Unit 4 · Topic 1.2 · Further integration
This topic has two halves. First you estimate an area with rectangles and trapezoids, which is what you do when you only have a table of values or a rough shape. Then you find areas exactly with definite integrals, including the part that QCAA tests nearly every year and students regularly get wrong: a curve that dips below the $x$-axis. Tick off each skill once you can do it without help.
Part 1
Split the region under the curve into strips of width $\delta x$ and stand a rectangle on each one. The only decision is how tall each rectangle is.
Find each rectangle's area and add them up. That is the sum $\sum f(x_i)\,\delta x_i$ the syllabus talks about. If the curve is increasing, left endpoints give an underestimate and right endpoints give an overestimate. If it is decreasing, it is the other way round.
Practise
For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step. The sample paper questions have no marking guide, so they have no tags.
Determine the approximate area (units2) between the curve and the $x$-axis for $0\le x\le8$ by calculating the area of the shaded rectangles.
Each rectangle is 2 units wide. Read the heights off the grid: 2, 4 and 9.
$$A\approx2(2)+2(4)+2(9)=30$$
(A) is $2+4+9$, which forgets that the strips are 2 wide.
The region bounded by the $x$-axis and the curve of $f(x)=x(4-x)$ represents the plan of a garden bed. All measurements are in metres.
Estimate the area of the garden bed using sums of the form $\sum_i f(x_i)\,\delta x_i$ where $x_1=1$, $x_2=2$, $x_3=3$ and $x_4=4$.
The strips are 1 m wide, so $\delta x_i=1$. Find each height, then add:
$$f(1)+f(2)+f(3)+f(4)=3+4+3+0=10$$
These are right endpoints, so the last rectangle has zero height. That is fine.
Part (b) of this question is in Part 3.
The approximate area under the curve $y=2x^{2}$ between $x=1$ and $x=4$ is found using rectangles (of width equal to one unit) as shown in the diagram.
What is the approximate area found using these rectangles?
Look at the diagram: each rectangle's top right corner touches the curve, so these are right endpoints, $x=2$, 3 and 4.
$$A\approx1\times\big(2(2)^{2}+2(3)^{2}+2(4)^{2}\big)=8+18+32=58$$
The other options are traps. (D) uses left endpoints, and (C) is the exact area, $\int_1^4 2x^2\,dx=42$.
Part 2
Instead of flat-topped rectangles, join the points on the curve with straight lines to make trapezoids. The result is the average of the left and right endpoint estimates, and it is usually much more accurate. The formula sheet gives it as
$$\int_{a}^{b}f(x)\,dx\approx\frac{w}{2}\Big[f(x_0)+2\big(f(x_1)+f(x_2)+\dots+f(x_{n-1})\big)+f(x_n)\Big],\quad w=\frac{b-a}{n}$$
I find it easier to remember in words: the first and last heights count once, and every height in the middle counts twice.
$$A\approx\frac{w}{2}\big[\text{first}+2(\text{middle})+\text{last}\big]$$
Whether it overestimates or underestimates depends on which way the curve bends. It does not depend on whether it is increasing. If the curve bends upwards (concave up), each straight top sits above the curve and the rule overestimates. If it bends downwards (concave down), it underestimates. Try both in the interactive below.
Try it yourself
Both curves below are increasing between $x=0$ and $x=4$. Switch between them with the trapezoidal rule selected and watch the error change sign.
Practise
State the trapezoidal rule and use it with six strips to determine an approximate value of the definite integral for the curve of $f(x)=4(x-3)^{2}$ from $x=0$ to $x=3$. Show all substitutions made into the rule.
State the rule (copy it from the formula sheet):
$$\int_{a}^{b}f(x)\,dx\approx\frac{w}{2}\Big[f(x_0)+2\big(f(x_1)+\dots+f(x_{n-1})\big)+f(x_n)\Big]$$
The width is $w=\frac{3-0}{6}=0.5$.
So the $x$-values are 0, 0.5, 1, 1.5, 2, 2.5 and 3, and the heights are 36, 25, 16, 9, 4, 1 and 0.
$$\int_{0}^{3}4(x-3)^{2}\,dx\approx\frac{0.5}{2}\Big[36+2(25+16+9+4+1)+0\Big]=0.25(146)=36.5$$
The exact value is 36. The estimate is slightly too big because the parabola is concave up.
The approximate area under the curve $f(x)=\sqrt{2x+1}$ between $x=0$ and $x=4$ using the trapezoidal rule with four strips is
Four strips from 0 to 4 means $w=1$. The heights are $f(0)=1$, $f(1)=\sqrt3$, $f(2)=\sqrt5$, $f(3)=\sqrt7$ and $f(4)=3$.
$$A\approx\frac{1}{2}\Big[1+2\big(\sqrt3+\sqrt5+\sqrt7\big)+3\Big]=2+\sqrt3+\sqrt5+\sqrt7$$
The $\frac12$ multiplies everything in the square bracket, including the middle terms. (B) only halves the first and last heights, and (C) forgets the $\frac12$ altogether.
The rate that water fills an empty vessel is given by $\frac{dV}{dt}=0.25e^{0.25t}$ (in litres per hour), $0\le t\le8\ln(6)$, where $t$ is time (in hours).
a) Determine the function that represents the volume of water in the vessel (in litres). [2 marks]
The vessel is full when $t=8\ln(6)$.
b) Determine the volume of water, to the nearest litre, the vessel can hold when full. [2 marks]
The table shows the approximate rate the water flows into the vessel at certain times.
| $t$ | $\frac{dV}{dt}$ |
|---|---|
| 0 | 0.25 |
| 1 | 0.32 |
| 2 | 0.41 |
| 3 | 0.53 |
c) Use information from the table and the trapezoidal rule to determine the approximate volume of water in the vessel after three hours. [2 marks]
a) Integrate, then use "empty" to find $c$. At $t=0$, $V=0$:
$$V=\int0.25e^{0.25t}\,dt=e^{0.25t}+c,\qquad 0=e^{0}+c\ \Rightarrow\ c=-1$$
b) $V(8\ln6)=e^{2\ln6}-1=e^{\ln36}-1=36-1=35$.
c) The volume after three hours is $\int_0^3\frac{dV}{dt}\,dt$. Use the table with $w=1$:
$$V(3)\approx\frac{1}{2}\Big[0.25+2(0.32+0.41)+0.53\Big]=\frac{1}{2}(2.24)=1.12$$
The exact value from (a) is $e^{0.75}-1\approx1.117$. The rate is increasing, yet the rule still overestimates, because the curve is concave up.
Using the trapezoidal rule with an interval size of 1, the approximate value of the integral $\displaystyle\int_{0}^{3}0.5^{x}\,dx$ is
With $w=1$ the heights are $0.5^0=1$, 0.5, 0.25 and 0.125.
$$\frac{1}{2}\Big[1+2(0.5+0.25)+0.125\Big]=\frac{1}{2}(2.625)=1.3125$$
(B) is the exact value of the integral. If you put the integral straight into your calculator, you answer a different question from the one asked.
A field is divided into five sections as shown. The width of each section is 1 metre. The perpendicular height, in metres, of each section is given in the diagram. The area of the field was approximated using the trapezoidal rule and found to be 11.12 m2.
a) Determine the height marked $x$ on the diagram. [2 marks]
b) Determine the area of the field, given the shape of the field is modelled by the function $f(x)=\dfrac{4x}{x^{2}+1}+1$, $0\le x\le5$. [1 mark]
a) This is the rule run backwards. Substitute everything you know with $w=1$, set it equal to 11.12 and solve:
$$\frac{1}{2}\Big[1+2(3+2.6+2.2+x)+1.76\Big]=11.12$$
$$2.76+2(7.8+x)=22.24\ \Rightarrow\ 7.8+x=9.74\ \Rightarrow\ x=1.94$$
b) The field is the area under $f$ from 0 to 5, and $f$ is above the axis the whole way. On your calculator:
$$\int_{0}^{5}\left(\frac{4x}{x^{2}+1}+1\right)dx\approx11.52$$
Using the trapezoidal rule with four sections, the approximate area under the curve $y=x^{2}+5$ between $x=0$ and $x=2$ is
Four sections from 0 to 2 gives $w=0.5$. The heights at 0, 0.5, 1, 1.5 and 2 are 5, 5.25, 6, 7.25 and 9.
$$\frac{0.5}{2}\Big[5+2(5.25+6+7.25)+9\Big]=0.25(51)=12.75$$
(A) is the exact area, $\frac{38}{3}\approx12.67$. The trapezoidal estimate is a little bigger because the parabola is concave up.
Part 3
When a continuous function $f(x)$ is above the $x$-axis from $x=a$ to $x=b$, the area between the curve and the axis is exactly the definite integral:
$$A=\int_{a}^{b}f(x)\,dx$$
Below the axis, the integral comes out negative, but an area cannot be negative. So sketch the curve first, every time, and check whether any part of it dips below the axis between your bounds. There are three cases.
For this curve, the total area is
$$A=A_1+A_2=\int_{a}^{c}f(x)\,dx+\left|\int_{c}^{b}f(x)\,dx\right|$$
If you integrate straight from $a$ to $b$, the negative piece cancels part of the positive piece and you get $A_1-A_2$. Multiple choice questions nearly always offer that number as a wrong answer.
Watch the integral cancel
Practise
Determine the area (units2) between the graph of $y=4x^{3}$ and the $x$-axis from $x=-1$ to $x=2$.
$y=4x^3$ is below the axis for $x<0$ and above it for $x>0$, so split at 0:
$$\left|\int_{-1}^{0}4x^{3}\,dx\right|+\int_{0}^{2}4x^{3}\,dx=\left|\Big[x^{4}\Big]_{-1}^{0}\right|+\Big[x^{4}\Big]_{0}^{2}=|-1|+16=17$$
(B) is what you get from integrating straight from $-1$ to 2.
Calculate the total enclosed area between the graph of $y=x^{2}-x-6$ and the $x$-axis from $x=1$ to $x=5$.
$x^2-x-6=(x-3)(x+2)$, so the curve crosses the axis at $x=3$. It is below the axis from 1 to 3 and above it from 3 to 5.
$$\left|\int_{1}^{3}\left(x^{2}-x-6\right)dx\right|+\int_{3}^{5}\left(x^{2}-x-6\right)dx\approx7.33+12.67=20.00$$
(A), (B) and (C) are the three numbers you would get by forgetting to split, or by stopping halfway.
The region bounded by the $x$-axis and the curve of $f(x)=x(4-x)$ represents the plan of a garden bed. All measurements are in metres. (The graph is with part (a) in Part 1.)
Use a definite integral to determine the area of the garden bed.
The bed runs between the $x$-intercepts, 0 and 4, and the curve is above the axis there. Expand before you integrate:
$$\int_{0}^{4}\left(4x-x^{2}\right)dx=\Big[2x^{2}-\tfrac{1}{3}x^{3}\Big]_{0}^{4}=32-\tfrac{64}{3}=\tfrac{32}{3}$$
Part (a)'s estimate of 10 m2 was a little low, which makes sense: the rectangles missed the bits of curve above their tops.
The area between the curve $y=9-x^{2}$ and the $x$-axis is
No bounds are given, so use the $x$-intercepts: $9-x^2=0$ gives $x=\pm3$.
$$\int_{-3}^{3}\left(9-x^{2}\right)dx=\Big[9x-\tfrac{1}{3}x^{3}\Big]_{-3}^{3}=(27-9)-(-27+9)=36$$
(B) is half of it, from 0 to 3 only.
Consider the function $f(x)=e^{x}\sin(x)$, $0\le x\le2\pi$.
a) State the exact values of the $x$-intercepts of the graph of $f(x)$. [2 marks]
b) Write an expression for the area enclosed between the graph of $f(x)$ and the $x$-axis. [2 marks]
c) Determine the area enclosed between the graph of $f(x)$ and the $x$-axis to the nearest square unit. [1 mark]
a) $e^x$ is never zero, so $f(x)=0$ only when $\sin x=0$.
In the domain, that is $x=0$, $\pi$ and $2\pi$.
b) $f$ is positive on $(0,\pi)$, where $\sin x>0$, and negative on $(\pi,2\pi)$. Split at $\pi$:
$$A=\int_{0}^{\pi}e^{x}\sin x\,dx+\left|\int_{\pi}^{2\pi}e^{x}\sin x\,dx\right|$$
c) On your calculator the two pieces are about 12.07 and $-279.32$. Take the absolute value of the second and add.
Let $R$ be the region enclosed by the graph of $y=xe^{x}$, the $x$-axis, and the lines $x=-1$ and $x=1$. The area of $R$ is closest to
$xe^x$ has the same sign as $x$, so it is below the axis from $-1$ to 0. Split at 0:
$$\left|\int_{-1}^{0}xe^{x}\,dx\right|+\int_{0}^{1}xe^{x}\,dx\approx0.26+1=1.26$$
(A) is the integral straight from $-1$ to 1, which subtracts the piece below the axis instead of adding it.
Practise
This is the area version of Unit 4 Topic 1.1, Part 2. Integrate with the unknowns left in as letters, set each area equal to the value you are given, and solve.
The graph of $y=f(x)$, where $f(x)$ is the quadratic function $f(x)=ax^{2}+bx+4$, is shown. Two regions of the area between the graph of $y=f(x)$ and the $x$-axis are shaded.
Region $P$ has an area of $\frac{13}{6}$ units2 and Region $Q$ has an area of $\frac{43}{6}$ units2. Determine the values of $a$ and $b$.
Both regions are above the axis, so each area is just the integral. The antiderivative is $F(x)=\frac{a}{3}x^{3}+\frac{b}{2}x^{2}+4x$, and $F(0)=0$.
$$P:\ F(0)-F(-1)=\frac{a}{3}-\frac{b}{2}+4=\frac{13}{6}\qquad Q:\ F(1)-F(0)=\frac{a}{3}+\frac{b}{2}+4=\frac{43}{6}$$
Subtract $P$ from $Q$: $b=\frac{30}{6}=5$. Add them: $\frac{2a}{3}+8=\frac{56}{6}$, so $\frac{2a}{3}=\frac{4}{3}$ and $a=2$.
Check: $f(x)=2x^2+5x+4$ has a negative discriminant ($25-32$), so it never touches the axis, which matches the graph.
The full walkthrough of 2021 Paper 1, Question 18 goes through it mark by mark.
Practise
A company makes windows using glass that has a mass of 5.6 kg per square metre. A customer orders an unusual window in a partial parabolic shape, as shown.
Determine the mass of the window.
Glass area = area under the parabola minus the triangle. Put the bottom left corner at the origin. The parabola has roots 0 and 8 and height 12 at $x=4$, so $y=kx(8-x)$ with $16k=12$:
$$y=\tfrac{3}{4}x(8-x),\qquad\int_{0}^{8}\tfrac{3}{4}x(8-x)\,dx=64\ \text{m}^{2}$$
The triangle has base 8 and sides 7 and 9. If its top vertex is at $(p,h)$, then $p^2+h^2=49$ and $(8-p)^2+h^2=81$. Subtracting gives $p=2$ and $h=\sqrt{45}$, so its area is $\frac12(8)\sqrt{45}\approx26.83$ m2.
Glass: $64-26.83=37.17$ m2. Mass: $37.17\times5.6\approx208$ kg.
The full walkthrough of 2023 Paper 2, Question 18 goes through it mark by mark.
A farmer has a paddock with straight fences on two sides (fence A and fence B) perpendicular to each other and bounded by a creek on the other side. Fence B is 4 kilometres long. The creek boundary can be modelled using
$$d=\ln(5x+e)-1$$
where $d$ is the perpendicular distance in kilometres from fence A to the creek and $x$ is the distance in kilometres along fence A from the point O.
The farmer wants to divide his paddock area in half with a straight fence, parallel to fence B.
Determine where the farmer should locate the fence.
First find where fence B is. It is 4 km long, so the creek is 4 km from fence A there:
$$\ln(5x+e)-1=4\ \Rightarrow\ 5x+e=e^{5}\ \Rightarrow\ x=\frac{e^{5}-e}{5}\approx29.14$$
The paddock is the area under the creek from O to fence B. The creek starts at O, because $d(0)=\ln e-1=0$.
$$\int_{0}^{29.14}\big(\ln(5x+e)-1\big)\,dx\approx89.59\ \text{km}^{2}$$
The new fence goes at $x=k$, where the area from 0 to $k$ is half of that, 44.80 km2. Solve on your calculator:
$$\int_{0}^{k}\big(\ln(5x+e)-1\big)\,dx=44.80\ \Rightarrow\ k\approx17.24$$
This is reasonable. It is more than halfway along (14.57 km), because the paddock is narrower near O.
Part 4
To find the area between two curves, subtract the bottom function from the top one and integrate. If $f(x)$ is on top and $g(x)$ is underneath for $a\le x\le b$,
$$A=\int_{a}^{b}\big(f(x)-g(x)\big)\,dx$$
It does not matter if one or both curves go below the $x$-axis. "Top minus bottom" is always the height of the region. What matters is knowing which curve is on top, so you have to sketch both (or graph them on your calculator) before you write anything down. If the curves cross each other, the points of intersection are usually your bounds. If they do not cross, the question gives you the bounds, as vertical lines like $x=0$ and $x=1$.
Harder questions make you work out one or both functions first, or give you the area and ask for an unknown. QCAA has not yet asked either of those for area between curves, but I would expect them, so treat them the same way as Part 3. The window question above is already an area between curves in disguise: the parabola on top and the triangle's sides underneath.
Watch top minus bottom
Practise
The curved lines represent graphs of the equations $y=x^{2}-4x+8$ and $y=10\cos(x+10)$.
a) Determine the coordinates of the points of intersection A and B. [1 mark]
b) State an integral expression representing the area enclosed by the two graphs. [2 marks]
c) Determine the area enclosed by the two graphs. [1 mark]
a) Graph both on your calculator, in radians, and use the intersection tool: A$(1.441,\ 4.312)$ and B$(3.473,\ 6.168)$.
b) Between A and B the cosine curve (dashed) is on top:
$$A=\int_{1.441}^{3.473}\Big(10\cos(x+10)-\left(x^{2}-4x+8\right)\Big)\,dx$$
c) Evaluate it on your calculator. Keep the bounds unrounded (store them) so the answer does not drift.
Consider the functions $f(x)=x^{2}$ and $g(x)=4x$.
a) Determine the $x$-coordinates of the points of intersection of the graphs of the two functions. [2 marks]
b) Use the results from Question 13a) to calculate the area enclosed by the graphs of $f(x)$ and $g(x)$. [3 marks]
a) $x^2=4x$ gives $x(x-4)=0$, so $x=0$ or $x=4$. Do not divide both sides by $x$, or you lose $x=0$.
b) Test a point in between: at $x=1$, $g=4$ and $f=1$, so the line is on top.
$$\int_{0}^{4}\left(4x-x^{2}\right)dx=\Big[2x^{2}-\tfrac{1}{3}x^{3}\Big]_{0}^{4}=32-\tfrac{64}{3}=\tfrac{32}{3}$$
It is the same integral as the garden bed in Part 3. That is no coincidence: $4x-x^2$ is $x(4-x)$.
The graphs of $f(x)=e^{x}$ and $g(x)=x^{2}-1$ are shown.
The area of the shaded section bounded by these graphs between the lines $x=0$ and $x=1$ is
$e^x$ is on top. Watch the brackets: subtracting $x^2-1$ gives $-x^2+1$.
$$\int_{0}^{1}\left(e^{x}-x^{2}+1\right)dx=\Big[e^{x}-\tfrac{1}{3}x^{3}+x\Big]_{0}^{1}=\left(e-\tfrac{1}{3}+1\right)-1=e-\tfrac{1}{3}$$
Practice
Eight questions of my own, covering every skill. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to. Sketch the curve before you start every one of them.
Using four rectangles of equal width, with heights taken at the right end of each strip, an estimate of the area under $y=x^{2}$ from $x=1$ to $x=3$ is
Each strip is $\delta x=\frac{3-1}{4}=0.5$ wide, and the right ends are $x=1.5,\ 2,\ 2.5,\ 3$.
$$(2.25+4+6.25+9)\times0.5=10.75$$
The curve is increasing, so right-end rectangles stick out above it and give an overestimate. The exact area is $\frac{26}{3}\approx8.67$, option (B). Option (A) uses left ends, and (D) has forgotten the width.
Use the trapezoidal rule with three strips to determine an approximate value of $\displaystyle\int_{1}^{4}\frac{12}{x}\,dx$. Show all substitutions made into the rule.
| $x$ | $1$ | $2$ | $3$ | $4$ |
|---|---|---|---|---|
| $f(x)=\frac{12}{x}$ | $12$ | $6$ | $4$ | $3$ |
$$\int_{1}^{4}\frac{12}{x}\,dx\approx\frac{w}{2}\Big[f(x_0)+2\big(f(x_1)+f(x_2)\big)+f(x_3)\Big]=\frac{1}{2}\Big[12+2(6+4)+3\Big]=17.5$$
$y=\frac{12}{x}$ is concave up here, so the trapezoids sit above the curve and 17.5 is a slight overestimate. The exact value is $12\ln4\approx16.6$.
Determine the area between the curve $y=\sin x$ and the $x$-axis from $x=0$ to $x=\pi$.
$$\int_{0}^{\pi}\sin x\,dx=\Big[-\cos x\Big]_{0}^{\pi}=-\cos\pi+\cos0=1+1=2$$
Both bounds needed exact values: $\cos\pi=-1$ and $\cos0=1$.
Calculate the area enclosed by the graph of $y=x^{2}-4x$ and the $x$-axis.
$$x^{2}-4x=x(x-4)=0\quad\Rightarrow\quad x=0,\ 4$$
$$\int_{0}^{4}\left(x^{2}-4x\right)dx=\Big[\tfrac{x^{3}}{3}-2x^{2}\Big]_{0}^{4}=\frac{64}{3}-32=-\frac{32}{3}$$
Calculate the total area enclosed between the graph of $y=e^{x}-3$ and the $x$-axis from $x=0$ to $x=2$.
$$\int_{0}^{1.0986}\left(e^{x}-3\right)dx\approx-1.2958,\qquad\int_{1.0986}^{2}\left(e^{x}-3\right)dx\approx1.6849$$
$$A\approx1.2958+1.6849\approx2.98$$
Integrating straight from 0 to 2 gives about 0.389. That is wrong, because the part below the axis cancels some of the part above. This is the same trap as the animation in Part 3.
The area enclosed between $y=\cos x$ and the $x$-axis from $x=0$ to $x=\pi$ is
$\cos x$ is positive from 0 to $\frac{\pi}{2}$ and negative from $\frac{\pi}{2}$ to $\pi$, so split at $\frac{\pi}{2}$.
$$\int_{0}^{\frac{\pi}{2}}\cos x\,dx=1,\qquad\int_{\frac{\pi}{2}}^{\pi}\cos x\,dx=-1,\qquad A=1+|-1|=2$$
Option (A) is what you get from integrating straight from 0 to $\pi$, where the two parts cancel. Option (B) only counts the part above the axis.
Consider the functions $f(x)=x$ and $g(x)=x^{2}-2x$.
$$x^{2}-2x=x\quad\Rightarrow\quad x^{2}-3x=0\quad\Rightarrow\quad x(x-3)=0\quad\Rightarrow\quad x=0,\ 3$$
$$A=\int_{0}^{3}\Big(x-\left(x^{2}-2x\right)\Big)dx=\int_{0}^{3}\left(3x-x^{2}\right)dx=\Big[\tfrac{3x^{2}}{2}-\tfrac{x^{3}}{3}\Big]_{0}^{3}=\frac{27}{2}-9=\frac{9}{2}$$
Part of the parabola is below the $x$-axis, but that does not matter here. Top minus bottom is always the height of the region.
Determine the area enclosed between the curves $y=3-x^{2}$ and $y=e^{x}-1$.
$$A=\int_{-1.9646}^{1.0580}\Big(\left(3-x^{2}\right)-\left(e^{x}-1\right)\Big)dx\approx6.43$$
Write down the integral you entered as well as the number. The bounds and the top-minus-bottom expression each earn marks in questions like this.
Keep practising
The window question has its own complex unfamiliar walkthrough. Try it again from a blank page first. The hard part is not the integral, it is deciding where to put the axes.
Past questions and marking-guide steps are from the 2020–2025 QCAA Mathematical Methods external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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