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Integration

Practice exam-style IB Math AI questions for Integration, aligned with the syllabus and grouped by topic.

Verified by Karim
Verified by Karim
Paper
Difficulty
Status
Level
Question 1
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A function FF satisfies

F(x)=6x28x+5F'(x)=6x^2-8x+5

and F(1)=7F(1)=7.

A

Find the general form of F(x)F(x).

[2]
B

Determine the value of CC.

[1]
C

Hence find F(3)F(3).

[2]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

The cross-section of a drainage channel is modelled by the region between the curve y=x2+6xy=-x^2+6x and the xx-axis, for 0x60\leq x\leq6. Both xx and yy are measured in metres.

A

Write down an integral that represents the cross-sectional area of the channel.

[1]
B

Find the cross-sectional area of the channel.

[2]
C

Find the mean depth of the channel across its width.

[2]
Question 3
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A flower bed is modelled by the region between the curve

y=(x1)(5x)y=(x-1)(5-x)

and the xx-axis. Both coordinates are measured in metres.

A

Find the two xx-intercepts of the curve.

[1]
B

Write down an integral representing the area of the flower bed.

[1]
C

Find the area of the flower bed.

[2]
Question 4
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

The gradient of a curve is given by

dydx=9x2+4\frac{\mathrm{d}y}{\mathrm{d}x}=9x^2+4

The curve passes through the point (1,2)(-1,2).

A

Find an expression for yy in terms of xx and an arbitrary constant CC.

[2]
B

Determine the value of CC.

[1]
C

Find the value of yy when x=2x=2.

[1]
Question 5
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

Water enters a tank at a rate of

r(t)=0.6t22.4t+5r(t)=0.6t^2-2.4t+5

litres per minute, where tt is the time in minutes after a pump is switched on. Initially, the tank contains 1212 litres of water.

A

Write down an integral representing the volume of water that enters the tank during the first 66 minutes.

[1]
B

Calculate the volume of water that enters during the first 66 minutes.

[2]
C

Find the volume of water in the tank after 66 minutes.

[1]
D

Calculate the average rate at which water enters the tank during these 66 minutes.

[2]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The velocity of a cyclist was recorded every 22 seconds from t=0t=0 to t=8t=8 seconds. The results are shown in the table.

time [s]

velocity [m s^-1]

0

3.2

2

4.1

4

5.0

6

4.6

8

3.5

A

The recorded velocities, in order, are 3.23.2, 4.14.1, 5.05.0, 4.64.6 and 3.5 m s13.5\ \text{m s}^{-1}. Use the trapezoidal rule to estimate the distance travelled during the 88 seconds.

[3]
B

A tracking device recorded the actual distance as 33.4 m33.4\ \text{m}. Calculate the percentage error in the trapezoidal estimate, stating whether it is an overestimate or an underestimate.

[2]
Question 7
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The area under the curve y=x2+1y=x^2+1, from x=0x=0 to x=2x=2, is to be approximated using four intervals of equal width.

A

Use the trapezoidal rule with ordinates at x=0x=0, 0.50.5, 11, 1.51.5 and 22 to estimate the area.

[2]
B

Use integration to find the exact area.

[2]
C

Explain why the trapezoidal rule gives an overestimate in this case.

[1]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A function FF is defined for x>0x>0 and satisfies

F(x)=3x4xF'(x)=3\sqrt{x}-\frac{4}{x}

It is given that F(1)=5F(1)=5.

A

Find the general form of F(x)F(x).

[2]
B

Determine the value of CC.

[1]
C

Hence find F(4)F(4).

[2]
Question 9
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The rate of change of a quantity QQ is modelled by

dQdt=6cos(3t)\frac{\mathrm{d}Q}{\mathrm{d}t}=6\cos(3t)

for 0tπ60\leq t\leq\dfrac{\pi}{6}, where tt is measured in seconds.

A

Find an antiderivative of 6cos(3t)6\cos(3t).

[2]
B

Find the exact change in QQ over the given interval.

[1]
C

Find the average value of dQdt\dfrac{\mathrm{d}Q}{\mathrm{d}t} over the interval, giving your answer exactly.

[2]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The region bounded by the curve y=4xy=\sqrt{4-x}, the xx-axis and the yy-axis is rotated through 2π2\pi radians about the xx-axis.

A first-quadrant coordinate diagram showing a decreasing curve from the positive y-axis to the positive x-axis. The region under the curve and between the coordinate axes is shaded, and the x-axis is identified as the axis of rotation.
A

Write down an integral representing the volume of the solid formed.

[2]
B

Find the exact volume of the solid and give its value to three significant figures.

[3]
Question 11
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A surveyor measures the width of a lake at 2525 metre intervals along a straight baseline of length 150150 metres. The measurements are shown in the table.

Distance along baseline [m]

Lake width [m]

0

0

25

18

50

27

75

31

100

24

125

15

150

0

A

The measured widths, in order, are 00, 1818, 2727, 3131, 2424, 1515 and 00 metres. Use the trapezoidal rule to estimate the surface area of the lake.

[4]
B

A satellite estimate gives the surface area as 3010 m23010\ \text{m}^2. Calculate the percentage by which the trapezoidal estimate is less than the satellite estimate.

[2]
Question 12
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The mass of a substance in a container increases at a rate

r(t)=2tet2+1r(t)=2t e^{t^2+1}

grams per hour, where 0t10\leq t\leq1. Initially, the container holds 1010 grams of the substance.

A

Using the substitution u=t2+1u=t^2+1, find an antiderivative of r(t)r(t).

[2]
B

Find the exact increase in mass during the first hour.

[1]
C

Calculate the mass in the container after one hour.

[2]
Question 13
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

For x0x\geq0, a rate is modelled by

r(x)=123x+2r(x)=\frac{12}{3x+2}
A

Find r(x)dx\displaystyle\int r(x)\,\mathrm{d}x.

[2]
B

Show that the accumulated amount from x=0x=0 to x=ax=a is

4ln(3a+22)4\ln\left(\frac{3a+2}{2}\right)
[2]
C

Hence determine aa if the accumulated amount is 4ln54\ln5.

[2]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The curve y=(x1)(x4)y=(x-1)(x-4) is considered on the interval 0x50\leq x\leq5.

Parabola $y=(x-1)(x-4)$ on $0\leq x\leq5$.
A

Find the value of 05(x1)(x4)dx\displaystyle\int_0^5(x-1)(x-4)\,\mathrm{d}x.

[2]
B

Explain why the answer to part (a) is not the total geometric area between the curve and the xx-axis.

[1]
C

Find the total geometric area between the curve and the xx-axis for 0x50\leq x\leq5.

[3]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

A region is bounded by the curve

x=ey/2x=e^{y/2}

the yy-axis and the horizontal lines y=0y=0 and y=ln4y=\ln4. The region is rotated through 2π2\pi radians about the yy-axis.

A

Write down an integral representing the volume of the solid formed.

[2]
B

Calculate the exact volume of the solid.

[3]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The curve x=y24x=y^2-4 and the yy-axis enclose a finite region.

Sideways parabola x = y^2 - 4 and the y-axis.
A

Find the yy-coordinates of the points where the curve intersects the yy-axis.

[1]
B

Write down a definite integral for the signed quantity obtained by integrating xx with respect to yy between these points.

[1]
C

Hence find the area of the enclosed region.

[3]
Question 17
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The height of a decorative canopy above a horizontal walkway is modelled by

h(x)=0.02x3+0.18x2+0.5h(x)=-0.02x^3+0.18x^2+0.5

where 0x80\leq x\leq 8, and xx and hh are measured in metres.

Canopy height curve over 0≤x≤8.
A
I.

Find an antiderivative of h(x)h(x).

[2]
II.

Calculate the area of the vertical cross-section beneath the canopy.

[2]
B
I.

Use the trapezoidal rule with four equal intervals to estimate the cross-sectional area.

[3]
II.

Find the percentage error in the trapezoidal estimate and state whether it is an overestimate or an underestimate. If you did not obtain an area in part (a)(ii), use 14.24 m214.24\ \text{m}^2.

[3]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

During a fundraising event, the total amount collected, in hundreds of dollars, is Q(t)Q(t), where tt is the number of hours after the event begins. The rate of collection is modelled by

dQdt=0.3t21.8t+4.5\frac{dQ}{dt}=0.3t^2-1.8t+4.5

for 0t60\leq t\leq6. At the start of the event, Q=25Q=25.

A
I.

Find the general form of Q(t)Q(t).

[2]
II.

Use the initial condition to determine a formula for Q(t)Q(t).

[2]
B
I.

Calculate the amount collected by the end of the six hours, giving your answer in dollars.

[2]
II.

Find the average rate of collection during the six-hour event, in dollars per hour. If you did not obtain Q(6)Q(6), use 41.241.2 hundreds of dollars.

[2]
C

Determine when the amount collected first reaches 35003500 dollars.

[2]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The vertical cross-section of a theatre backdrop is bounded by the xx-axis, the lines x=0x=0 and x=20x=20, and the curve

y=0.04x2+0.8x+1y=-0.04x^2+0.8x+1

where both coordinates are measured in metres.

Parabolic backdrop profile over 0 to 20 m.
A
I.

Write down a definite integral representing the area of the backdrop.

[1]
II.

Find the area of the backdrop.

[3]
B
I.

Use the trapezoidal rule with four equal intervals to estimate the area.

[2]
II.

Calculate the percentage error in this estimate. If you did not obtain the exact area, use 73.333 m273.333\ldots\ \text{m}^2.

[2]
C

Painting the backdrop costs 18.50 dollars per square metre. Find the total cost, to the nearest dollar, using the exact area from part (a).

[2]
Question 20
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The electrical power produced by a small wind turbine is recorded every 0.50.5 hours during a three-hour test. The results are shown in the table.

Time [h]

Power [kW]

0.0

1.2

0.5

1.8

1.0

2.5

1.5

3.1

2.0

2.7

2.5

2.0

3.0

1.4

A
I.

The recorded powers, in order, are 1.21.2, 1.81.8, 2.52.5, 3.13.1, 2.72.7, 2.02.0 and 1.41.4 kilowatts. Use the trapezoidal rule to estimate the energy produced.

[3]
II.

State why the units of the estimate are kilowatt-hours rather than kilowatts.

[1]
B
I.

A battery has a stated capacity of 8 kWh8\ \text{kWh}, of which 85%85\% is usable. Determine whether the usable capacity is sufficient to store the estimated energy.

[2]
II.

Explain one limitation of using the trapezoidal estimate to select the battery.

[1]
C

An alternative model for the power is p(t)=0.4t2+1.2t+1.3p(t)=-0.4t^2+1.2t+1.3. Calculate the energy predicted by this model for 0t30\leq t\leq3 and compare it with the trapezoidal estimate.

[2]
Question 21
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A survey team measures the width of a coastal wetland at 4040 metre intervals along a baseline of length 240240 metres. The measured widths are shown in the table.

Distance along baseline [m]

Wetland width [m]

0

0

40

22

80

35

120

41

160

33

200

18

240

0

A
I.

State the number of intervals and their common width.

[1]
II.

The recorded widths, in order, are 00, 2222, 3535, 4141, 3333, 1818 and 00 metres. Use the trapezoidal rule to estimate the area of the wetland.

[3]
B
I.

A satellite model estimates the wetland area to be 5720 m25720\ \text{m}^2. Calculate the percentage by which the trapezoidal estimate exceeds the satellite estimate.

[2]
II.

Suggest one reason why the two estimates may differ.

[1]
C

The two endpoint widths are known exactly, but each interior width may have an error of up to 2 m2\ \text{m}. Find the resulting lower and upper bounds for the trapezoidal estimate.

[3]
Question 22
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A manufacturer cuts a panel whose height is modelled by

p(x)=0.001x30.06x2+0.9x+2p(x)=0.001x^3-0.06x^2+0.9x+2

for 0x200\leq x\leq20, where both xx and p(x)p(x) are measured in metres. The panel lies between the curve and the xx-axis. For part (b), a vertical cut at x=ax=a divides the panel into two regions of equal area.

Positive cubic panel profile on its domain, used to find area and equal-area cut.
A
I.

Find an antiderivative of p(x)p(x).

[2]
II.

Calculate the area of the panel.

[2]
B

A vertical cut at x=ax=a is to divide the panel into two pieces of equal area.

I.

Write down an equation that can be solved to find aa.

[1]
II.

Hence determine the value of aa.

[2]
C
I.

The manufactured panel is vertically scaled by a factor of 0.950.95, while its width is unchanged. Find the area of the scaled panel.

[2]
II.

The material costs 12.4012.40 dollars per square metre. Calculate the cost of the scaled panel.

[2]
Question 23
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

The number of people in an exhibition hall is N(t)N(t), where tt is measured in minutes after the doors open. The net rate of change is modelled by

dNdt=0.02t3+0.3t2+2\frac{dN}{dt}=-0.02t^3+0.3t^2+2

for 0t100\leq t\leq10. Initially, there are 120120 people in the hall.

A
I.

Find a formula for N(t)N(t).

[3]
II.

Find the number of people in the hall after 1010 minutes.

[1]
B
I.

Determine when the number of people first reaches 160160.

[2]
II.

Find the average net rate of change in the number of people during the first 1010 minutes. If you did not obtain N(10)N(10), use 190190.

[1]
C

Justify why the model predicts that the number of people is increasing throughout the stated interval.

[2]
Question 24
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A curve has gradient

dydx=6x2\frac{dy}{dx}=6x-2

and passes through the point (0,2)(0,2). The region between the curve, the xx-axis and the lines x=0x=0 and x=2x=2 is denoted by RR.

Upward-opening curve on 0≤x≤2 with sample ordinates.
A
I.

Find the equation of the curve.

[3]
II.

Show that the curve lies above the xx-axis for all real values of xx.

[1]
B

Find the area of RR.

[3]
C
I.

Use the trapezoidal rule with four equal intervals to estimate the area of RR.

[2]
II.

Explain why the trapezoidal estimate is greater than the exact area.

[1]
Question 25
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A treatment system removes a contaminant from a vessel at the rate

r(t)=40tet2/4r(t)=40t e^{-t^2/4}

grams per hour, where t0t\geq0. Initially, the vessel contains 120120 grams of the contaminant.

A
I.

Using the substitution u=t24u=-\frac{t^2}{4}, find an antiderivative of r(t)r(t).

[3]
II.

Show that the mass removed during the first TT hours is

80(1eT2/4)80\left(1-e^{-T^2/4}\right)

[1]
B
I.

Calculate the mass remaining after two hours.

[2]
II.

State the limiting mass remaining according to the model as tt\to\infty.

[1]
C

Determine when the mass remaining first reaches 6060 grams.

[3]
Question 26
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The profile of a rotationally symmetric wooden ornament is modelled by

y=2+x,0x9y=2+\sqrt{x},\qquad 0\leq x\leq9

where xx and yy are measured in centimetres. The region under the curve is rotated through 2π2\pi radians about the xx-axis.

A labelled profile diagram showing the curve y equals 2 plus square root of x from x equals 0 to x equals 9, the region beneath it, and an arrow indicating rotation about the horizontal axis.
A
I.

Write down an integral representing the volume of the ornament.

[2]
II.

Expand the integrand and find an antiderivative.

[2]
B

Hence find the volume of the ornament, giving both an exact answer and an answer to three significant figures.

[2]
C

A cylindrical hole of radius 1 cm1\ \text{cm} is drilled along the xx-axis through the entire 9 cm9\ \text{cm} length of the ornament, centred on the axis of rotation.

I.

Find the volume remaining after the hole is drilled.

[2]
II.

Calculate the percentage of the original volume removed.

[2]
Question 27
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A function FF is defined for x>0x>0 and satisfies

F(x)=x1/2+3xF'(x)=x^{-1/2}+\frac{3}{x}

It is given that F(1)=2F(1)=2.

A
I.

Find the general form of F(x)F(x).

[2]
II.

Use the boundary condition to determine F(x)F(x).

[2]
B
I.

Find the exact value of 14F(x)dx\displaystyle\int_1^4F'(x)\,dx.

[2]
II.

Hence find the average value of F(x)F'(x) on 1x41\leq x\leq4.

[1]
C

Determine the value of xx for which F(x)=10F(x)=10.

[2]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A filtration system removes a contaminant at a rate
r(t)=12e0.3t+3r(t)=12e^{-0.3t}+3
where rr is measured in milligrams per minute and tt is the time, in minutes, after the system starts. The term 33 represents a constant background removal rate.

Removal rate model with horizontal background rate 3.
A
I.

Find an antiderivative of r(t)r(t).

[2]
II.

Hence show that the mass removed during the first TT minutes is
M(T)=40(1e0.3T)+3TM(T)=40(1-e^{-0.3T})+3T

[2]
B

Calculate the mass removed during the first 88 minutes.

[2]
C
I.

Find the total mass that will eventually be removed by the exponentially decreasing component of the rate.

[2]
II.

Determine when 90%90\% of the mass in part (c)(i) has been removed. If you did not obtain 4040 in part (c)(i), use 4040.

[2]
III.

Explain why the total mass removed by the complete rate model does not approach a finite limit.

[2]
Question 29
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The rate of change of the volume of water in a tidal basin is modelled by
q(t)=6sin(πt6),0t12q(t)=6\sin\left(\frac{\pi t}{6}\right),\qquad 0\leq t\leq12
where qq is measured in thousands of cubic metres per hour. Positive values represent inflow and negative values represent outflow.

Graph of q(t) over one full 12-hour sinusoidal cycle, showing inflow for positive values and outflow for negative values.
A
I.

Find an antiderivative of q(t)q(t).

[2]
II.

Calculate the signed change in volume over the 1212 hours.

[2]
B

Find the total volume of water that passes through the basin entrance, counting inflow and outflow as positive.

[3]
C
I.

Show that the increase in the basin volume from time 00 to time tt is
A(t)=36π(1cos(πt6))A(t)=\frac{36}{\pi}\left(1-\cos\left(\frac{\pi t}{6}\right)\right)

[2]
II.

Determine the two times when the basin contains half of its maximum additional volume. If you did not obtain A(t)A(t), use the expression given in part (c)(i).

[3]
Question 30
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The inside profile of a rotationally symmetric vessel is modelled by
x=0.8+0.12y2,0y5x=0.8+0.12y^2,\qquad 0\leq y\leq5
where xx and yy are measured in metres. The region between the profile and the yy-axis is rotated through 2π2\pi radians about the yy-axis.

A vertical cross-section of a vessel with the y-axis as its axis of rotation and the curved profile x = 0.8 + 0.12y squared shown from y = 0 to y = 5.
A
I.

Write down an integral for the volume of the vessel.

[2]
II.

Find the exact volume and its value to three significant figures.

[3]
B

A cylindrical vessel of height 55 metres has the same volume. Determine its radius.

[3]
C

A second vessel has profile x=a+0.1y2x=a+0.1y^2 for 0y40\leq y\leq4 and volume 10π m310\pi\ \text{m}^3. Determine the positive value of aa.

[4]
Question 31
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The accumulated quantity of a dissolved mineral is modelled from the rate
r(t)=18tt2+4,t0r(t)=\frac{18t}{t^2+4},\qquad t\geq0
where rr is measured in grams per hour.

Graph of the accumulation rate r(t) against time t.
A
I.

Using the substitution u=t2+4u=t^2+4, find an antiderivative of r(t)r(t).

[2]
II.

Show that the quantity accumulated by time TT is
Q(T)=9ln(T2+44)Q(T)=9\ln\left(\frac{T^2+4}{4}\right)

[2]
B

Determine when the accumulated quantity first reaches 2020 grams.

[3]
C
I.

Find the average rate of accumulation from t=0t=0 until the time found in part (b). If you did not obtain a time, use 5.745.74 hours.

[2]
II.

Determine whether Q(T)Q(T) approaches a finite limit as TT\to\infty and justify your answer.

[3]
Question 32
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The electrical power generated by a solar installation is modelled by
P(t)=5+3cos(π(t2)6),0t12P(t)=5+3\cos\left(\frac{\pi(t-2)}6\right),\qquad0\leq t\leq12
where PP is measured in kilowatts and tt is measured in hours.

Power vs time graph for a solar installation over 12 hours with a 5 kW baseline.
A
I.

Find the total electrical energy generated during the 1212 hours.

[2]
II.

State the average power over this interval.

[2]
B

Find the intervals during which the generated power exceeds 5 kW5\ \text{kW}.

[3]
C
I.

Calculate the energy generated above the baseline power of 5 kW5\ \text{kW}.

[3]
II.

Hence find the average power during the times when the power exceeds 5 kW5\ \text{kW}.

[2]
Question 33
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The curve

y=2sinx1y=2\sin x-1

is considered on the interval 0x2π0\leq x\leq2\pi.

One period of y = 2 sin x - 1 on 0 ≤ x ≤ 2π, with intercepts marked.
A
I.

Find the two values of xx at which the curve intersects the xx-axis.

[2]
II.

State the interval on which the curve is above the xx-axis.

[1]
B
I.

Find the signed integral 02π(2sinx1)dx\displaystyle\int_0^{2\pi}(2\sin x-1)\,dx.

[2]
II.

Explain why the answer to part (b)(i) is not the total geometric area.

[2]
C

Find the total geometric area between the curve and the xx-axis for 0x2π0\leq x\leq2\pi.

[4]
Question 34
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A designer models the outer profile of a lampshade by

x=1+2ey/4,0y8x=1+2e^{-y/4},\qquad 0\leq y\leq8

where xx and yy are measured in centimetres. The region between the curve and the yy-axis is rotated through 2π2\pi radians about the yy-axis.

Exponential lampshade profile with y=8 boundary.
A
I.

Write down an integral representing the volume of the lampshade model.

[2]
II.

Show that the integrand can be written as 1+4ey/4+4ey/21+4e^{-y/4}+4e^{-y/2}.

[2]
B

Hence calculate the volume of the model.

[3]
C
I.

A cylindrical model has radius 3 cm3\ \text{cm} and height 8 cm8\ \text{cm}. Find its volume.

[1]
II.

Calculate the percentage by which the exponential-profile volume is less than the cylindrical volume.

[2]
Question 35
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The curve

y=x34xy=x^3-4x

is considered on the interval 3x3-3\leq x\leq3.

Odd cubic y=x^3-4x on -3≤x≤3, with its three x-intercepts marked.
A
I.

Find the xx-intercepts of the curve.

[2]
II.

State the symmetry of the graph.

[1]
B
I.

Find 33(x34x)dx\displaystyle\int_{-3}^{3}(x^3-4x)\,dx.

[1]
II.

Explain why this value does not imply that the geometric area is zero.

[2]
C

Find the total geometric area between the curve and the xx-axis for 3x3-3\leq x\leq3.

[4]
Question 36
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For 0xπ0\leq x\leq\pi, define

f(x)=4sinx3cosxf(x)=\frac{4\sin x}{3-\cos x}

The accumulated value from 00 to bb is denoted by

A(b)=0bf(x)dxA(b)=\int_0^b f(x)\,dx

Curve of f(x)=4sinx/(3-cosx) on [0,π].
A
I.

Using the substitution u=3cosxu=3-\cos x, find an antiderivative of f(x)f(x).

[3]
II.

Find the exact value of A(π)A(\pi).

[1]
B

Let bb be a value in 0bπ0\leq b\leq\pi such that A(b)=12A(π)A(b)=\frac12A(\pi).

I.

Show that cosb=322\cos b=3-2\sqrt2.

[3]
II.

Hence find bb.

[1]
C

Explain why the value of bb in part (b) is unique on 0bπ0\leq b\leq\pi.

[2]
Question 37
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A decorative solid is formed by rotating the region under
y=xex/2,0xay=\sqrt{x}e^{-x/2},\qquad 0\leq x\leq a
about the xx-axis, where a>0a>0.

Curve y = sqrt(x)e^(-x/2) on x >= 0.
A
I.

Show that the volume is
V(a)=π0axexdxV(a)=\pi\int_0^a xe^{-x}\,dx

[2]
II.

Use technology or an appropriate antiderivative to show that
V(a)=π(1(a+1)ea)V(a)=\pi\left(1-(a+1)e^{-a}\right)

[3]
B

Calculate the volume when a=4a=4.

[2]
C
I.

State the limiting volume as aa\to\infty.

[1]
II.

Determine the value of aa for which the volume is 95%95\% of its limiting value. If you did not obtain the limiting volume, use π\pi.

[3]
III.

Explain how a solid extending indefinitely in the positive xx-direction can have a finite volume.

[1]
Question 38
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The value of
I=02ex2dxI=\int_0^2e^{-x^2}\,dx
is approximated using the trapezoidal rule. Values of ex2e^{-x^2} are generated using a GDC.

xx

ex2e^{-x^2}

0.00

1.000000

0.25

0.939413

0.50

0.778801

0.75

0.569783

1.00

0.367879

1.25

0.209611

1.50

0.105399

1.75

0.046771

2.00

0.018316

A
I.

Use four equal intervals to find the trapezoidal estimate T4T_4.

[3]
II.

Use eight equal intervals to find the trapezoidal estimate T8T_8.

[2]
B

A high-accuracy numerical integration gives I=0.882081I=0.882081. Calculate the percentage error in T8T_8.

[2]
C
I.

The leading trapezoidal error is assumed to be proportional to h2h^2. Show that reducing the interval width by a factor of 22 should reduce this error by a factor of 44.

[2]
II.

Hence derive and use an improved estimate
IT8+T8T43I\approx T_8+\frac{T_8-T_4}{3}

[3]
Question 39
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The curve
x=y34yx=y^3-4y
and the yy-axis enclose two finite regions.

Curve x = y^3 - 4y with its y-axis intersections.
A
I.

Find the yy-coordinates of the intersections with the yy-axis.

[2]
II.

Find the signed integral 22(y34y)dy\int_{-2}^{2}(y^3-4y)\,dy and explain its value.

[3]
B

Find the total geometric area enclosed by the curve and the yy-axis.

[3]
C
I.

The two regions are rotated about the yy-axis. Find the exact total volume formed.

[3]
II.

For the family x=y3a2yx=y^3-a^2y, where a>0a>0, the corresponding total enclosed area is AA. Deduce the volume of revolution in terms of AA.

[2]
Question 40
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A storage tank is modelled by rotating the region under
y=k(1x2a2),axay=k\left(1-\frac{x^2}{a^2}\right),\qquad -a\leq x\leq a
about the xx-axis, where a>0a>0 and k>0k>0.

A symmetric parabolic tank profile above the x-axis with intercepts at x = -a and x = a and maximum height k at x = 0, indicating rotation about the x-axis.
A
I.

Find the area AA of the generating region.

[2]
II.

Show that the tank volume is
V=16πak215V=\frac{16\pi ak^2}{15}

[3]
B

For a tank with a=3a=3 and k=3k=3, calculate its volume.

[2]
C
I.

For tanks with a fixed generating area AA, express VV in terms of AA and aa.

[3]
II.

Interpret how the volume changes as aa increases while AA remains fixed.

[2]
Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The rate at which a medicine is absorbed is modelled by

r(t)=Atekt2,t0r(t)=At e^{-kt^2},\qquad t\geq0

where tt is measured in hours, rr is measured in milligrams per hour, k=0.5 h2k=0.5\ \text{h}^{-2}, and A>0A>0 has units mg h2\text{mg h}^{-2}. The total amount eventually absorbed is 8080 milligrams.

Absorption rate of the medicine over time.
A
I.

Show that the total amount eventually absorbed is A2k\frac{A}{2k} milligrams.

[3]
II.

Hence state the value of AA, including its units.

[2]
B

Determine the time by which 75%75\% of the medicine has been absorbed.

[3]
C
I.

Use the trapezoidal rule with interval width 0.50.5 hours to estimate the amount absorbed during the first 22 hours.

[2]
II.

The exact amount absorbed during the first 22 hours is 80(1e2)80(1-e^{-2}). Calculate the percentage error in the trapezoidal estimate and state its direction.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The vertical width of a proposed tunnel was measured at horizontal positions x=0x=0, 2.52.5, 55, 7.57.5 and 1010 metres. The corresponding widths were 00, 4.64.6, 6.16.1, 4.74.7 and 00 metres. A mathematical model for the width is
w(x)=kx(10x),0x10w(x)=kx(10-x),\qquad0\leq x\leq10

x [m]

width [m]

0

0

2.5

4.6

5

6.1

7.5

4.7

10

0

A
I.

Use the trapezoidal rule to estimate the cross-sectional area from the measurements.

[2]
II.

Use the central measurement to determine kk.

[2]
B

Find the cross-sectional area predicted by the model and compare it with the trapezoidal estimate.

[3]
C
I.

The model region is rotated about the xx-axis. Find the volume of the resulting solid.

[3]
II.

For the general model w(x)=kx(Lx)w(x)=kx(L-x) on 0xL0\leq x\leq L, show that its volume of revolution VV and cross-sectional area AA satisfy
V=6πA25LV=\frac{6\pi A^2}{5L}

[3]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For c>1c>1, consider the positive function
fc(x)=sinxc+cosx,0xπf_c(x)=\frac{\sin x}{c+\cos x},\qquad0\leq x\leq\pi

Positive family of f_c(x) on [0,π] for several c>1.
A
I.

Using the substitution u=c+cosxu=c+\cos x, find an antiderivative of fc(x)f_c(x).

[2]
II.

Show that the area under the curve is
A(c)=ln(c+1c1)A(c)=\ln\left(\frac{c+1}{c-1}\right)

[3]
B

Determine cc if the area under the curve is ln2\ln2.

[2]
C
I.

For c=2c=2, find the value aa such that the area from x=0x=0 to x=ax=a is half the total area.

[3]
II.

State what happens to A(c)A(c) as cc\to\infty.

[1]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The difference between two decaying concentrations is modelled by
d(x)=exe2x,x0d(x)=e^{-x}-e^{-2x},\qquad x\geq0

Curve of d(x)=e^-x-e^-2x for x>=0.
A
I.

Show that d(x)0d(x)\geq0 for x0x\geq0.

[2]
II.

Find the total area under d(x)d(x) for x0x\geq0.

[2]
B

Show that the area accumulated from 00 to TT is
A(T)=12(1eT)2A(T)=\frac12(1-e^{-T})^2

[3]
C
I.

Determine when 90%90\% of the total area has accumulated.

[3]
II.

The entire region is rotated about the xx-axis. Find the exact volume formed.

[2]
Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For 0<a<40<a<4, the curve
y=x2ay=x^2-a
is considered on the interval 2x2-2\leq x\leq2.

Curve

Parameter range

xx interval

y=x2ay=x^2-a

0<a<40<a<4

2x2-2\leq x\leq2

A
I.

Find the signed integral S(a)=22(x2a)dxS(a)=\int_{-2}^{2}(x^2-a)\,dx.

[2]
II.

Determine the value of aa for which the signed integral is zero.

[3]
B

Show that the total geometric area is
G(a)=1634a+83a3/2G(a)=\frac{16}{3}-4a+\frac{8}{3}a^{3/2}

[4]
C
I.

Find the geometric area when the signed integral is zero.

[2]
II.

For this same value of aa, the region between the curve and the xx-axis is rotated about the xx-axis. Find the exact volume formed.

[2]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

For h>0h>0, a region is bounded by the yy-axis and
x=y(hy),0yhx=\sqrt{y(h-y)},\qquad0\leq y\leq h

A right semicircular profile in the xy-plane, bounded by the y-axis and x = square root of y times h minus y, with endpoints at y = 0 and y = h.
A
I.

Show that the curve is part of a circle and state its radius.

[2]
II.

Hence find the area AA of the region.

[2]
B

The region is rotated about the yy-axis. Show that its volume is
V=πh36V=\frac{\pi h^3}{6}

[3]
C
I.

Find AA and VV when h=6h=6.

[2]
II.

Eliminate hh to express VV in terms of AA.

[2]
Question 47
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The profile of a rotationally symmetric component is modelled in two sections by

y=1+x2,0x2y=1+\frac{x}{2},\qquad 0\leq x\leq2

and

y=2cos(π(x2)6),2x5y=2\cos\left(\frac{\pi(x-2)}{6}\right),\qquad 2\leq x\leq5

The coordinates are measured in centimetres. The region under the profile is rotated through 2π2\pi radians about the xx-axis.

Piecewise profile of the component, showing the line and cosine sections.
A
I.

Verify that the two sections meet at x=2x=2.

[2]
II.

State the radius of the component at x=5x=5.

[1]
B
I.

Write down the volume as the sum of two definite integrals.

[3]
II.

Evaluate the volume of the first section exactly.

[1]
III.

Evaluate the volume of the second section exactly. You may use cos2u=12(1+cos2u)\cos^2u=\frac12(1+\cos2u).

[2]
C

Hence find the total volume of the component, giving an exact answer and an answer to three significant figures.

[2]
D

The second section is replaced by a cone of base radius 2 cm2\ \text{cm} and height 3 cm3\ \text{cm}. Calculate the percentage by which the original second-section volume exceeds the cone's volume.

[1]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The integral
I=01exdxI=\int_0^1e^x\,dx
is approximated using the trapezoidal rule with NN equal intervals. The resulting estimate is denoted by TNT_N.

Graph of y=e^x on [0,1] with trapezoidal chords for N=4 and N=8.
A
I.

Find the exact value of II.

[1]
II.

Find T4T_4 and T8T_8.

[4]
B

Explain why both estimates in part (a)(ii) are greater than the exact value.

[2]
C
I.

Show that
TN=1N[1+e2+ee1/Ne1/N1]T_N=\frac1N\left[\frac{1+e}{2}+\frac{e-e^{1/N}}{e^{1/N}-1}\right]

[3]
II.

Using the expression in part (c)(i), determine the least value of NN for which the percentage error in TNT_N is less than 0.01%0.01\%.

[3]

Differentiation

Kinematics