Skip to content
  1. IB Question Bank
  2. Maths AA
  3. Calculus
Topic 5.09 · SL and HL

Definite integrals (analytical, area under/between curves): notes and practice questions

Summary
  • A definite integral calculates the area under a curve f(x)f(x) between x=ax = a and x=bx = b:

∫abf(x) dx=F(b)−F(a) \int_a^b f(x) \, dx = F(b) - F(a)
where F(x)F(x) is the antiderivative of f(x)f(x).

  • The area between two curves f(x)f(x) and g(x)g(x) over [a,b][a, b]:

∫ab(f(x)−g(x)) dx \int_a^b \big(f(x) - g(x)\big) \, dx
If f(x)≥g(x)f(x) \geq g(x) in [a,b][a, b], this gives the net area.

How it is examined

When ff crosses the axis, the area needs splitting at the root and the negative piece made positive; integrating straight through gives the signed value and scores poorly. Between curves, the order (upper minus lower) matters and the intersections have to be found first. 6 to 8 marks.

Given in the booklet

Only A=∫aby dxA = \displaystyle\int_a^b y\,\mathrm{d}x, the area between a curve and the xx-axis. There is no "area between two curves" formula in the AA booklet, so ∫(yupper−ylower) dx\int (y_{\text{upper}} - y_{\text{lower}})\,\mathrm{d}x is recall.

Key ideas
  • Definite integrals, including analytical approach.
  • Areas of a region enclosed by a curve y=f(x)y = f(x) and the xx-axis, where f(x)f(x) can be positive or negative, without the use of technology.
  • Areas between curves.
At HL

Extended at AHL 5.17 (areas about the yy-axis, volumes of revolution).

Linking questions

  • The difference from SL 5.5 is that ff may go negative and no technology is allowed, which is what makes this a Paper 1 subtopic.

Practice questions

78 questions · 43 medium · 35 hard
Showing 20 of 20

Question 1

MediumPaper 1 · no calculator5 marks

Find the value of ∫193x−2xdx\int_{1}^{9} \frac{3x-2}{\sqrt{x}} dx.

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 calculator20 marks
(a)

The function ff is defined by f(x)=exsinh⁡xf(x) = e^x \sinh x, where x∈Rx \in \mathbb{R}.

Find the Maclaurin series for f(x)f(x) up to and including the x3x^3 term.

[4]
(b)

Hence, find an approximate value for ∫01ex2sinh⁡(x2)dx\int_0^1 e^{x^2} \sinh(x^2)dx.

[4]
(c)(i)

The function gg is defined by g(x)=excosh⁡xg(x) = e^x \cosh x, where x∈Rx \in \mathbb{R}.

Show that g′′(x)=2g′(x)g''(x) = 2g'(x).

[3]
(c)(ii)

Hence, find the values of g′′′(0)g'''(0) and g(4)(0)g^{(4)}(0).

[2]
(d)

Using the result from part (c), find the Maclaurin series for g(x)g(x) up to and including the x4x^4 term.

[4]
(e)

Hence, or otherwise, determine the value of lim⁡x→02excosh⁡x−2−2x−2x2x3\lim_{x \to 0} \frac{2e^x \cosh x - 2 - 2x - 2x^2}{x^3}.

[3]

Question 4

HardPaper 1 · no calculator20 marks
(a)

The acceleration, a ms−2a \text{ ms}^{-2}, of a particle moving in a straight line at time tt seconds, t≥0t \ge 0, is given by a=−2v−8a = -2v - 8, where v ms−1v \text{ ms}^{-1} is the particle's velocity. At t=0t=0, the particle is at the origin O and has an initial velocity v0 ms−1v_0 \text{ ms}^{-1}, where v0>0v_0 > 0.

By solving an appropriate differential equation, show that the particle's velocity at time tt is given by v(t)=(v0+4)e−2t−4v(t) = (v_0 + 4)e^{-2t} - 4.

[6]
(b)(i)

The particle moves in the positive direction until it reaches its maximum displacement from O at time TT. Show that e2T=v0+44e^{2T} = \frac{v_0+4}{4}.

[2]
(b)(ii)

Find an expression for the maximum displacement, smaxs_{\text{max}}, in terms of v0v_0.

[5]
(c)

Let v(T−k)v(T-k) represent the particle's velocity kk seconds before it reaches smaxs_{\text{max}}, where 0<k<T0 < k < T. By using the result from part (b)(i), show that v(T−k)=4(e2k−1)v(T-k) = 4(e^{2k} - 1).

[2]
(d)

Similarly, let v(T+k)v(T+k) represent the particle's velocity kk seconds after it reaches smaxs_{\text{max}}. Deduce a similar expression for v(T+k)v(T+k) in terms of kk.

[2]
(e)

Hence, show that the speed of the particle kk seconds before it reaches smaxs_{\text{max}} is greater than or equal to its speed kk seconds after it reaches smaxs_{\text{max}}.

[3]

Question 5

MediumPaper 1 · no calculator7 marks

A solid is formed by rotating the curve with equation x2=8yx^2 = 8y for y≥0y \ge 0 by 360∘360^\circ about the yy-axis. This forms a solid paraboloid.

A cylindrical hole of radius 44 is drilled through the center of the paraboloid, along the yy-axis. The resulting solid is a ring of height hh.

This information is shown in the following diagrams.

Diagram showing a cross-section of a paraboloid with a cylindrical hole and a 3D view of the resulting ring

The volume of the ring is 144π144\pi.

Find the value of hh.

Question 6

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 7

MediumPaper 1 · no calculator15 marks
(a)

Consider the function ff defined by f(x)=x3−6x2+8xf(x) = x^3 - 6x^2 + 8x.

Find the xx-intercepts of the graph of y=f(x)y=f(x).

[3]
(b)

The graph of y=f(x)y=f(x) for 0≤x≤40 \le x \le 4 is shown below. The graph encloses two regions with the xx-axis, shaded in the diagram.

Graph of y = x^3 - 6x^2 + 8x from x=0 to x=4, showing two regions bounded by the x-axis. The first region from x=0 to x=2 is above the axis, the second from x=2 to x=4 is below the axis.

Find the total area of the shaded regions.

[6]
(c)

The total surface area of a closed right cylinder is 8, equal to the total shaded area found in part (b). The cylinder has a height of 4−ππ\frac{4-\pi}{\pi}.

Diagram of a cylinder with radius r and height h.

Find the radius, rr, of the cylinder.

[4]
(d)

Hence, find the volume of the cylinder.

[2]

Question 8

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 9

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 10

HardPaper 1 · no calculator15 marks
(a)

A drone takes off from a platform. Its height, hh metres, above the platform after tt seconds is given by h(t)=6t−t2h(t) = 6t - t^2, for 0≤t≤80 \le t \le 8. This is shown in the following diagram.

Graph of height h versus time t for the drone, showing a parabola opening downwards with vertex in the first quadrant and passing through the origin

The drone lands back on the platform when t=pt=p.

Find the value of pp.

[2]
(b)(i)

The drone reaches its maximum height when t=qt=q.

Find the value of qq.

[3]
(b)(ii)

Find the drone's maximum height above the platform.

[2]
(c)

Find the drone's vertical distance from the platform when t=8t=8.

[2]
(d)

The total vertical distance travelled by the drone in the first 8 seconds is given by dd.

Find the value of dd.

[2]
(e)

A second drone, Drone B, takes off from the same platform. Its velocity is given by vB(t)=8−2tv_B(t) = 8 - 2t, for t≥0t \ge 0.

When t=kt = k, the total vertical distance travelled by Drone B is equal to dd.

Find the value of kk.

[4]

Question 11

MediumPaper 1 · no calculator5 marks
(a)

A continuous random variable YY has probability density function gg defined by

g(y)={14c,−c≤y≤3c0,otherwiseg(y)=\begin{cases} \frac{1}{4c}, & -c \le y \le 3c \\ 0, & \text{otherwise} \end{cases}

where cc is a positive real number.

(a) State E(Y)E(Y) in terms of cc.

[1]
(b)

(b) Use integration to find Var(Y)Var(Y) in terms of cc.

[4]

Question 12

HardPaper 1 · no calculator7 marks
(a)

Consider the functions f(x)=sin⁡xf(x) = \sin x and g(x)=cos⁡2xg(x) = \cos 2x, where 0≤x≤π0 \le x \le \pi.

The graphs of ff and gg are shown in the following diagram.

Graph of sin(x) and cos(2x) intersecting, with shaded region R

The graphs intersect at points P and Q. The region enclosed by the two graphs is shaded and labelled R.

(a) Find the xx-coordinates of P and Q.

[3]
(b)

(b) Find the area of R.

[4]

Question 13

MediumPaper 1 · no calculator6 marks
(a)

The expression 5x−2x3\frac{5x-2}{\sqrt[3]{x}} can be written in the form 5xp−2xq5x^p - 2x^q. Write down the value of pp and the value of qq.

[2]
(b)

Hence, find the value of ∫185x−2x3 dx\int_1^8 \frac{5x-2}{\sqrt[3]{x}} \, dx.

[4]

Question 14

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 15

MediumPaper 1 · no calculator4 marks

Find the area of the region enclosed by the curves y=1xy = \frac{1}{x}, y=1x2y = \frac{1}{x^2} and the lines x=1x=1 and x=2x=2.

Question 16

HardPaper 1 · no calculator16 marks
(a)

A particle moves in a straight line. Its velocity, v ms−1v \,\text{ms}^{-1}, at time tt seconds is given by v(t)=−t3+6t2−9tv(t) = -t^3 + 6t^2 - 9t, for 0≤t≤50 \le t \le 5. The particle is at the origin at t=0t=0.

The graph of vv is shown in the following diagram.

Graph of velocity v against time t, showing a cubic function starting at (0,0), going down to a minimum between t=0 and t=3, then up to touch the t-axis at t=3, and then continuing downwards.

(a) Find the displacement of the particle from the origin at t=3t=3.

[4]
(b)

(b) Find an expression for the acceleration of the particle.

[2]
(c)

(c) The particle is momentarily at rest at t=0t=0 and again at t=kt=k. Find the greatest speed of the particle in the interval 0≤t≤k0 \le t \le k.

[5]
(d)

(d) Find the greatest speed of the particle for 0≤t≤50 \le t \le 5.

[2]
(e)

(e) Write down an expression that represents the distance travelled by the particle while its speed is increasing. Do not evaluate the expression.

[3]

Question 17

MediumPaper 1 · no calculator12 marks
(a)(i)

Consider the expression f(x)=11+4xf(x) = \frac{1}{1+4x}, where ∣x∣<14|x| < \frac{1}{4}.

(a) (i) Show that the first four terms in the binomial expansion of f(x)f(x) are 1−4x+16x2−64x31 - 4x + 16x^2 - 64x^3.

[3]
(a)(ii)

(ii) Hence, find an approximation for ∫11+4xdx\int \frac{1}{1+4x} dx.

[2]
(b)(i)

(b) (i) Use your result from part (a)(ii) to find an approximate value for ∫01/811+4xdx\int_0^{1/8} \frac{1}{1+4x} dx.

[3]
(b)(ii)

(ii) Hence, find an approximation for ln⁡(1.5)\ln(1.5) in the form pq\frac{p}{q} where p,q∈Zp, q \in \mathbb{Z}.

[4]

Question 18

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 19

MediumPaper 2 · calculator7 marks
(a)

A small reconnaissance drone is performing a vertical ascent and descent. Its vertical velocity, v ms−1v \text{ ms}^{-1}, at time tt seconds, for 0≤t≤100 \le t \le 10, is modelled by the function v(t)=tsin⁡t−2.5v(t) = t \sin t - 2.5.

The following diagram shows the graph of vv.

Graph of vertical velocity v(t) of a drone

(a) Find the smallest value of tt for which the drone is momentarily stationary.

[2]
(b)

(b) Find the total vertical distance travelled by the drone during the first 10 seconds.

[3]
(c)

(c) Find the vertical acceleration of the drone when t=8t=8 seconds.

[2]

Question 20

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]

58 more Definite integrals (analytical, area under/between curves) questions in the app

Every answer is marked mark by mark, IB-style, and the AI tutor helps when you are stuck.

Where marks are lost

  • Using your own wrong value after failing a "show that".
Free. Every IB subject.
No card, no trial that runs out. Just a free account.
  • 50 marked answers a month
    Marked mark by mark, IB-style
  • Hints and mark schemes
    On every part of every question
  • 3,000+ questions
    All 6 subjects, SL and HL, mapped to the syllabus
  • Progress that adapts
    Your Study Profile picks what to practise next

Practise this topic as a session

Pick a difficulty and paper, and FourtyFive tracks your progress on this topic as you go.

or with email
FAQ

Questions,
answered.

Can't find what you're looking for? Email our student team.

What does Definite integrals (analytical, area under/between curves) cover in IB Maths AA?

A definite integral calculates the area under a curve f(x) between x = a and x = b:. $$. \int_a^b f(x) \, dx = F(b) - F(a).

Is Definite integrals (analytical, area under/between curves) SL or HL?

Both. SL and HL students study Definite integrals (analytical, area under/between curves), and HL goes further: Extended at AHL 5.17 (areas about the y-axis, volumes of revolution).

How do I revise Definite integrals (analytical, area under/between curves) for IB Maths AA?

Start from the core idea: a definite integral calculates the area under a curve f(x) between x = a and x = b:. In the exam: when f crosses the axis, the area needs splitting at the root and the negative piece made positive; integrating straight through gives the signed value and scores poorly. Between curves, the order (upper minus lower) matters and the intersections have to be found first. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Definite integrals (analytical, area under/between curves)?

FourtyFive has 78 Definite integrals (analytical, area under/between curves) 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.

Is FourtyFive free for Definite integrals (analytical, area under/between curves) practice?

Yes. A free account gives you 50 marked answers a month, and you do not need a card to sign up.

Can I handwrite Definite integrals (analytical, area under/between curves) answers on an iPad?

Yes. In the FourtyFive iPad app you write your working by hand with Apple Pencil, the way you would on paper, and it is marked the same way.

Start with the IB question
bank built for you.

Free to start, no card needed. Thousands of syllabus-mapped questions, AI Examiner marking, your weakest topics first.