 2$. For $a = b = r$ this is the familiar $\pi r^2$.

Worked Example 9 — One arch of a cycloid

A point on a rolling wheel traces $x = t - \sin t$, $y = 1 - \cos t$. Find the area under one arch, $0 \leqslant t \leqslant 2\pi$.

①$\dfrac{dx}{dt} = 1 - \cos t$, so the integrand is $(1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t$.

②$\cos^2 t = \tfrac{1}{2}(1 + \cos 2t)$, giving $\tfrac{3}{2} - 2\cos t + \tfrac{1}{2}\cos 2t$.

③$\Big[\tfrac{3}{2}t - 2\sin t + \tfrac{1}{4}\sin 2t\Big]_0^{2\pi} = 3\pi$

The area is exactly three times the area of the rolling circle (radius $1$, area $\pi$).

4 Quick Reference

Limit of a sum

$\int_a^b y\,dx = \lim_{\delta x\to 0}\sum y\,\delta x$

Upper / lower sums

Rectangles over and under the curve squeeze the integral.

Between curves

$\int(\text{top} - \text{bottom})\,dx$ between the intersections.

Which is top?

Test a value between the limits.

Below the axis

Integral is negative; split at roots for total area.

Parametric

$\int y\,\dfrac{dx}{dt}\,dt$ with $t$-limits.

Negative answer?

$x$ decreasing in $t$: swap limits or take modulus.

Symmetry

Find one piece and multiply — but check the signs.

5 Practice Questions

Question 1

Write $\displaystyle\lim_{\delta x\to 0}\sum_{x=0}^{3}(2x + 1)\,\delta x$ as an integral and evaluate it.

▶ Show solution

$\displaystyle\int_0^3(2x+1)\,dx = \Big[x^2 + x\Big]_0^3 = 9 + 3 = 12$

Question 2

Evaluate $\displaystyle\lim_{\delta x\to 0}\sum_{x=1}^{e}\frac{1}{x}\,\delta x$.

▶ Show solution

$= \displaystyle\int_1^e\frac{1}{x}\,dx = \Big[\ln x\Big]_1^e = 1 - 0 = 1$

Question 3

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

▶ Show solution

$x^2 = x + 2 \Rightarrow (x-2)(x+1) = 0$, so $x = -1$ or $2$. The line is on top (at $x = 0$: $2 > 0$).

$\displaystyle\int_{-1}^2(x + 2 - x^2)\,dx = \Big[\tfrac{1}{2}x^2 + 2x - \tfrac{1}{3}x^3\Big]_{-1}^2$

$= \left(2 + 4 - \tfrac{8}{3}\right) - \left(\tfrac{1}{2} - 2 + \tfrac{1}{3}\right) = \tfrac{10}{3} + \tfrac{7}{6} = \tfrac{9}{2}$

Question 4

Find the total area enclosed between $y = x^3 - x$ and the $x$-axis.

▶ Show solution

$x^3 - x = x(x-1)(x+1)$: roots $-1, 0, 1$.

$\displaystyle\int_0^1(x^3 - x)\,dx = \tfrac{1}{4} - \tfrac{1}{2} = -\tfrac{1}{4}$, an area of $\tfrac{1}{4}$ below the axis.

By rotational symmetry the part for $-1 \leqslant x \leqslant 0$ has area $\tfrac{1}{4}$ above the axis. Total area $= \tfrac{1}{2}$.

Question 5

Find the exact area of the region bounded by $y = e^x$, $y = e^{-x}$ and the line $x = 1$.

▶ Show solution

The curves meet at $x = 0$. For $0 < x \leqslant 1$, $e^x > e^{-x}$.

$\displaystyle\int_0^1(e^x - e^{-x})\,dx = \Big[e^x + e^{-x}\Big]_0^1 = \left(e + e^{-1}\right) - 2 = e + \dfrac{1}{e} - 2$

Question 6

Show that the curves $y = \sin 2x$ and $y = \sin x$ meet at $x = \tfrac{\pi}{3}$, and find the area enclosed between them for $0 \leqslant x \leqslant \tfrac{\pi}{3}$.

▶ Show solution

$\sin 2x = \sin x \Rightarrow 2\sin x\cos x - \sin x = 0 \Rightarrow \sin x(2\cos x - 1) = 0$, so $x = 0$ or $\cos x = \tfrac{1}{2}$, $x = \tfrac{\pi}{3}$ ✓

At $x = \tfrac{\pi}{6}$: $\sin 2x = \tfrac{\sqrt{3}}{2} > \tfrac{1}{2}$, so $\sin 2x$ is on top.

$\displaystyle\int_0^{\pi/3}(\sin 2x - \sin x)\,dx = \Big[-\tfrac{1}{2}\cos 2x + \cos x\Big]_0^{\pi/3}$

$= \left(\tfrac{1}{4} + \tfrac{1}{2}\right) - \left(-\tfrac{1}{2} + 1\right) = \tfrac{1}{4}$

Question 7

Find the area enclosed by $y = \sqrt{x}$ and $y = \tfrac{1}{2}x$.

▶ Show solution

$\sqrt{x} = \tfrac{1}{2}x \Rightarrow x = \tfrac{1}{4}x^2 \Rightarrow x = 0$ or $4$. At $x = 1$: $1 > \tfrac{1}{2}$, so $\sqrt{x}$ is on top.

$\displaystyle\int_0^4\left(x^{1/2} - \tfrac{1}{2}x\right)dx = \Big[\tfrac{2}{3}x^{3/2} - \tfrac{1}{4}x^2\Big]_0^4 = \tfrac{16}{3} - 4 = \tfrac{4}{3}$

Question 8

A curve has parametric equations $x = t^3$, $y = t^2$. Find the area between the curve, the $x$-axis and the line $x = 8$, for $x \geqslant 0$.

▶ Show solution

$x = 0 \Rightarrow t = 0$; $x = 8 \Rightarrow t = 2$. $\dfrac{dx}{dt} = 3t^2$.

$\displaystyle\int_0^2 t^2 \cdot 3t^2\,dt = \Big[\tfrac{3}{5}t^5\Big]_0^2 = \tfrac{96}{5}$

Question 9

A curve has $x = \ln t$, $y = t^2$, $t > 0$. Find the area under the curve from $t = 1$ to $t = 2$ and verify it using the Cartesian equation.

▶ Show solution

$\dfrac{dx}{dt} = \dfrac{1}{t}$, so $\displaystyle\int_1^2 t^2 \cdot \frac{1}{t}\,dt = \int_1^2 t\,dt = \Big[\tfrac{1}{2}t^2\Big]_1^2 = \tfrac{3}{2}$

Check: $t = e^x$ so $y = e^{2x}$, and $x$ runs from $0$ to $\ln 2$.

$\displaystyle\int_0^{\ln 2}e^{2x}\,dx = \tfrac{1}{2}\left(e^{2\ln 2} - 1\right) = \tfrac{1}{2}(4 - 1) = \tfrac{3}{2}$ ✓

Question 10

A curve has parametric equations $x = t^2$, $y = 4t - t^3$, for $0 \leqslant t \leqslant 2$.

(a) Find the coordinates of the points where the curve meets the $x$-axis. (b) Show that the area of the region enclosed by the curve and the $x$-axis is $\displaystyle\int_0^2(8t^2 - 2t^4)\,dt$. (c) Find this area exactly. (d) Find a Cartesian equation for the curve and use it to confirm your answer to (c).

▶ Show solution

(a) $y = t(4 - t^2) = t(2-t)(2+t) = 0$; in range, $t = 0$ or $t = 2$. Points $(0, 0)$ and $(4, 0)$.

(b) For $0 < t < 2$, $y > 0$ and $x$ increases from $0$ to $4$. $\dfrac{dx}{dt} = 2t$.

Area $= \displaystyle\int_0^2(4t - t^3)(2t)\,dt = \int_0^2(8t^2 - 2t^4)\,dt$ ✓

(c) $\Big[\tfrac{8}{3}t^3 - \tfrac{2}{5}t^5\Big]_0^2 = \tfrac{64}{3} - \tfrac{64}{5} = \tfrac{128}{15}$

(d) $t = \sqrt{x}$ (as $t \geqslant 0$), so $y = 4x^{1/2} - x^{3/2}$, $0 \leqslant x \leqslant 4$.

$\displaystyle\int_0^4\left(4x^{1/2} - x^{3/2}\right)dx = \Big[\tfrac{8}{3}x^{3/2} - \tfrac{2}{5}x^{5/2}\Big]_0^4 = \tfrac{64}{3} - \tfrac{64}{5} = \tfrac{128}{15}$ ✓

Areas and Integration as a Limit · A Level Maths Year 2 · Created with MathJax