2024 Paper 1, Question 1
Determine $\int x^{4}\,dx$.
Add one to the power, then divide by the new power: $\frac{x^{5}}{5}+c$. (A) is the derivative, which is the most common slip on this question.
Answer
(D) $\frac{1}{5}x^{5}+c$
Unit 3 · Topic 4.1 · Introduction to integration
Rules for integration
Integration runs differentiation backwards. If you know a function's derivative, integrating gets you back to the function. The rules are quick to learn, but the exam rewards students who have practised applying them, so there are plenty of past questions below. Tick off each skill once you can do it without help.
Part 1
The basic rules
Antidifferentiation, also called integration, is the reverse of differentiation. The indefinite integral of $f(x)$ is written $\int f(x)\,dx$, which you read as "the integral of $f(x)$ with respect to $x$".
Every time you integrate, add a constant of integration, $+c$. Differentiating gets rid of any constant, so when you go backwards you cannot know what it was. Leaving it off costs you a mark.
Powers of
$$\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+c,\quad n\ne-1$$
Exponentials
$$\int e^{x}\,dx=e^{x}+c$$
The one power that does not fit
$$\int \frac{1}{x}\,dx=\ln x+c,\quad x>0$$
Trigonometric functions
$$\int \sin x\,dx=-\cos x+c$$ $$\int \cos x\,dx=\sin x+c$$
Sums
$$\int \big(f(x)+g(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx$$
Constant multiples
$$\int k\,f(x)\,dx=k\int f(x)\,dx$$
Watch the power rule
There is no product rule for integration, so you cannot integrate two factors separately and multiply. Expand brackets, split fractions and rewrite things like $\frac{20}{x^{3}}$ as $20x^{-3}$ before you start. Most of the questions below need a step like this.
Practise
Exam questions: the basic rules
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.
2025 Paper 2, Question 1
Determine $\int 10.4x^{3}\,dx$.
The constant stays where it is. The power goes up to 4, so divide by 4: $\frac{10.4}{4}=2.6$. (D) multiplies by 4 instead of dividing.
Answer
(A) $2.6x^{4}+c$
2023 Paper 1, Question 4
If the gradient of the function $f(x)$ is given by $\dfrac{20}{x^{3}}$, then $f(x)$ is equal to
"The gradient" means $f'(x)$, so integrate. Rewrite it as a power first: $20x^{-3}$. Adding one to $-3$ gives $-2$. It is not $-4$.
$$f(x)=\frac{20x^{-2}}{-2}+c=-10x^{-2}+c$$
Answer
(C) $-\dfrac{10}{x^{2}}+c$
2020 Paper 1, Question 2
Determine $\displaystyle\int \frac{e^{x}+1}{e^{x}}\,dx$.
Split the fraction before you integrate:
$$\frac{e^{x}+1}{e^{x}}=1+e^{-x}$$
The 1 integrates to $x$. For $e^{-x}$, the inside is $-x$, so divide by $-1$ (more on this in Part 2). That sign is the difference between (A) and (B).
Answer
(A) $x-e^{-x}+c$
2025 Paper 1, Question 11
Determine the following integrals.
(a) $\displaystyle\int \frac{1}{x}\,dx$[1 mark]
(b) $\displaystyle\int e^{4x}\,dx$[1 mark]
(c) $\displaystyle\int x\left(x^{3}+8\right)dx$[3 marks]
(a) Straight from the rules: $\ln x+c$.
1 markApplying the rule for $\frac{1}{x}$.
(b) The inside is $4x$, so divide by 4: $\frac{1}{4}e^{4x}+c$.
1 markIntegrating the exponential. Writing it as $\frac{e^{4x}}{4}$ is fine too.
(c) You cannot integrate a product one factor at a time, so expand first:
$$x\left(x^{3}+8\right)=x^{4}+8x$$
1 markExpanding the brackets. Both terms have to be right for this mark.
$$\int\left(x^{4}+8x\right)dx=\frac{x^{5}}{5}+4x^{2}+c$$
1 markIntegrating the $x^{4}$ term.
1 markIntegrating the $8x$ term. $\frac{8x^{2}}{2}$ is accepted, so you do not lose a mark for not simplifying.
Answer
(a) $\ln x+c$ · (b) $\frac{1}{4}e^{4x}+c$ · (c) $\frac{x^{5}}{5}+4x^{2}+c$
Part 2
A linear expression inside
When the inside of a function is linear, like $ax+b$, integrate as normal and then divide by $a$. That is because when you differentiate, the chain rule multiplies by $a$, so going backwards has to divide by it.
Brackets to a power
$$\int (ax+b)^{n}\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+c$$
Exponentials
$$\int e^{ax+b}\,dx=\frac{1}{a}e^{ax+b}+c$$
Logarithms
$$\int \frac{1}{ax+b}\,dx=\frac{1}{a}\ln(ax+b)+c$$
Trigonometric functions
$$\int \sin(ax+b)\,dx=-\frac{1}{a}\cos(ax+b)+c$$ $$\int \cos(ax+b)\,dx=\frac{1}{a}\sin(ax+b)+c$$
Watch the chain rule go backwards
The exponential and trig versions are not on your formula sheet, so learn them. If you are ever unsure, differentiate your answer. You should get back exactly what you started with.
Practise
Exam questions: a linear expression inside
2021 Paper 1, Question 3
Determine $\int 10e^{4x}\,dx$.
The inside is $4x$, so divide by 4: $10\times\frac{1}{4}e^{4x}=\frac{5}{2}e^{4x}$. (A) treats $e^{4x}$ like a power of $x$, which it is not. (B) multiplies by 4, which is differentiating.
Answer
(C) $\dfrac{5}{2}e^{4x}+c$
2025 Paper 1, Question 9
Determine $\displaystyle\int \frac{1}{(7x-4)}\,dx$, where $x>\frac{4}{7}$.
Use $\int\frac{1}{ax+b}\,dx=\frac{1}{a}\ln(ax+b)+c$ with $a=7$. Only the coefficient of $x$ matters. The $-4$ plays no part in the fraction out the front. The condition $x>\frac{4}{7}$ just keeps the inside of the log positive.
Answer
(A) $\frac{1}{7}\ln(7x-4)+c$
2020 Paper 1, Question 3
Determine $2\displaystyle\int (4x+6)^{3}\,dx$.
The power goes up to 4, so divide by 4. The inside is $4x+6$, so divide by $a=4$ as well. Then multiply by the 2 out the front:
$$2\times\frac{(4x+6)^{4}}{4\times4}=\frac{(4x+6)^{4}}{8}$$
Answer
(D) $\dfrac{(4x+6)^{4}}{8}+c$
2020 Paper 1, Question 9
Determine $\displaystyle\int \frac{x+1}{x^{2}+2x}\,dx$.
The inside here is not linear, so none of the rules above fit directly. On a multiple choice question you can always work backwards: differentiate each option and see which one gives you the original.
Differentiating (C) with the chain rule:
$$\frac{d}{dx}\left[\tfrac{1}{2}\ln\left(x^{2}+2x\right)\right]=\frac{1}{2}\cdot\frac{2x+2}{x^{2}+2x}=\frac{x+1}{x^{2}+2x}$$
It works because the numerator, $x+1$, is half the derivative of the denominator.
Answer
(C) $\frac{1}{2}\ln\left(x^{2}+2x\right)+c$
Part 3
Finding a function from its derivative
You will often be given $f'(x)$ and an initial condition, a point the function has to pass through, and be asked for $f(x)$. The method is the same every time:
- Integrate $f'(x)$, including $+c$.
- Identify the initial condition.
- Substitute it into $f(x)$ and solve for $c$.
Harder questions hide the initial condition in the words. "An empty vessel" means the volume is $0$ at $t=0$. "There are 100 at 7 pm" means $N(0)=100$ when $t$ is measured from 7 pm.
Try it yourself
Which curve is the right one?
Here $f'(x)=2x-2$, so $f(x)=x^{2}-2x+c$. Every value of $c$ gives a curve with exactly the same gradient. They are the same parabola, slid up or down. The initial condition $f(3)=1$ picks out just one of them. Move $c$ until your curve passes through the orange point.
The thin grey curves are the other members of the family, one for each whole-number value of .
Practise
Exam questions: finding a function from its derivative
2024 Paper 2, Question 3
The derivative of the function $f(x)$ is given by $f'(x)=\sin(2x)$. It is known that $f\left(\frac{\pi}{2}\right)=4$. Determine $f(x)$.
Integrate, dividing by 2 for the inside: $f(x)=-\frac{1}{2}\cos(2x)+c$. Only (C) has the $-\frac{1}{2}$, but check the constant anyway.
Substitute $x=\frac{\pi}{2}$: $-\frac{1}{2}\cos\pi+c=\frac{1}{2}+c=4$, so $c=3.5$.
Answer
(C) $-\frac{1}{2}\cos(2x)+3.5$
2021 Paper 2, Question 7
Determine $f(x)$, given $f'(x)=6x^{2}+\dfrac{1}{x^{2}}+\dfrac{1}{x}$ and $f(1)=5$.
Write the middle term as $x^{-2}$, which integrates to $-x^{-1}$. The last term is $\frac{1}{x}$, which integrates to $\ln x$. Several options get one of these wrong.
$$f(x)=2x^{3}-\frac{1}{x}+\ln x+c$$
Then $f(1)=2-1+0+c=5$, so $c=4$.
Answer
(B) $f(x)=2x^{3}-\dfrac{1}{x}+\ln(x)+4$
2024 Paper 1, Question 13
(a) $F(x)=\displaystyle\int\left(4e^{2x}+\sin(2x)\right)dx$. Use integration to determine $F(x)$, if $F(0)=5$.[3 marks]
(b) If $\dfrac{dy}{dx}=\left(\dfrac{3x^{7}-2x}{x^{4}}\right)^{2}$, determine $y$.[3 marks]
(a) Both terms have a linear inside of $2x$, so divide each by 2:
$$F(x)=2e^{2x}-\frac{1}{2}\cos(2x)+c$$
1 markIntegrating the exponential term.
1 markIntegrating the trig term, minus sign included.
Substitute $F(0)=5$. Remember $e^{0}=1$ and $\cos0=1$: $2-\frac{1}{2}+c=5$, so $c=\frac{7}{2}$.
1 markFinding the constant. $3.5$ is fine. Follow-through applies if your integral was wrong.
(b) Simplify the fraction first, then expand the square. The middle term is $2\times3x^{3}\times\left(-2x^{-3}\right)=-12$:
$$\frac{dy}{dx}=\left(3x^{3}-2x^{-3}\right)^{2}=9x^{6}-12+4x^{-6}$$
1 markExpanding the squared bracket.
1 markSimplifying into separate terms, so each one is a power of $x$ you can integrate.
Now integrate term by term. There is no initial condition here, so the $+c$ stays.
$$y=\frac{9x^{7}}{7}-12x-\frac{4x^{-5}}{5}+c$$
1 markIntegrating every term.
Answer
(a) $F(x)=2e^{2x}-\frac{1}{2}\cos(2x)+\frac{7}{2}$ · (b) $y=\frac{9}{7}x^{7}-12x-\frac{4}{5}x^{-5}+c$
2025 Paper 2, Question 11
(a) Determine the equation of $f(x)$ if $f'(x)=\cos(x)-4\sin(2x+\pi)$ and $f(\pi)=2$.[4 marks]
(b) Use your result from Question 11(a) to determine the value of $x$ where $f(x)=4.8$, for $2\le x\le 4$.[1 mark]
(c) Use your result from Question 11(a) to determine the minimum value of $f(x)$, for $2\le x\le 4$.[1 mark]
(a) Integrate each term. For the second, the inside is $2x+\pi$, so divide by 2. The minus signs need care: $-4\times\left(-\frac{1}{2}\cos(2x+\pi)\right)=+2\cos(2x+\pi)$.
$$f(x)=\sin x+2\cos(2x+\pi)+c$$
1 markIntegrating the cosine term.
1 markIntegrating the sine term, including the divide by 2.
Substitute $f(\pi)=2$: $\sin\pi+2\cos3\pi+c=0-2+c=2$, so $c=4$.
1 markSubstituting $x=\pi$ and $f(\pi)=2$ into your $f(x)$.
1 markFinding $c=4$. You do not need to write the whole function out again for this mark.
(b) This is Paper 2, so use your calculator. Graph $f(x)$ and $y=4.8$ over $2\le x\le4$ and find the intersection: $x\approx2.34$.
1 markThe solution in the domain. Any number of decimal places is accepted.
(c) Use the calculator's minimum tool on the same domain: $f(x)\approx1.94$ at $x\approx3.27$. The end points give $f(2)\approx6.22$ and $f(4)\approx3.53$, which are both bigger.
1 markThe minimum value, to any number of decimal places.
Answer
(a) $f(x)=\sin x+2\cos(2x+\pi)+4$ · (b) $x\approx2.34$ · (c) $1.94$
Part 4
Modelling with integrals
In modelling questions you are given a rate, such as litres per hour or people per year, and asked about the amount. Integrate the rate to get the amount, then find $c$ from a starting value hidden in the story.
Units are a good check. If the rate is in litres per hour and $t$ is in hours, your integral is in litres.
Practise
Exam questions: modelling
2022 Paper 1, Question 14
The rate that water fills an empty vessel is given by $\dfrac{dV}{dt}=0.25e^{0.25t}$ (in litres per hour), $0\le t\le 8\ln(6)$, 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.
| $t$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $\dfrac{dV}{dt}$ | $0.25$ | $0.32$ | $0.41$ | $0.53$ |
(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, dividing by 0.25 for the inside: $V=e^{0.25t}+c$. The hidden initial condition is the word "empty", so $V(0)=0$. That gives $1+c=0$ and $c=-1$.
1 markIntegrating the rate. The $+c$ is not needed for this mark.
1 markFinding $c=-1$ from the empty vessel.
(b) Substitute $t=8\ln6$. The log laws do the rest: $e^{0.25\times8\ln6}=e^{2\ln6}=e^{\ln36}=36$, so $V=36-1=35$.
1 markSimplifying the exponential term to 36.
1 markThe volume in litres.
(c) The trapezoidal rule is from Unit 4, but it is often tacked on like this. With strips of width 1:
$$V(3)\approx\frac{1}{2}\Big[0.25+2(0.32+0.41)+0.53\Big]=\frac{1}{2}(2.24)=1.12$$
1 markSetting up the trapezoidal rule with the values from the table.
1 markThe approximate volume.
That agrees with the exact value from part (a), $e^{0.75}-1\approx1.12$.
Answer
(a) $V=e^{0.25t}-1$ · (b) 35 L · (c) about 1.12 L
2020 Paper 2, Question 12
The rates of change in population for two cities are given by
$$\text{City A: }A'(t)=\frac{45}{t+1},\qquad \text{City B: }B'(t)=105e^{0.03t}$$
where $t$ is the number of years since 2018 and both $A'(t)$ and $B'(t)$ are measured in people per year. At the beginning of 2018, City A had a population of 5000, and City B had a population of 3500.
(a) Determine the population models for both cities.[3 marks]
(b) Use the information in 12(a) to predict the population of City B at the beginning of 2028.[1 mark]
(c) Use the information in 12(a) to predict the year in which the population of both cities will be the same.[3 marks]
(a) City A: $A(t)=45\ln(t+1)+c$. At $t=0$, $\ln1=0$, so $c=5000$.
1 markThe general model for City A, from integrating.
City B: divide by 0.03 for the inside. $\frac{105}{0.03}=3500$, so $B(t)=3500e^{0.03t}+c$. At $t=0$, $3500+c=3500$, so $c=0$.
1 markThe general model for City B.
1 markBoth models with their constants found from the 2018 populations.
(b) The beginning of 2028 is $t=10$: $B(10)=3500e^{0.3}\approx4725$.
1 markThe population of City B. 4724 and 4725 are both accepted, but a decimal is not, because people come in whole numbers.
(c) Solve $45\ln(t+1)+5000=3500e^{0.03t}$ on your calculator by graphing both and finding the intersection: $t\approx12.7$. Then turn that back into a year: 12.7 years after the start of 2018 is during 2030.
1 markShowing your approach, either the equation or the two graphs and their intersection.
1 markSolving to get $t\approx12.7$.
1 markTurning that into the year, 2030. July or August 2030 is fine too.
Answer
(a) $A(t)=45\ln(t+1)+5000$, $B(t)=3500e^{0.03t}$ · (b) about 4725 people · (c) 2030
2020 Paper 1, Question 20
At the end of the first stage of its growth cycle, a species of tree has a height of 5 metres and a trunk radius of 15 cm.
In the second stage of its growth cycle, the tree stays at this height for the next 10 years. However, the growth rate of the trunk radius (in cm per year) varies over the 10 years and is given by the function
$$r(t)=1.5+\sin\left(\frac{\pi t}{5}\right)$$
Assume the density (mass per unit volume) of the tree trunk is approximately 1 g/cm³ and the tree trunk is in the shape of a cylinder.
Determine the ratio of the trunk's mass at the end of the second stage to its mass at the end of the first stage.
Careful with the notation: $r(t)$ here is the growth rate. It is not the radius. Call the radius $R(t)$, so $R'(t)=1.5+\sin\left(\frac{\pi t}{5}\right)$. Integrate, dividing by $\frac{\pi}{5}$ for the inside:
$$R(t)=1.5t-\frac{5}{\pi}\cos\left(\frac{\pi t}{5}\right)+c$$
1 markAn integrated expression for the radius. A definite integral from 0 to 10 works too.
The hidden initial condition is the start of the second stage: $R(0)=15$. Since $\cos0=1$, $-\frac{5}{\pi}+c=15$, so $c=15+\frac{5}{\pi}$.
After 10 years, $\cos2\pi=1$, so the $\frac{5}{\pi}$ terms cancel: $R(10)=15-\frac{5}{\pi}+15+\frac{5}{\pi}=30$ cm. The radius has doubled.
1 markThe radius. A value or a function is accepted.
Mass is density times volume, and the volume of a cylinder is $\pi R^{2}h$. The height stays at 5 m, which is 500 cm. Keep the units the same as the radius.
1 markUsing the cylinder formula with your radius to set up the volume.
$$V_{0}=\pi(15)^{2}(500)=112\,500\pi\text{ cm}^{3},\qquad V_{10}=\pi(30)^{2}(500)=450\,000\pi\text{ cm}^{3}$$
1 markBoth volumes, at the start and after 10 years.
The density is 1 g/cm³, so each mass in grams equals its volume, and the ratio is
$$\frac{450\,000\pi}{112\,500\pi}=4$$
1 markThe ratio of the masses.
Answer
The ratio is $4:1$
1 markLogical organisation: define your variables, and use linking statements that show where each given fact is used.
The full walkthrough of 2020 Paper 1, Question 20 goes through it mark by mark.
2022 Paper 2, Question 19
Flying foxes enter and leave a fruit-growing region every evening. The rate at which the flying foxes enter the region is modelled by the function
$$A(t)=42\sin\left(0.03t-\frac{\pi}{3}\right)+71,\quad 0\le t\le 240$$
The rate at which the flying foxes leave the region is modelled by the function
$$L(t)=42\sin\left(0.04t-\frac{\pi}{3}\right)+42,\quad 0\le t\le 240$$
Both $A(t)$ and $L(t)$ are measured in animals per minute and $t$ is measured in minutes after 7 pm. There are 100 flying foxes in the region at 7 pm.
Determine the maximum number of flying foxes in the region and the time that this occurs.
The number in the region, $N(t)$, changes at the rate "in minus out": $N'(t)=A(t)-L(t)$. Integrate, dividing by 0.03 and 0.04 for the insides:
$$N(t)=-1400\cos\left(0.03t-\frac{\pi}{3}\right)+1050\cos\left(0.04t-\frac{\pi}{3}\right)+29t+c$$
Use $N(0)=100$. Since $\cos\left(-\frac{\pi}{3}\right)=\frac{1}{2}$, you get $-700+525+c=100$, so $c=275$.
1 markThe function for the number of foxes. Integrating each rate on its own and subtracting is fine too.
The number is greatest when foxes stop arriving faster than they leave, so solve $A(t)=L(t)$ on your calculator: $t\approx177.7$. $N'(t)$ changes from positive to negative there, so it is a maximum. The end point is lower: $N(240)\approx5171$.
1 markA method to find the maximum, such as graphing $N(t)$ or solving $A(t)=L(t)$.
Then $N(177.7)\approx7034$, and 177.7 minutes after 7 pm is about 9:58 pm.
1 markThe turning point, about $(177.7,\ 7034)$.
1 markThe number as a whole number of foxes, with the time. 178 minutes after 7 pm is accepted too.
Answer
About 7034 flying foxes, at about 9:58 pm
The full walkthrough of 2022 Paper 2, Question 19 goes through it mark by mark.
Game plan
Your game plan for integrating
Work through these whenever you see an integral sign or a rate you need to undo.
- 1Simplify first. Expand brackets, split fractions and rewrite roots and reciprocals as powers.
- 2Integrate term by term. If the inside is $ax+b$, divide by $a$.
- 3Add $+c$.
- 4Look for an initial condition, stated or hidden in the words, and substitute it to find $c$.
- 5Check by differentiating your answer. You should get back what you started with.
Practice
Practice questions
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. For any integral, you can check your answer by differentiating it.
Question 1 of 8
Which of the following is not an antiderivative of $2x$?
Differentiate each option. (A), (B) and (C) all give $2x$, because the constant disappears when you differentiate. That is why every integral needs a $+c$. (D) differentiates to $4x$, so it is not an antiderivative of $2x$.
Answer
(D) $2x^{2}$
Question 2 of 8
Determine $\displaystyle\int\left(3\sqrt{x}-\frac{4}{x^{2}}+\frac{2}{x}\right)dx$, for $x>0$.
1Rewrite the first two terms as powers of $x$. Leave $\frac{2}{x}$ as it is, because it needs the log rule.
$$\int\left(3x^{\frac{1}{2}}-4x^{-2}+\frac{2}{x}\right)dx$$
2Integrate one term at a time. Add one to each power and divide by the new power.
$$3\times\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-4\times\frac{x^{-1}}{-1}+2\ln x+c$$
3Tidy up. Dividing by $\frac{3}{2}$ is the same as multiplying by $\frac{2}{3}$, and the two minus signs make a plus.
Answer
$2x^{\frac{3}{2}}+\dfrac{4}{x}+2\ln x+c$
Question 3 of 8
Determine $\displaystyle\int\frac{\left(x^{2}+1\right)^{2}}{x^{2}}\,dx$.
1You cannot integrate the top and bottom separately. Expand the top first.
$$\frac{x^{4}+2x^{2}+1}{x^{2}}$$
2Split the fraction, so each term is divided by $x^{2}$.
$$x^{2}+2+x^{-2}$$
3Now integrate term by term.
$$\frac{x^{3}}{3}+2x+\frac{x^{-1}}{-1}+c$$
Answer
$\dfrac{x^{3}}{3}+2x-\dfrac{1}{x}+c$
Question 4 of 8
Determine $\displaystyle\int\sin(3x)\,dx$.
The integral of $\sin$ is $-\cos$, and the linear inside $3x$ means you divide by 3. Check by differentiating: the chain rule multiplies by 3, which cancels the $\frac{1}{3}$ and gives back $\sin(3x)$.
Option (A) multiplies by 3 instead of dividing, and (C) has lost the minus sign.
Answer
(B) $-\frac{1}{3}\cos(3x)+c$
Question 5 of 8
Determine the following integrals.
(a) $\displaystyle\int\left(e^{2x}+1\right)^{2}dx$[2 marks]
(b) $\displaystyle\int\frac{6}{3x-1}\,dx$, for $x>\frac{1}{3}$[2 marks]
(a) Expand the square first. Remember $\left(e^{2x}\right)^{2}=e^{4x}$, by the index laws.
$$\int\left(e^{4x}+2e^{2x}+1\right)dx$$
Each exponential has a linear power, so divide by its coefficient: $\frac{e^{4x}}{4}$, then $\frac{2e^{2x}}{2}=e^{2x}$, then $x$.
(b) Take the 6 out the front. The inside is $3x-1$, so divide by 3.
$$6\times\frac{1}{3}\ln(3x-1)+c=2\ln(3x-1)+c$$
Answer
(a) $\dfrac{e^{4x}}{4}+e^{2x}+x+c$ · (b) $2\ln(3x-1)+c$
Question 6 of 8
Determine $\displaystyle\int\frac{8}{(2x+1)^{3}}\,dx$.
1Rewrite the fraction as a bracket to a negative power.
$$\int 8(2x+1)^{-3}\,dx$$
2Use the bracket rule. Add one to the power to get $-2$, then divide by the new power and by the 2 from the inside.
$$8\times\frac{(2x+1)^{-2}}{2\times(-2)}+c=-2(2x+1)^{-2}+c$$
Check by differentiating. The power comes down, $-2\times-2=4$, and the chain rule multiplies by 2, which gives back $8(2x+1)^{-3}$.
Answer
$-\dfrac{2}{(2x+1)^{2}}+c$
Question 7 of 8
The rate of change in the number of members of a gym is given by $\dfrac{dN}{dt}=12e^{0.2t}$ members per week, where $t$ is the time in weeks since the gym opened. Initially, the gym had 100 members. Determine the number of members five weeks after the gym opened. Express your answer in terms of $e$.
1You are given a rate and asked for an amount, so integrate.
2The inside is $0.2t$, so divide by 0.2. $\frac{12}{0.2}=60$.
$$N=\int 12e^{0.2t}\,dt=60e^{0.2t}+c$$
3Find $c$. "Initially 100 members" means $N(0)=100$, and $e^{0}=1$.
$$60+c=100\quad\Rightarrow\quad c=40$$
4Substitute $t=5$. Since $0.2\times5=1$, the exponential is just $e$.
$$N(5)=60e+40$$
Answer
$60e+40$ members
Question 8 of 8
A town's population changes at the rate $P'(t)=200e^{0.05t}-150$ people per year, where $t$ is the number of years since the start of 2020. At the start of 2020 the population was 12 000. Determine a model for the population, and the year in which it first reaches 20 000.
1Integrate the rate. For the exponential, divide by 0.05: $\frac{200}{0.05}=4000$.
$$P(t)=4000e^{0.05t}-150t+c$$
2Use $P(0)=12\,000$.
$$4000+c=12\,000\quad\Rightarrow\quad c=8000$$
3Solve $4000e^{0.05t}-150t+8000=20\,000$ on your calculator, by graphing both sides and finding the intersection.
$$t\approx27.97$$
4Turn that into a year. 27.97 years after the start of 2020 is late in 2047.
Do not forget the $-150t$ term. A model with only the exponential grows far too quickly.
Answer
$P(t)=4000e^{0.05t}-150t+8000$, and the population reaches 20 000 during 2047
Keep practising
Ready for something harder?
The tree trunk and flying fox questions both have their own complex unfamiliar walkthroughs. In each one the hard part is spotting what to integrate, so try them from a blank page first.
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