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Topic 5.19 · HL only

Integration by parts (+ repeated by parts): notes and practice questions

Summary
  • Integration by parts uses the product rule in reverse to integrate products of functions.
  • The formula is ∫udvdxdx=uv−∫vdudxdx\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx.
  • Choose uu to simplify upon differentiation and dvdv to be easily integrable, often following the LIATE order (Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential).
  • For polynomial ×\times exponential/trig, let uu be the polynomial to reduce its power.
  • For logarithms or inverse trig functions, let uu be that function and dv=dxdv=dx to eliminate it.
  • Cyclic integrals (exponential ×\times trig) require two applications and algebraic solving.
  • For definite integrals, evaluate uvuv at the limits and subtract the integrated remaining part.

How it is examined

The substitution rule is a gift for question generation: if the integral is not the reverse chain rule shape, the question must supply the substitution. Changing the limits when substituting in a definite integral, rather than substituting back, is the standard examiner complaint. ∫exsin⁡x dx\int e^x \sin x\,\mathrm{d}x needs the by-parts twice and then rearranging for the original integral, which is a full question on its own. 6 to 8 marks.

Given in the booklet

The by-parts formula ∫udvdxdx=uv−∫vdudxdx\displaystyle\int u\frac{\mathrm{d}v}{\mathrm{d}x}\mathrm{d}x = uv - \int v\frac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x is given.

Key ideas
  • Integration by substitution.
  • Integration by parts.
  • Repeated integration by parts.

Linking questions

  • ∫ln⁡x dx\int \ln x\,\mathrm{d}x and ∫arcsin⁡x dx\int \arcsin x\,\mathrm{d}x are by parts with dv=dx\mathrm{d}v = \mathrm{d}x, which is the trick worth teaching and worth asking.

Practice questions

15 questions · 3 medium · 12 hard
Showing 15 of 15

Question 1

MediumPaper 1 · no calculator7 marks

Solve the differential equation xdydx+y=xcos⁡xx\frac{dy}{dx} + y = x \cos x, for x>0x > 0, given that y=3y = 3 when x=πx = \pi.

Give your answer in the form y=f(x)y = f(x).

Question 2

HardPaper 1 · no calculator9 marks
(a)

Find ∫x2sin⁡(2x)dx\int x^2 \sin(2x) dx.

[6]
(b)

Hence, find the exact value of ∫0π4x2sin⁡(2x)dx\int_0^{\frac{\pi}{4}} x^2 \sin(2x) dx.

[3]

Question 3

MediumPaper 1 · no calculator6 marks

Find ∫(ln⁡x)2 dx\int (\ln x)^2 \, dx.

Question 4

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 5

MediumPaper 1 · no calculator5 marks

Find ∫arctan(2x)dx∫ arctan(2x) \text{d}x

Question 6

HardPaper 1 · no calculator9 marks
(a)

Find ∫x2cos⁡(2x)dx\int x^2 \cos(2x) dx.

[6]
(b)

Hence, find the exact value of ∫0π4x2cos⁡(2x)dx\int_0^{\frac{\pi}{4}} x^2 \cos(2x) dx.

[3]

Question 7

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 8

HardPaper 1 · no calculator7 marks

Find ∫x3exdx\int x^3 e^x dx.

Question 9

HardPaper 1 · no calculator11 marks
(a)

Find ∫xsec⁡2x dx\int x \sec^2 x \, dx.

[6]
(b)

The region RR is enclosed by the curve y=xsec⁡2xy = x \sec^2 x, the xx-axis, and the lines x=0x=0 and x=π3x=\frac{\pi}{3}. Show that the area of RR is π33−ln⁡2\frac{\pi\sqrt{3}}{3} - \ln 2.

[5]

Question 10

HardPaper 1 · no calculator9 marks

Find ∫e2xcos⁡(x)dx\int e^{2x} \cos(x) dx.

Question 11

HardPaper 2 · calculator19 marks
(a)

A pharmaceutical company is testing a new drug. The concentration of the drug, CC, in the bloodstream of a patient, in micrograms per millilitre (μg/mL\mu\text{g/mL}), tt hours after administration, is modelled by the function C(t)=5te−0.2tC(t) = 5te^{-0.2t}, for 0≤t≤150 \le t \le 15.

Sketch the graph of C(t)C(t) for 0≤t≤150 \le t \le 15, clearly indicating the coordinates of the initial concentration point AA, the maximum concentration point MM, and the concentration point BB at t=15t=15 hours.

[4]
(b)

State the range of the concentration C(t)C(t) during the observed period.

[1]
(c)

Find the equation of the straight line connecting the initial concentration point AA and the concentration point BB at t=15t=15 hours.

[3]
(d)

Show that the rate of change of the drug concentration is given by C′(t)=(5−t)e−0.2tC'(t) = (5-t)e^{-0.2t}.

[2]
(e)

At a certain time, the rate of change of the drug concentration is parallel to the line AB. Find the equation of the tangent line to the graph of C(t)C(t) at this time. Give all coefficients in your equation correct to 33 significant figures.

[4]
(f)

Calculate the area of the region enclosed by the graph of C(t)C(t) and the line AB.

[5]

Question 12

HardPaper 1 · no calculator38 marks
(a)

Find the general solution to the following differential equation. (a)

dydx=y2x2−1\frac{dy}{dx} = \frac{y^2}{x^2-1}

[5]
(b)

(b)

dydx=y2sin⁡xcos⁡x\frac{dy}{dx} = y^2 \sin x \cos x

[5]
(c)

(c) Find the particular solution to the differential equation (x2+4)dydx=xy(x^2+4) \frac{dy}{dx} = xy, given the initial condition y(0)=1y(0)=1.

[6]
(d)

(d)

exdydx=1ye^x \frac{dy}{dx} = \frac{1}{y}

[4]
(e)

(e)

dydx+2y=xe−x\frac{dy}{dx} + 2y = xe^{-x}

[6]
(f)

(f) xdydx−3y=x5x \frac{dy}{dx} - 3y = x^5 for x>0x>0.

[6]
(g)

(g) dydx+ycot⁡x=cos⁡x\frac{dy}{dx} + y \cot x = \cos x for 0<x<π0 < x < \pi.

[6]

Question 13

HardPaper 1 · no calculator17 marks
(a)

The lifetime, TT (in years), of a certain electronic component is modelled by a continuous random variable with probability density function f(t)f(t) given by

f(t)={kt2e−t/5t≥00otherwisef(t) = \begin{cases} kt^2 e^{-t/5} & t \ge 0 \\ 0 & \text{otherwise} \end{cases}

where kk is a positive constant.

(a) Show that k=1250k = \frac{1}{250}.

[5]
(b)

(b) Find the mode of the distribution of TT.

[4]
(c)

(c) Find the mean lifetime of the component, E(T)E(T).

[4]
(d)

(d) Find the variance of TT.

[4]

Question 14

HardPaper 1 · no calculator12 marks
(a)

A continuous random variable XX has probability density function

f(x)={kxe−xx≥00otherwisef(x) = \begin{cases} kx e^{-x} & x \ge 0 \\ 0 & \text{otherwise} \end{cases}

Show that k=1k = 1.

[4]
(b)

Find the variance of XX.

[8]

Question 15

HardPaper 1 · no calculator7 marks

Solve the differential equation xdydx+y=xcos⁡(x)x\frac{dy}{dx} + y = x\cos(x), for x>0x > 0.

Given that y=2πy = \frac{2}{\pi} when x=πx = \pi, find the solution in the form y=f(x)y = f(x).

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What does Integration by parts (+ repeated by parts) cover in IB Maths AA?

Integration by parts uses the product rule in reverse to integrate products of functions. The formula is ∫ u (dv)/(dx) dx = uv - ∫ v (du)/(dx) dx. Choose u to simplify upon differentiation and dv to be easily integrable, often following the LIATE order (Logarithmic, Inverse Trig, Algebraic, Trigonometric, Exponential).

Is Integration by parts (+ repeated by parts) SL or HL?

Integration by parts (+ repeated by parts) is HL only. SL students are not examined on it.

How do I revise Integration by parts (+ repeated by parts) for IB Maths AA?

Start from the core idea: integration by parts uses the product rule in reverse to integrate products of functions. In the exam: the substitution rule is a gift for question generation: if the integral is not the reverse chain rule shape, the question must supply the substitution. Changing the limits when substituting in a definite integral, rather than substituting back, is the standard examiner complaint. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Integration by parts (+ repeated by parts)?

FourtyFive has 15 Integration by parts (+ repeated by parts) questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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