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Topic 5.08 · SL and HL

Indefinite integrals + u-substitution: notes and practice questions

Summary
  • Indefinite integral: The reverse process of differentiation, represented as:

∫f(x) dx=F(x)+C \int f(x) \, dx = F(x) + C
where F′(x)=f(x)F'(x) = f(x) and CC is the constant of integration.

  • **uu-substitution**: A method for simplifying integrals by substituting u=g(x)u = g(x) and du=g′(x)dxdu = g'(x)dx. Used when f(x)f(x) includes a composite function.

How it is examined

The 1a\frac{1}{a} from a linear composite, and the constant of integration, are the two marks students give away. ln⁡∣x∣\ln|x| without the modulus is a real deduction. 3 to 6 marks, Paper 1.

Given in the booklet

The four standard integrals (∫1x\int \frac{1}{x}, ∫sin⁡x\int \sin x, ∫cos⁡x\int \cos x, ∫ex\int e^x) are given at SL 5.10, and ∫xn\int x^n at SL 5.5. The linear-composite forms are not given, so the 1a\frac{1}{a} factor is recall.

Key ideas
  • Indefinite integral of xnx^n (n∈Qn \in \mathbb{Q}), sin⁡x\sin x, cos⁡x\cos x, 1x\dfrac{1}{x} and exe^x.
  • The composites of any of these with the linear function ax+bax + b.
  • Integration by inspection (reverse chain rule) or by substitution for expressions of the form ∫k g′(x)f(g(x)) dx\displaystyle\int k\,g'(x)f(g(x))\,\mathrm{d}x.
At HL

Extended at AHL 5.15 and AHL 5.16.

Linking questions

  • The ∫kg′(x)f(g(x)) dx\int k g'(x) f(g(x))\,\mathrm{d}x shape is the only substitution SL is asked to spot unaided. At HL (AHL 5.16) any other substitution will be provided.

Practice questions

49 questions · 2 easy · 28 medium · 19 hard
Showing 20 of 20

Question 1

EasyPaper 1 · no calculator5 marks

Consider the derivative of a function h′(x)=−5(x+1)2+2exh'(x) = - \frac{5}{(x + 1)^{2}} + 2e^{x}. Find the equation of h(x)h(x) given that h(0)=9h(0) = 9.

Question 2

MediumPaper 1 · no calculator4 marks

The gradient of a curve is given by dydx=e2x−2\frac{dy}{dx} = e^{2x-2}. The curve passes through the point (1,4)(1, 4).

(a) Find the equation of the curve.

Question 3

HardPaper 1 · no calculator17 marks
(a)

By using an appropriate substitution, show that ∫sin⁡(x) dx=2sin⁡(x)−2xcos⁡(x)+C\int \sin(\sqrt{x}) \, dx = 2\sin(\sqrt{x}) - 2\sqrt{x} \cos(\sqrt{x}) + C.

[6]
(b)

The following diagram shows part of the curve y=sin⁡(x)y = \sin(\sqrt{x}) for x≥0x \ge 0.

Graph of y = sin(sqrt(x) ) showing x-intercepts and regions R1, R2, R3

The curve intersects the x-axis at x1,x2,x3,…x_1, x_2, x_3, \dots.

The nth x-intercept of the curve, xnx_n, is given by xn=n2π2x_n = n^2 \pi^2, where n∈Z+n \in \mathbb{Z}^+.

Write down an expression for xn+1x_{n+1}.

[1]
(c)

The regions bounded by the curve and the x-axis are denoted by R1,R2,R3,…R_1, R_2, R_3, \dots as shown on the diagram.

Calculate the area of region RnR_n.

Give your answer in the form (an+b)π(an+b)\pi, where a,b∈Z+a, b \in \mathbb{Z}^+.

[7]
(d)

Hence, show that the areas of the regions R1,R2,R3,…R_1, R_2, R_3, \dots form an arithmetic sequence.

[3]

Question 4

EasyPaper 1 · no calculator4 marks

Given that dydx=e2x−3\frac{dy}{dx} = e^{2x-3}, and y=5y = 5 when x=32x = \frac{3}{2}, find yy in terms of xx.

Question 5

MediumPaper 1 · no calculator5 marks

The gradient of the tangent to a curve y=f(x)y = f(x) is given by f′(x)=2e2x−1f'(x) = 2e^{2x-1}. The curve passes through the point (12,5)(\frac{1}{2}, 5).

(a) Find f(x)f(x).

Question 6

HardPaper 1 · no calculator15 marks
(a)(i)

Expand and simplify (1+a)3(1+a)^3 in ascending powers of aa.

[2]
(a)(ii)

By using a suitable substitution for aa, show that 1+3cos⁡(2x)+3cos⁡2(2x)+cos⁡3(2x)=8cos⁡6(x)1+3\cos(2x)+3\cos^2(2x)+\cos^3(2x) = 8\cos^6(x).

[4]
(b)(i)

Consider the function g(x)=4sin⁡(x)(1+3cos⁡(2x)+3cos⁡2(2x)+cos⁡3(2x))g(x) = 4\sin(x)(1+3\cos(2x)+3\cos^2(2x)+\cos^3(2x) ).

Show that ∫0pg(x) dx=327(1−cos⁡7p)\int_0^p g(x) \, dx = \frac{32}{7}(1-\cos^7 p), where pp is a positive real constant.

[4]
(b)(ii)

It is given that ∫pπ2g(x) dx=128\int_p^{\frac{\pi}{2}} g(x) \, dx = \frac{1}{28}, where 0≤p≤π20 \le p \le \frac{\pi}{2}. Find the value of pp.

[5]

Question 7

MediumPaper 1 · no calculator4 marks

Given that dydx=sec⁡2(x2)\frac{dy}{dx} = \sec^2\left(\frac{x}{2}\right), and y=5y = 5 when x=π2x = \frac{\pi}{2}, find yy in terms of xx.

Question 8

HardPaper 1 · no calculator20 marks
(a)

Consider the family of integrals defined by In=∫xne−x dxI_n = \int x^n e^{-x} \, dx for n∈N0n \in \mathbb{N}_0, where N0={0,1,2,...}\mathbb{N}_0 = \{0, 1, 2, ...\}.

(a) By using integration by parts, show that In=−xne−x+nIn−1I_n = -x^n e^{-x} + n I_{n-1} for n≥1n \ge 1.

[3]
(b)

(b) Hence, find an explicit expression for ∫x3e−x dx\int x^3 e^{-x} \, dx.

[4]
(c)

(c) The region RR is enclosed by the graph of y=x3/2e−x/2y = x^{3/2} e^{-x/2} and the xx-axis for x≥0x \ge 0. The region RR is rotated by 2π2\pi radians about the xx-axis. Find the volume of the solid generated.

[5]
(d)

(d) Show that lim⁡x→∞xne−x=0\lim_{x \to \infty} x^n e^{-x} = 0 for any n∈Nn \in \mathbb{N}.

[3]
(e)(i)

Consider the function h(x)=xe−xh(x) = x e^{-x}.

(e) (i) Find the Maclaurin series for h(x)h(x) up to and including the term in x4x^4.

[3]
(e)(ii)

(ii) Hence, find the value of the fourth derivative of h(x)h(x) at x=0x=0, i.e. h(4)(0)h^{(4)}(0).

[2]

Question 9

MediumPaper 1 · no calculator4 marks

Given that f′(x)=e2x+1f'(x) = e^{2x+1}, and the point (−12,3)(-\frac{1}{2}, 3) lies on the graph of f(x)f(x), find f(x)f(x).

Question 10

HardPaper 1 · no calculator8 marks

Consider the homogeneous differential equation dydx=y2−x22xy\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}, for x>0x > 0 and y>0y > 0.

It is given that y=3y = \sqrt{3} when x=1x = 1.

By using the substitution y=vxy = vx, show that the solution to the differential equation is x2+y2=4xx^2 + y^2 = 4x.

Question 11

MediumPaper 1 · no calculator8 marks
(a)

The continuous random variable XX has probability density function

f(x)={k1+4x2,0≤x≤120,otherwise.f(x) = \begin{cases} \frac{k}{1+4x^2}, & 0 \le x \le \frac{1}{2} \\ 0, & \text{otherwise.} \end{cases}

(a) Find the value of kk.

[4]
(b)

(b) Find E(X)E(X).

[4]

Question 12

HardPaper 1 · no calculator13 marks
(a)

Consider the function ff defined by f(x)=6x2(x3+1)2f(x)=\frac{6x^2}{(x^3+1)^2} for x≥0x \ge 0. The graph of ff is shown in the following diagram.

Graph of f(x) showing a curve starting at the origin, rising to a maximum, and then approaching the x-axis as x increases

Show that f′(x)=12x(1−2x3)(x3+1)3f'(x)=\frac{12x(1-2x^3)}{(x^3+1)^3}.

[4]
(b)

Find ∫f(x)dx\int f(x)dx.

[4]
(c)

Consider a function g(x)g(x) defined for x≥0x \ge 0. The derivative of gg is such that g′(x)=f′(x)g'(x) = f'(x), for all x≥0x \ge 0.

Let RR be the region enclosed by the graph of ff, the graph of gg, the line x=0x = 0 and the line x=2x = 2. The area of RR is 163\frac{16}{3}.

Find the two possible expressions for g(x)g(x).

[5]

Question 13

MediumPaper 1 · no calculator7 marks

By using the substitution u=exu = e^x, find ∫exe2x−2ex−8dx\int \frac{e^x}{e^{2x} - 2e^x - 8} \mathrm{d}x.

Question 14

HardPaper 2 · calculator20 marks
(a)

The rate of change of a certain quantity RR with respect to a variable xx is given by R′(x)=1x(M−x)R'(x)=\frac{1}{x(M-x)}, x∈Rx \in \mathbb{R}, x≠0x \neq 0, x≠Mx \neq M where MM is a positive constant.

The expression for R′(x)R'(x) can be written in the form Ax+BM−x\frac{A}{x} + \frac{B}{M-x}, where A,B∈RA, B \in \mathbb{R}.

Find AA and BB in terms of MM.

[3]
(b)

Hence, find an expression for R(x)R(x).

[3]
(c)

The concentration of a certain chemical product, CC (in mol/L), in a reaction vessel at time tt (in minutes) can be modelled by the differential equation dCdt=C(L−C)8L\frac{dC}{dt} = \frac{C(L-C)}{8L}, where LL is the maximum possible concentration and C(0)=0.2C(0) = 0.2 mol/L is the initial concentration.

By solving the differential equation, show that C=0.2L(L−0.2)e−t8+0.2C = \frac{0.2 L}{(L-0.2)e^{-\frac{t}{8}}+0.2}.

[8]
(d)

At t=12t=12 minutes, the concentration of the product has reached 0.60.6 mol/L.

Find the value of LL, giving your answer correct to four significant figures.

[3]
(e)

Find the value of tt when the rate of change of the concentration is at its maximum.

[3]

Question 15

MediumPaper 1 · no calculator6 marks

The following diagram shows part of the graph of y=sin⁡x3+cos⁡xy = \frac{\sin x}{3 + \cos x} for x≥0x \ge 0.

Graph of y = sin(x)/(3+cos(x) ) with shaded region R

The shaded region RR is bounded by the curve, the x-axis, the y-axis and the line x=ax = a.

The area of RR is ln⁡(43)\ln\left(\frac{4}{3}\right).

Find the value of aa.

Question 16

HardPaper 2 · calculator21 marks
(a)

The growth of a bacterial colony, BB, in a petri dish can be modelled by the logistic differential equation

dBdt=kB(1−BN)\frac{\text{d}B}{\text{d}t} = k B \left(1 - \frac{B}{N}\right)

where tt is the time measured in hours and k,Nk, N are positive constants.

The constant NN represents the maximum number of bacteria the petri dish can sustain indefinitely due to limited nutrients.

In the context of this bacterial growth model, interpret the meaning of dBdt\frac{\text{d}B}{\text{d}t}.

[1]
(b)

Show that d2Bdt2=k2B(1−BN)(1−2BN)\frac{\text{d}^2B}{\text{d}t^2} = k^2B\left(1-\frac{B}{N}\right)\left(1-\frac{2B}{N}\right).

[4]
(c)

Hence show that the bacterial colony will grow at its maximum rate when B=N2B = \frac{N}{2}. Justify your answer.

[5]
(d)

Hence determine the maximum value of dBdt\frac{\text{d}B}{\text{d}t} in terms of kk and NN.

[2]
(e)

Let B0B_0 be the initial number of bacteria.

By solving the logistic differential equation, show that its solution can be expressed in the form

kt=ln⁡(B(N−B0)B0(N−B))kt = \ln\left(\frac{B(N-B_0)}{B_0(N-B)}\right).

[7]
(f)

After 5 hours, the number of bacteria is 2B02B_0. It is known that N=3B0N = 3B_0.

Find the value of kk for this bacterial growth model.

[2]

Question 17

MediumPaper 1 · no calculator4 marks

Given that f′(x)=3e1−2xf'(x) = 3e^{1-2x} and f(12)=5f(\frac{1}{2}) = 5, find f(x)f(x).

Question 18

HardPaper 2 · calculator24 marks
(a)(i)

A team of engineers is designing a new roller coaster ride. The path of a certain section of the ride can be modelled by the function g(x)=x2+3x−10x−4g(x)=\frac{x^2 + 3x - 10}{x-4}, where xx is the horizontal distance in metres from the starting point and g(x)g(x) is the vertical height in metres. The domain of the function is x∈R,x≠4x \in \mathbb{R}, x\neq 4.

Find the coordinates where the path of the roller coaster crosses the horizontal ground (x-axis).

[3]
(a)(ii)

Find the coordinates where the path of the roller coaster crosses the vertical axis (y-axis).

[1]
(b)

Write down the equation of the vertical asymptote of the graph of gg.

[1]
(c)

The oblique asymptote of the graph of gg can be written as y=ax+by = ax + b where a,b∈Za, b \in \mathbb{Z}.

Find the value of aa and the value of bb.

[4]
(d)

Sketch the graph of gg for −20≤x≤20-20 \le x \le 20, clearly indicating the points of intersection with each axis and any asymptotes.

[3]
(e)(i)

Engineers want to analyse the inverse of the roller coaster's height function, h(x)=1g(x)h(x) = \frac{1}{g(x)}.

Express h(x)h(x) in partial fractions.

[7]
(e)(ii)

Hence find the exact value of ∫01h(x)dx\int_{0}^{1} h(x) dx, expressing your answer as a single logarithm.

[5]

Question 19

MediumPaper 1 · no calculator6 marks

The function ff is defined as f(x)=arctan⁡x1+x2f(x) = \sqrt{\frac{\arctan x}{1+x^2}}, where x≥0x \ge 0.

Consider the shaded region R enclosed by the graph of ff, the xx-axis and the line x=1x = 1, as shown in the following diagram.

Graph of y=f(x) with shaded region R

The shaded region R is rotated by 2π2\pi radians about the xx-axis to form a solid.

Show that the volume of the solid is π332\frac{\pi^3}{32}.

Question 20

HardPaper 2 · calculator19 marks
(a)

(a) In a controlled biological experiment, the rate of change of the population PP of a certain microorganism with respect to time tt is modeled by the differential equation t2dPdt=P2−2tP+2t2t^2 \frac{dP}{dt} = P^2 - 2tP + 2t^2, where t>0t > 0 is in hours and PP is in thousands of organisms. It is known that at t=1t = 1 hour, the population is P=4P = 4 thousand.

Use Euler's method, with a step length of 0.1, to find an approximate value of PP when t=1.4t = 1.4.

[4]
(b)

(b) Use the substitution P=vtP = vt to show that tdvdt=v2−3v+2t\frac{dv}{dt} = v^2 - 3v + 2.

[3]
(c)(i)

(c.i) By solving the differential equation from part (b), and given that P>2tP > 2t, show that P=6t−2t23−2tP = \frac{6t - 2t^2}{3 - 2t}.

[10]
(c)(ii)

(c.ii) Find the actual value of PP when t=1.4t = 1.4.

[1]
(c)(iii)

(c.iii) Using the graph of P=6t−2t23−2tP = \frac{6t - 2t^2}{3 - 2t}, suggest a reason why the approximation given by Euler's method in part (a) is not a good estimate to the actual value of PP at t=1.4t = 1.4.

[1]

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What does Indefinite integrals + u-substitution cover in IB Maths AA?

Indefinite integral: The reverse process of differentiation, represented as:. $$. \int f(x) \, dx = F(x) + C.

Is Indefinite integrals + u-substitution SL or HL?

Both. SL and HL students study Indefinite integrals + u-substitution, and HL goes further: Extended at AHL 5.15 and AHL 5.16.

How do I revise Indefinite integrals + u-substitution for IB Maths AA?

Start from the core idea: indefinite integral: The reverse process of differentiation, represented as:. In the exam: the (1)/(a) from a linear composite, and the constant of integration, are the two marks students give away. ln|x| without the modulus is a real deduction. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Indefinite integrals + u-substitution?

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