Unit 4 · Topic 1.1 · Further integration
In Unit 3 you found antiderivatives, which are functions. Now you put numbers in for the bounds, and the integral becomes a single value. That value is also the total change in a quantity when you know its rate of change, which is where most of the exam marks for this topic are. Tick off each skill once you can do it without help.
Part 1
The fundamental theorem of calculus says that if $F$ is an antiderivative of $f$, then
$$\int_{a}^{b}f(x)\,dx=F(b)-F(a)$$
This is called a definite integral, because it gives a number rather than a function. You do not need $+c$: it would appear in both $F(b)$ and $F(a)$ and cancel out. The numbers $a$ and $b$ are the bounds (or terminals), and the function you are integrating, $f(x)$, is the integrand.
In working, write the antiderivative in square brackets with the bounds on the right, then substitute. For example:
$$\int_{1}^{3}2x\,dx=\Big[x^{2}\Big]_{1}^{3}=9-1=8$$
Watch the bounds go in
Where those exact values come from
Try it yourself
Split the interval from 1 to 2 into $n$ thin strips of width $\delta x$ and add up the rectangles, $\sum f(x_i)\,\delta x$. As $n$ grows, the sum gets closer to the exact value of the integral. That is what the integral sign means: a long, thin S for "sum".
Practise
Determine the value of $\displaystyle\int_{1}^{2}3x^{2}\,dx$.
$$\int_{1}^{2}3x^{2}\,dx=\Big[x^{3}\Big]_{1}^{2}=8-1=7$$
The most common slip is to forget to subtract $F(1)$, which gives 8. That is not an option here, but it often is.
Determine $\displaystyle\int_{1}^{3}\frac{1}{2x}\,dx$.
Take the $\frac{1}{2}$ out the front first, then use $\int\frac{1}{x}\,dx=\ln x$:
$$\frac{1}{2}\int_{1}^{3}\frac{1}{x}\,dx=\frac{1}{2}\Big[\ln x\Big]_{1}^{3}=\frac{1}{2}(\ln3-\ln1)=\frac{1}{2}\ln3$$
Remember that $\ln1=0$.
Determine $\displaystyle\int_{a}^{b}2\cos(x)\,dx$, where $a=\frac{\pi}{3}$ and $b=\frac{\pi}{2}$.
$$\int_{\pi/3}^{\pi/2}2\cos x\,dx=\Big[2\sin x\Big]_{\pi/3}^{\pi/2}=2(1)-2\left(\frac{\sqrt{3}}{2}\right)=2-\sqrt{3}$$
(D) is what you get if you subtract the wrong way round, $F(a)-F(b)$. The upper bound always goes first.
Determine $\displaystyle\int_{0}^{3}\pi\sin\left(\frac{\pi}{3}x\right)dx$.
Use $\int\sin(ax)\,dx=-\frac{1}{a}\cos(ax)$ with $a=\frac{\pi}{3}$. The $\pi$ out the front cancels with $\frac{3}{\pi}$:
$$\Big[-3\cos\left(\tfrac{\pi}{3}x\right)\Big]_{0}^{3}=-3\cos\pi-\big(-3\cos0\big)=3+3=6$$
Take care with the double negative. Losing it gives 0 or $-6$.
Calculate $\displaystyle\int_{-1}^{7}\frac{dx}{\sqrt{2+x}}$.
Rewrite as a power, $(2+x)^{-\frac{1}{2}}$, then use the power rule with a linear inside:
$$\int_{-1}^{7}(2+x)^{-\frac{1}{2}}\,dx=\Big[2\sqrt{2+x}\Big]_{-1}^{7}=2\sqrt{9}-2\sqrt{1}=6-2=4$$
Part 2
Sometimes you are given the value of the integral and one of the bounds, or a constant in the integrand, is unknown. Work the integral out as usual, leaving the unknown as a letter. Then set it equal to the given value and solve. Check your answer against any condition in the question, such as $b>1$.
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.
Determine the value of $b$ given $\displaystyle\int_{a}^{b}3x^{2}\,dx=117$ and $\displaystyle\int_{a}^{b-1}3x^{2}\,dx=56$ for $b>1$.
Evaluate each integral with the unknowns left in:
$$b^{3}-a^{3}=117\qquad(b-1)^{3}-a^{3}=56$$
Subtract the second equation from the first, and $a^{3}$ disappears:
$$b^{3}-(b-1)^{3}=61\ \Rightarrow\ 3b^{2}-3b+1=61\ \Rightarrow\ b^{2}-b-20=0$$
This factorises as $(b-5)(b+4)=0$, so $b=5$ or $b=-4$.
Reject $b=-4$, because $b>1$.
As a check, $a^{3}=125-117=8$, so $a=2$, and $\int_{2}^{4}3x^{2}\,dx=64-8=56$.
The full walkthrough of 2022 Paper 1, Question 17 goes through it mark by mark.
Part 3
If you know the rate at which something changes, integrating that rate over a time period gives the total change in that period. It is the fundamental theorem again, with $Q$ for the quantity:
$$\int_{a}^{b}Q'(t)\,dt=Q(b)-Q(a)$$
For example, if you know the rate at which water flows from a tap in litres per minute, integrating that rate from $t=3$ to $t=7$ gives the number of litres that flowed out between minute 3 and minute 7. Velocity works the same way: integrating it gives the change in displacement.
"Marginal" is the economists' word for the derivative. If $R(x)$ is the revenue in dollars from selling $x$ units, with $R(50)=2000$ and $R(51)=2043$, then the marginal revenue at $x=50$ is about $43 per unit. You will see marginal revenue, marginal cost and marginal profit. To get the total revenue (or cost, or profit) over a range of units, integrate the marginal function over that range.
Watch the area add up
Practise
Oil is leaking from a tanker at the rate of $r(t)=9000e^{-0.2t}$ litres per hour, where $t$ is in hours. Determine how much oil leaks from the tanker (to the nearest litre) from time $t=0$ to time $t=10$.
The total leaked is the integral of the rate over the 10 hours. On Paper 2 you can enter it straight into your calculator:
$$\int_{0}^{10}9000e^{-0.2t}\,dt=45000\left(1-e^{-2}\right)\approx38910$$
A marble moves in one direction in a straight line with velocity $v=2\ln(t+1)$ (in metres per second) where $t$ is time (in seconds) since the marble passed through the origin. Determine the distance from the origin the marble has rolled after 4 seconds.
The marble only moves one way, so distance equals displacement. Integrate the velocity from 0 to 4 on your calculator:
$$\int_{0}^{4}2\ln(t+1)\,dt\approx8.09$$
(C) is $v(4)=2\ln5$, the velocity at 4 seconds. It is not the distance.
The velocity (m s−1) at time $t$ (s) of an object is given by $v(t)=0.4t^{2}+3t$ for $t\ge0$. The change in displacement (m) of the object from four to five seconds is
$$\int_{4}^{5}\left(0.4t^{2}+3t\right)dt=\Big[\tfrac{0.4}{3}t^{3}+1.5t^{2}\Big]_{4}^{5}=\tfrac{0.4}{3}(61)+1.5(9)\approx21.63$$
(D) is the integral from 0 to 5. "From four to five seconds" means the lower bound is 4.
Part 4
In modelling questions there is often more than one rate, such as something flowing in and something flowing out. Find the total change for each over the same time period, then combine them. Check the units and the time period before you integrate.
Practise
The energy produced by solar panels on a house is stored in a battery. Energy consumed in the house comes out of the battery.
The rate of energy produced by the solar panels during daylight hours on a particular day is approximately modelled by $R(t)=2.5\sin\left(\frac{\pi}{12}(t-6)\right)$, $6\le t\le18$, where $t$ is the number of hours after midnight. $R(t)=0$ for any hours after sunset and before sunrise.
The rate of energy consumed in the house is $\frac{4}{\pi}$ units per hour.
Determine the total energy change in the battery during daylight hours, assuming that sunrise is at 6:00 am and sunset is at 6:00 pm.
Daylight runs from $t=6$ to $t=18$. Energy goes in from the panels and out to the house, so find each total, then subtract.
Energy in: integrate $R(t)$, using $\int\sin(ax+b)\,dx=-\frac{1}{a}\cos(ax+b)$ with $a=\frac{\pi}{12}$:
$$\int_{6}^{18}2.5\sin\left(\tfrac{\pi}{12}(t-6)\right)dt=\Big[-\tfrac{30}{\pi}\cos\left(\tfrac{\pi}{12}(t-6)\right)\Big]_{6}^{18}=-\tfrac{30}{\pi}(\cos\pi-\cos0)=\tfrac{60}{\pi}$$
Energy out: the rate is constant, so it is rate times time, $\frac{4}{\pi}\times12=\frac{48}{\pi}$.
Total change in the battery: $\frac{60}{\pi}-\frac{48}{\pi}=\frac{12}{\pi}$ units. It is positive, so the battery gains energy over the day, which makes sense because the panels produce more than the house uses at the peak.
The full walkthrough of 2025 Paper 1, Question 16 goes through it mark by mark.
Game plan
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.
Using four rectangles of equal width, with heights taken at the left end of each strip, an estimate of $\displaystyle\int_{0}^{2}x^{2}\,dx$ is
Four strips from 0 to 2 means each is $\delta x=0.5$ wide. The left ends are $x=0,\ 0.5,\ 1,\ 1.5$.
$$\sum f(x_i)\,\delta x=(0+0.25+1+2.25)\times0.5=1.75$$
Option (B) is the exact value, $\frac{8}{3}$. It is not an estimate. Option (C) uses the right ends instead, and (D) has forgotten to multiply by the width.
The estimate in Question 1 was 1.75, and the exact value of $\displaystyle\int_{0}^{2}x^{2}\,dx$ is $\frac{8}{3}$. Without calculating, explain whether using 8 left-end rectangles would give a larger or smaller estimate.
Evaluate $\displaystyle\int_{1}^{4}\left(3\sqrt{x}-2\right)dx$.
$$\Big[2x^{\frac{3}{2}}-2x\Big]_{1}^{4}$$
$$\left(2(8)-8\right)-\left(2(1)-2\right)=8-0=8$$
$\displaystyle\int_{0}^{\ln2}e^{2x}\,dx$ is equal to
The antiderivative is $\frac{1}{2}e^{2x}$. At the top, $e^{2\ln2}=e^{\ln4}=4$, and at the bottom $e^{0}=1$.
$$\frac{1}{2}(4-1)=\frac{3}{2}$$
Option (A) has forgotten to divide by 2, and (C) has left out the bottom bound.
Evaluate $\displaystyle\int_{0}^{\frac{\pi}{3}}\sin(2x)\,dx$.
$$\Big[-\tfrac{1}{2}\cos(2x)\Big]_{0}^{\frac{\pi}{3}}$$
$$-\tfrac{1}{2}\left(-\tfrac{1}{2}\right)-\left(-\tfrac{1}{2}\right)(1)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$$
Watch the signs. Two minus signs meet in each term, so it is worth writing every bracket.
Water drains from a tank at the rate of $r(t)=20e^{-0.1t}$ litres per minute, where $t$ is the time in minutes since the tap was opened. Determine how much water drains from the tank between $t=3$ and $t=5$.
$$\int_{3}^{5}20e^{-0.1t}\,dt\approx26.86$$
As a Technology-free check, $\Big[-200e^{-0.1t}\Big]_{3}^{5}=200\left(e^{-0.3}-e^{-0.5}\right)$, which gives the same value.
The marginal cost of making $x$ items is $C'(x)=50-0.4x$ dollars per item. Determine the increase in cost when production rises from 20 items to 50 items.
$$\int_{20}^{50}(50-0.4x)\,dx=\Big[50x-0.2x^{2}\Big]_{20}^{50}$$
$$(2500-500)-(1000-80)=2000-920=1080$$
There is no starting cost to find. The definite integral gives the change straight away.
Cars enter a car park at the rate $E(t)=120e^{-0.2t}$ cars per hour and leave at the rate $L(t)=20t$ cars per hour, where $t$ is the time in hours after 7 am, for $0\le t\le4$. Determine the change in the number of cars in the car park between 8 am and 10 am.
$$\int_{1}^{3}\left(120e^{-0.2t}-20t\right)dt\approx81.95$$
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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