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Integral Calculus

Practice exam-style IB Math AA questions for Integral Calculus, aligned with the syllabus and grouped by topic.

Paper
Difficulty
Status
Level
Question 1
SL • Paper 2
Easy
Calculator Permitted

The gradient of a curve is given by

dydx=6x24x+3x,x>0\frac{dy}{dx}=6x^2-4x+\frac{3}{\sqrt{x}}, \quad x>0

The curve passes through the point (4,18)(4,18).

A

Find yy in terms of xx.

[4]
Write your answer here...

0

Question 2
SL • Paper 1
Medium
Non Calculator

A function ff satisfies f(x)=3x122x+e2x1f'(x)=3x^{\frac{1}{2}}-\dfrac{2}{x}+e^{2x-1}, for x>0x>0, and f(1)=4f(1)=4.

A

Find f(x)f(x).

[5]
Write your answer here...

0

Question 3
SL • Paper 1
Medium
Non Calculator

The curves y=4x2y=4-x^2 and y=x+2y=x+2 enclose a finite region.

A

Find the xx-coordinates of the points of intersection of the two curves.

[2]
Write your answer here...
B

Find the area of the enclosed region.

[3]
Write your answer here...

0

Question 4
SL • Paper 1
Medium
Non Calculator
A

Evaluate 0π2sinx(2+cosx)3dx\displaystyle \int_0^{\frac{\pi}{2}} \sin x(2+\cos x)^3\,dx.

[4]
Write your answer here...

0

Question 5
SL • Paper 1
Medium
Non Calculator

A function ff satisfies f(x)=6x4sinxf''(x)=6x-4\sin x, f(0)=3f'(0)=3 and f(0)=1f(0)=1.

A

Find f(x)f'(x).

[3]
Write your answer here...
B

Hence find f(x)f(x).

[2]
Write your answer here...

0

Question 6
HL • Paper 1
Medium
Non Calculator
A

Evaluate 01xe2xdx\displaystyle \int_0^1 xe^{2x}\,dx.

[4]
Write your answer here...

0

Question 7
SL • Paper 2
Medium
Calculator Permitted

The function ff satisfies

f(x)=4cos(2x1)3x+2,x>2f'(x)=4\cos(2x-1)-\frac{3}{x+2}, \quad x>-2

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

A

Find an expression for f(x)f(x).

[4]
Write your answer here...
B

Find f(3)f(3).

[2]
Write your answer here...

0

Question 8
SL • Paper 2
Medium
Calculator Permitted

For 0x40\le x\le 4, the shaded region is bounded by the curve

y=3sin(0.6x)+2y=3\sin(0.6x)+2

and the xx-axis.

Graph of $y=3\sin(0.6x)+2$ on $0\le x\le 4$, with the $x$-axis ($y=0$) visible and the region between the curve and the $x$-axis shaded, including the vertical endpoint boundaries.
A

Explain why the area of the shaded region is equal to 04(3sin(0.6x)+2)dx\int_0^4 \left(3\sin(0.6x)+2\right)\,dx.

[2]
Write your answer here...
B

Find the area of the shaded region.

[3]
Write your answer here...

0

Question 9
SL • Paper 1
Medium
Non Calculator

The function ff is defined by f(x)=x24f(x)=x^2-4.

A

Write down the xx-intercepts of the graph of ff.

[1]
Write your answer here...
B

Find the total area enclosed by the graph of ff, the xx-axis and the lines x=1x=-1 and x=3x=3.

[5]
Write your answer here...

0

Question 10
SL • Paper 1
Medium
Non Calculator

Let f(x)=x(x2)f(x)=x(x-2), for 0x30\le x\le 3.

A

Find the value of 03f(x)dx\displaystyle \int_0^3 f(x)\,dx.

[2]
Write your answer here...
B

Find the total area between the graph of ff and the xx-axis for 0x30\le x\le 3.

[3]
Write your answer here...

0

Question 11
HL • Paper 1
Medium
Non Calculator

Consider the integral I=011x2dxI=\displaystyle \int_0^1 \sqrt{1-x^2}\,dx.

A

Using the substitution x=sinθx=\sin\theta, show that I=0π2cos2θdθI=\displaystyle \int_0^{\frac{\pi}{2}} \cos^2\theta\,d\theta.

[4]
Write your answer here...
B

Hence find the exact value of II.

[3]
Write your answer here...

0

Question 12
HL • Paper 1
Medium
Non Calculator
A

Find the exact value of 0π2x2cosxdx\displaystyle \int_0^{\frac{\pi}{2}} x^2\cos x\,dx.

[5]
Write your answer here...

0

Question 13
HL • Paper 1
Medium
Non Calculator

The region RR is bounded by the curve x=4y2x=4-y^2 and the yy-axis, for 0y20\le y\le 2.

A

Find the area of RR.

[2]
Write your answer here...
B

The region RR is rotated through 2π2\pi radians about the yy-axis. Find the volume of the solid formed.

[4]
Write your answer here...

0

Question 14
HL • Paper 1
Medium
Non Calculator

The curves y=xy=\sqrt{x} and y=xy=x enclose a region in the first quadrant.

A

Find the xx-coordinates of the points of intersection of the two curves.

[1]
Write your answer here...
B

The region is rotated through 2π2\pi radians about the xx-axis. Find the volume of the solid formed.

[4]
Write your answer here...

0

Question 15
SL • Paper 2
Medium
Calculator Permitted

Let

f(x)=x34x2+x+6f(x)=x^3-4x^2+x+6

The graph of ff crosses the xx-axis at x=1x=-1, x=2x=2 and x=3x=3.

Cubic graph of y=f(x).
A

Write down an integral expression for the total area enclosed by the graph of ff and the xx-axis between x=1x=-1 and x=3x=3.

[2]
Write your answer here...
B

Find this total area.

[4]
Write your answer here...

0

Question 16
SL • Paper 2
Medium
Calculator Permitted

The functions ff and gg are defined by

f(x)=e0.4x+1,g(x)=0.3x2+1.5f(x)=e^{0.4x}+1, \quad g(x)=0.3x^2+1.5

The graphs of ff and gg enclose two finite regions.

Exponential and quadratic curves with three crossings, forming two finite enclosed regions.
A

Find the xx-coordinates of all points of intersection of the two graphs.

[2]
Write your answer here...
B

Find the total area of the two finite regions enclosed by the graphs.

[4]
Write your answer here...

0

Question 17
HL • Paper 2
Medium
Calculator Permitted

Consider the integral

I=01x1+2xdxI=\int_0^1 \frac{x}{\sqrt{1+2x}}\,dx

Use the substitution u=1+2xu=1+2x.

A

Show that

I=1413u1uduI=\frac14\int_1^3 \frac{u-1}{\sqrt{u}}\,du
[3]
Write your answer here...
B

Hence find the exact value of II.

[3]
Write your answer here...

0

Question 18
HL • Paper 2
Medium
Calculator Permitted

Let

I=02xln(x+1)dxI=\int_0^2 x\ln(x+1)\,dx
A

Use integration by parts to find an exact expression for II.

[5]
Write your answer here...
B

Give the value of II to three significant figures.

[1]
Write your answer here...

0

Question 19
HL • Paper 2
Medium
Calculator Permitted

A curve is given by

x=y24y,0y4x=y^2-4y, \quad 0\le y\le 4

The region RR is enclosed by the curve and the yy-axis.

Curve x=y^2-4y and the y-axis boundary
A

Write down an integral expression for the area of RR.

[2]
Write your answer here...
B

Find the area of RR.

[2]
Write your answer here...
C

The region RR is rotated through 360360^\circ about the yy-axis. Find the volume of the solid formed.

[2]
Write your answer here...

0

Question 20
SL • Paper 1
Medium
Non Calculator

A differentiable function ff is defined for x>12x>-\frac{1}{2}. It is given that

f(x)=4(2x+1)332x+1f'(x)=4(2x+1)^3-\frac{3}{2x+1}

and f(0)=5f(0)=5.

A
I.

Find f(x)f(x) in terms of xx and an arbitrary constant CC.

[3]
Write your answer here...
II.

Hence determine f(x)f(x).

[2]
Write your answer here...
B
I.

Find the exact value of 01f(x)dx\displaystyle \int_0^1 f'(x)\,dx.

[3]
Write your answer here...
II.

Hence find f(1)f(1). If you did not obtain a value in part (b)(i), use 4032ln340-\frac{3}{2}\ln 3.

[2]
Write your answer here...

0

Question 21
SL • Paper 1
Medium
Non Calculator

For x0x\ge 0, define

F(x)=0x2t(t2+4)3dtF(x)=\int_0^x 2t(t^2+4)^3\,dt
A
I.

Find F(x)F(x) in terms of xx.

[3]
Write your answer here...
II.

Find F(2)F(2).

[2]
Write your answer here...
B
I.

Show that FF is an increasing function for x0x\ge 0.

[2]
Write your answer here...
II.

Hence determine the value of aa, where a0a\ge 0, such that 0a2t(t2+4)3dt=960\displaystyle \int_0^a 2t(t^2+4)^3\,dt=960. If you did not obtain F(2)F(2) in part (a)(ii), use 960960.

[3]
Write your answer here...

0

Question 22
SL • Paper 1
Medium
Non Calculator

A function ff satisfies

f(x)=6e3x2cos(2x)f'(x)=6e^{3x}-2\cos(2x)

and f(0)=1f(0)=1.

A

Part (a)

I.

Find f(x)f(x) in terms of xx and an arbitrary constant CC.

[2]
Write your answer here...
II.

Hence determine f(x)f(x).

[2]
Write your answer here...
B

Find f(π4)f\left(\frac{\pi}{4}\right).

[2]
Write your answer here...
C

Part (c)

I.

Find the exact value of 0π4f(x)dx\displaystyle \int_0^{\frac{\pi}{4}} f'(x)\,dx. If you did not obtain f(π4)f\left(\frac{\pi}{4}\right) in part (b), use 2e3π422e^{\frac{3\pi}{4}}-2.

[2]
Write your answer here...
II.

Determine the exact value of aa such that 0a(f(x)+2cos(2x))dx=14\displaystyle \int_0^a \left(f'(x)+2\cos(2x)\right)\,dx=14.

[3]
Write your answer here...

0

Question 23
HL • Paper 1
Medium
Non Calculator

Let J=e2xsinxdxJ=\displaystyle \int e^{2x}\sin x\,dx.

A

By using integration by parts twice, show that J=12e2xsinx14e2xcosx14JJ=\dfrac12e^{2x}\sin x-\dfrac14e^{2x}\cos x-\dfrac14J.

[4]
Write your answer here...
B

Hence find e2xsinxdx\displaystyle \int e^{2x}\sin x\,dx.

[2]
Write your answer here...

0

Question 24
SL • Paper 2
Medium
Calculator Permitted

A small wind turbine produces power at a rate

P(t)=4+1.8sin(0.4t)P(t)=4+1.8\sin(0.4t)

kilowatts, where tt is the time in hours after 06:00 and 0t120\leq t\leq 12. The total energy produced, in kilowatt-hours, is found by integrating P(t)P(t) with respect to tt.

Power output of the wind turbine over the 12-hour interval.
A
I.

Write down an integral expression for the total energy produced during the 12-hour period.

[1]
Write your answer here...
II.

Calculate this total energy.

[3]
Write your answer here...
B
I.

Find the average power produced during the 12-hour period.

[2]
Write your answer here...
II.

Explain why the energy produced from 06:00 to 18:00 is not equal to 12P(6)12P(6).

[1]
Write your answer here...
C

Determine the time after 06:00 at which half of the total energy has been produced.

[3]
Write your answer here...

0

Question 25
SL • Paper 2
Medium
Calculator Permitted

A curve CC has gradient function

dydx=3x2ex312cos(2x),x0\frac{dy}{dx}=3x^2e^{x^3-1}-2\cos(2x), \quad x\geq 0

The curve passes through the point (0,5)(0,5).

A
I.

Find yy in terms of xx.

[4]
Write your answer here...
II.

Verify that your expression satisfies the given gradient function.

[1]
Write your answer here...
B
I.

Find the value of yy when x=1.5x=1.5.

[2]
Write your answer here...
II.

The point on CC with xx-coordinate aa has yy-coordinate 1010, where 0<a<20<a<2. Find aa. If you did not obtain an expression for yy in part (a), use y=ex31sin(2x)+4.63y=e^{x^3-1}-\sin(2x)+4.63.

[3]
Write your answer here...

0

Question 26
SL • Paper 2
Medium
Calculator Permitted

The height, in metres, of an arch above a horizontal walkway is modelled by

h(x)=1+xe0.25x2,0x5h(x)=1+xe^{-0.25x^2}, \quad 0\leq x\leq 5

where xx is the horizontal distance in metres from one end of the arch.

Graph of the arch height function above the walkway.
A
I.

Find the area under the arch between x=0x=0 and x=5x=5.

[3]
Write your answer here...
II.

Find the average height of the arch on this interval.

[1]
Write your answer here...
B

Show that the area under the arch from x=0x=0 to x=cx=c is given by

c+2(1e0.25c2)c+2\left(1-e^{-0.25c^2}\right)
[3]
Write your answer here...
C
I.

Find the value of cc for which the area under the arch from x=0x=0 to x=cx=c is 4 m24\ \text{m}^2.

[2]
Write your answer here...
II.

State why there is only one such value of cc in the interval 0c50\leq c\leq 5.

[1]
Write your answer here...

0

Question 27
SL • Paper 2
Medium
Calculator Permitted

A curve has gradient function

dydx=2xx2+5,x0\frac{dy}{dx}=2x\sqrt{x^2+5}, \quad x\ge 0

The curve passes through (0,4)(0,4).

A

Find yy in terms of xx.

[4]
Write your answer here...
B

Find the value of xx for which y=20y=20.

[3]
Write your answer here...

0

Question 28
HL • Paper 2
Medium
Calculator Permitted

Consider the integral

I=01x2exdxI=\int_0^1 x^2e^{-x}\,dx
A

Use repeated integration by parts to find the exact value of II.

[6]
Write your answer here...
B

Write down the value of II to three significant figures.

[1]
Write your answer here...

0

Question 29
HL • Paper 2
Medium
Calculator Permitted

Let

I=e2xsinxdxI=\int e^{2x}\sin x\,dx
A

Use integration by parts twice to show that

I=e2x5(2sinxcosx)+CI=\frac{e^{2x}}{5}(2\sin x-\cos x)+C
[5]
Write your answer here...
B

Hence find 0πe2xsinxdx\int_0^{\pi} e^{2x}\sin x\,dx to three significant figures.

[2]
Write your answer here...

0

Question 30
HL • Paper 2
Medium
Calculator Permitted

The region bounded by the curve

y=ln(x+2)y=\ln(x+2)

the xx-axis and the line x=4x=4 is rotated through 360360^\circ about the xx-axis.

Graph of y=ln(x+2) with the line x=4.
A

Find the xx-intercept of the curve.

[1]
Write your answer here...
B

Write down an integral expression for the volume of the solid formed.

[2]
Write your answer here...
C

Find this volume.

[4]
Write your answer here...

0

Question 31
HL • Paper 3
Medium
Calculator Permitted

A decorative lamp shade is modelled by rotating the curve y=hsin(πxL),0xLy=h\sin\left(\frac{\pi x}{L}\right),\quad 0\le x\le L through 2π2\pi radians about the xx-axis, where h>0h>0 and L>0L>0.

Normalised lamp-shade arch about the x-axis.
A
I.

Show that 0Lsin2(πxL)dx=L2\int_0^L \sin^2\left(\frac{\pi x}{L}\right)\,dx=\frac{L}{2}

[3]
Write your answer here...
II.

Hence show that the volume of the lamp shade is V=πh2L2V=\dfrac{\pi h^2L}{2}.

[2]
Write your answer here...
B

lamp shade has volume 500 cm3500\text{ cm}^3 and L=10 cmL=10\text{ cm}. Find hh.

[2]
Write your answer here...
C

The value of hh from part (b) is kept fixed. Find the value of LL required to double the volume.

[1]
Write your answer here...
D

Find the area enclosed by the generating curve and the xx-axis in terms of hh and LL.

[2]
Write your answer here...

0

Question 32
SL • Paper 1
Hard
Non Calculator

The function ff is defined by

f(x)=(x+1)(x2)(x4),1x4f(x)=(x+1)(x-2)(x-4), \quad -1\le x\le 4
A

Part (a)

I.

Write down the zeros of ff.

[1]
Write your answer here...
II.

Determine the sign of f(x)f(x) on each of the intervals 1<x<2-1<x<2 and 2<x<42<x<4.

[2]
Write your answer here...
B

Part (b)

I.

Show that an antiderivative of ff is F(x)=x445x33+x2+8xF(x)=\frac{x^4}{4}-\frac{5x^3}{3}+x^2+8x.

[3]
Write your answer here...
II.

Find the value of 14f(x)dx\displaystyle \int_{-1}^{4} f(x)\,dx. If you did not show the antiderivative in part (b)(i), use F(x)=x445x33+x2+8xF(x)=\frac{x^4}{4}-\frac{5x^3}{3}+x^2+8x.

[2]
Write your answer here...
C

Find the total area enclosed by the graph of ff and the xx-axis for 1x4-1\le x\le 4.

[3]
Write your answer here...

0

Question 33
SL • Paper 1
Hard
Non Calculator

The functions ff and gg are defined by

f(x)=5x2,g(x)=x2+1f(x)=5-x^2, \qquad g(x)=x^2+1

The graphs of ff and gg enclose a finite region.

A
I.

Find the xx-coordinates of the points of intersection of the graphs.

[3]
Write your answer here...
II.

State which graph is above the other between the points of intersection.

[1]
Write your answer here...
B

Find the area of the finite region enclosed by the two graphs.

[4]
Write your answer here...
C

horizontal line y=ky=k, where 1<k<51<k<5, intersects the graph of ff at two points. The area enclosed by y=f(x)y=f(x) and y=ky=k is 43\frac{4}{3}. Find the value of kk.

[3]
Write your answer here...

0

Question 34
SL • Paper 1
Hard
Non Calculator

The function ff is defined by

f(x)=x1x,12x2f(x)=x-\frac{1}{x}, \quad \frac{1}{2}\le x\le 2
A
I.

Find the xx-intercept of the graph of ff on this interval.

[1]
Write your answer here...
II.

State the sign of f(x)f(x) on each side of this intercept in the interval 12x2\frac{1}{2}\le x\le 2.

[2]
Write your answer here...
B
I.

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

[2]
Write your answer here...
II.

Find the exact value of 122f(x)dx\displaystyle \int_{\frac{1}{2}}^2 f(x)\,dx.

[2]
Write your answer here...
C

Find the total area between the graph of ff and the xx-axis on the interval 12x2\frac{1}{2}\le x\le 2.

[3]
Write your answer here...

0

Question 35
HL • Paper 1
Hard
Non Calculator

Consider the definite integral

I=01x3(1+x2)2dxI=\int_0^1 \frac{x^3}{(1+x^2)^2}\,dx
A
I.

Use the substitution u=1+x2u=1+x^2 to show that

I=1212u1u2duI=\frac{1}{2}\int_1^2 \frac{u-1}{u^2}\,du
[4]
Write your answer here...
II.

Hence find the exact value of II.

[3]
Write your answer here...
B

The region under the curve y=x321+x2y=\dfrac{x^{\frac{3}{2}}}{1+x^2}, above the xx-axis, between x=0x=0 and x=1x=1, is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid formed. If you did not obtain a value for II, use I=12ln214I=\frac{1}{2}\ln2-\frac{1}{4}.

[3]
Write your answer here...

0

Question 36
HL • Paper 1
Hard
Non Calculator

The curve y=lnxy=\ln x is considered for 1xe1\le x\le e.

A
I.

Use integration by parts to show that

lnxdx=xlnxx+C\int \ln x\,dx=x\ln x-x+C
[4]
Write your answer here...
II.

Hence find the area between the curve, the xx-axis, and the lines x=1x=1 and x=ex=e.

[2]
Write your answer here...
B
I.

Use integration by parts and the result from part (a)(i) to show that

(lnx)2dx=x(lnx)22xlnx+2x+C\int (\ln x)^2\,dx=x(\ln x)^2-2x\ln x+2x+C
[4]
Write your answer here...
II.

The region between y=lnxy=\ln x, the xx-axis, and the lines x=1x=1 and x=ex=e is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid formed. If you did not show the result in part (b)(i), use it here.

[2]
Write your answer here...

0

Question 37
HL • Paper 1
Hard
Non Calculator

A curve is given by

x=ey1,0yln3x=e^y-1, \quad 0\le y\le \ln 3

The region RR is bounded by the curve, the yy-axis, and the lines y=0y=0 and y=ln3y=\ln 3.

A
I.

Find the coordinates of the endpoints of the curve on the boundary of RR.

[2]
Write your answer here...
II.

Find the area of RR.

[3]
Write your answer here...
B
I.

The region RR is rotated through 2π2\pi radians about the yy-axis. Write down an integral expression for the volume of the solid formed.

[2]
Write your answer here...
II.

Hence find the exact volume of the solid formed.

[3]
Write your answer here...

0

Question 38
HL • Paper 1
Hard
Non Calculator

Consider the integral

I=0111+xdxI=\int_0^1 \frac{1}{1+\sqrt{x}}\,dx
A
I.

Use the substitution u=xu=\sqrt{x} to show that

I=012u1+uduI=\int_0^1 \frac{2u}{1+u}\,du
[4]
Write your answer here...
II.

Hence evaluate II exactly.

[3]
Write your answer here...
B

The region under the curve y=11+xy=\dfrac{1}{\sqrt{1+\sqrt{x}}}, above the xx-axis, between x=0x=0 and x=1x=1, is rotated through 2π2\pi radians about the xx-axis. Find the exact volume of the solid formed. If you did not obtain a value for II, use I=22ln2I=2-2\ln2.

[3]
Write your answer here...

0

Question 39
SL • Paper 2
Hard
Calculator Permitted

The functions ff and gg are defined by

f(x)=6e0.3x+1,g(x)=0.8x+2,x0f(x)=6e^{-0.3x}+1, \quad g(x)=0.8x+2, \quad x\geq 0

The region RR is bounded by the yy-axis and the graphs of ff and gg.

Graphs of f and g for x >= 0.
A
I.

Find the xx-coordinate of the point of intersection of the two graphs.

[2]
Write your answer here...
II.

State which curve is above the other for the region RR.

[1]
Write your answer here...
B

Find the area of RR.

[4]
Write your answer here...
C
I.

vertical line x=kx=k divides RR into two regions of equal area. Write down an equation that kk satisfies.

[2]
Write your answer here...
II.

Hence find kk. If you did not obtain the area in part (b), use A=5.56A=5.56.

[2]
Write your answer here...

0

Question 40
SL • Paper 2
Hard
Calculator Permitted

Let

f(x)=2sinx1,0x2πf(x)=2\sin x-1, \quad 0\leq x\leq 2\pi

The graph of ff crosses the xx-axis twice on this interval.

Curve of f(x)=2sin x-1 on 0≤x≤2π.
A
I.

Find the two xx-intercepts of the graph of ff.

[2]
Write your answer here...
II.

Find the signed area 02πf(x)dx\displaystyle\int_0^{2\pi}f(x)\,\mathrm dx.

[2]
Write your answer here...
B

Find the total area between the graph of ff and the xx-axis for 0x2π0\leq x\leq 2\pi.

[4]
Write your answer here...
C
I.

Let kk be the value such that the total area between the graph and the xx-axis from x=0x=0 to x=kx=k is half of the total area found in part (b). Write down an equation for kk.

[2]
Write your answer here...
II.

Hence find kk. If you did not obtain the total area in part (b), use A=9.02A=9.02.

[2]
Write your answer here...

0

Question 41
HL • Paper 2
Hard
Calculator Permitted

The curve CC has equation

y=xe0.5x,0x5y=xe^{-0.5x}, \quad 0\leq x\leq 5

The region RR is bounded by CC, the xx-axis, and the lines x=0x=0 and x=5x=5.

Curve y=xe^-0.5x on 0<=x<=5.
A
I.

Use integration by parts to find the area of RR.

[4]
Write your answer here...
II.

Find the maximum height of the region RR.

[1]
Write your answer here...
B

The region RR is rotated through 2π2\pi radians about the xx-axis. Show that the volume of the solid formed is

V=π05x2exdxV=\pi\int_0^5x^2e^{-x}\,\mathrm dx
[2]
Write your answer here...
C
I.

Use repeated integration by parts to find 05x2exdx\displaystyle\int_0^5x^2e^{-x}\,\mathrm dx exactly.

[4]
Write your answer here...
II.

Hence find the volume of the solid formed.

[2]
Write your answer here...

0

Question 42
HL • Paper 2
Hard
Calculator Permitted

A plane region RR lies between the curve

x=y9y2,0y3x=y\sqrt{9-y^2}, \quad 0\leq y\leq 3

and the yy-axis.

Plane region bounded by the curve x = y sqrt(9 - y^2) for 0 <= y <= 3 and the y-axis.
A
I.

Write down an integral expression for the area of RR.

[1]
Write your answer here...
II.

Use the substitution u=9y2u=9-y^2 to find the area of RR.

[4]
Write your answer here...
B

The region RR is rotated through 2π2\pi radians about the yy-axis. Find the volume of the solid formed.

[4]
Write your answer here...
C
I.

Let y=ky=k divide the solid into two parts of equal volume. Write down an equation for kk.

[1]
Write your answer here...
II.

Hence find kk. If you did not obtain the volume in part (b), use V=32.4πV=32.4\pi.

[2]
Write your answer here...

0

Question 43
HL • Paper 2
Hard
Calculator Permitted

The region RR is bounded by the curve

y=ln(2x+1)y=\ln(2x+1)

the xx-axis, the yy-axis, and the line x=4x=4.

Curve with shaded region R and the boundary segment x = 4.
A
I.

Use integration by parts to find the exact area of RR.

[3]
Write your answer here...
II.

Write this area correct to three significant figures.

[1]
Write your answer here...
B

The region RR is rotated through 2π2\pi radians about the xx-axis. Write down an integral expression for the volume VV of the solid formed, and calculate VV.

[4]
Write your answer here...
C
I.

Let x=ax=a divide the solid into two parts of equal volume. Write down an equation for aa.

[2]
Write your answer here...
II.

Hence find aa. If you did not obtain the volume in part (b), use V=31.3V=31.3.

[2]
Write your answer here...

0

Question 44
HL • Paper 3
Hard
Calculator Permitted

For a>0a>0, define I(a)=012x1+ax2dxI(a)=\int_0^1 \frac{2x}{1+a x^2}\,dx This question investigates how I(a)I(a) varies with the parameter aa.

Decreasing curve of I(a)=ln(1+a)/a with a horizontal line at I(a)=0.5.
A
I.

Use the substitution u=1+ax2u=1+a x^2 to express I(a)I(a) as an integral with respect to uu.

[2]
Write your answer here...
II.

Hence show that I(a)=ln(1+a)aI(a)=\dfrac{\ln(1+a)}{a}.

[2]
Write your answer here...
B

Find the value of aa for which I(a)=0.5I(a)=0.5.

[2]
Write your answer here...
C

Prove that I(a)I(a) is a decreasing function of aa for a>0a>0.

[3]
Write your answer here...
D

Deduce the number of positive solutions of I(a)=mI(a)=m when 0<m<10<m<1.

[2]
Write your answer here...

0

Question 45
HL • Paper 3
Hard
Calculator Permitted

For nZ0n\in\mathbb{Z}_{\ge 0}, let

In=01xnexdxI_n=\int_0^1 x^n e^x\,dx

This question investigates a recurrence relation for InI_n.

nn

InI_n

4

9e240.46459e-24\approx0.4645

5

12044e0.3956120-44e\approx0.3956

6

265e7200.3447265e-720\approx0.3447

7

50401854e0.30555040-1854e\approx0.3055

8

14833e403200.274414833e-40320\approx0.2744

9

362880133497e0.2490362880-133497e\approx0.2490

10

1334971e36288000.22801334971e-3628800\approx0.2280

11

3991680014684681e0.210339916800-14684681e\approx0.2103

A
I.

Write down the exact value of I0I_0.

[1]
Write your answer here...
II.

Use integration by parts to show that In=enIn1I_n=e-nI_{n-1} for n1n\ge 1.

[4]
Write your answer here...
B

Hence find the exact values of I1I_1, I2I_2 and I3I_3.

[3]
Write your answer here...
C

Show that 0<In<en+10<I_n<\dfrac{e}{n+1} for n0n\ge 0.

[2]
Write your answer here...
D

Use the result in part (c) to find an integer NN such that In<0.01I_n<0.01 for all nNn\ge N.

[2]
Write your answer here...

0

Question 46
HL • Paper 3
Hard
Calculator Permitted

For 0<k<10<k<1, the curve y=exy=e^{-x} and the line y=ky=k enclose a finite region with the yy-axis. The region is denoted by RkR_k.

Illustrative graph for k₀=0.6 showing y=e^{-x}, y=k₀, and the shaded region R_{k₀} bounded by the y-axis.
A
I.

Find the xx-coordinate of the point where y=exy=e^{-x} meets y=ky=k.

[1]
Write your answer here...
II.

Show that the area of RkR_k is A(k)=1k+klnkA(k)=1-k+k\ln k.

[3]
Write your answer here...
B

Find A(k)A'(k) and interpret its sign for 0<k<10<k<1.

[2]
Write your answer here...
C

Find the value of kk for which the area of RkR_k is 0.250.25.

[2]
Write your answer here...
D

For this value of kk, find the volume generated when RkR_k is rotated through 2π2\pi radians about the xx-axis. If you did not obtain a value for kk, use k=0.383k=0.383.

[3]
Write your answer here...

0

Question 47
HL • Paper 3
Hard
Calculator Permitted

A region RR is bounded by the curve x=y2(3y),0y3x=y^2(3-y),\quad 0\le y\le 3 and the yy-axis. The curve lies to the right of the yy-axis on this interval.

Curve x=y^2(3-y) bounding region R.
A

Find the area of RR.

[3]
Write your answer here...
B

Find the volume generated when RR is rotated through 2π2\pi radians about the yy-axis.

[3]
Write your answer here...
C
I.

horizontal line y=cy=c divides RR into two regions of equal area. Write an equation satisfied by cc.

[1]
Write your answer here...
II.

Find cc.

[2]
Write your answer here...
D

Explain why integration with respect to yy is the natural method for this region.

[1]
Write your answer here...

0

Question 48
HL • Paper 3
Hard
Calculator Permitted

The region RR is bounded by the curve y=lnxy=-\ln x, the xx-axis, and the vertical lines x=e1x=e^{-1} and x=1x=1.

Shaded region R under $y=-\ln x$, bounded by $x=e^{-1}$ and $x=1$.
A

Find the exact area of RR.

[3]
Write your answer here...
B
I.

Use integration by parts to show that

(lnx)2dx=x((lnx)22lnx+2)+C\int (\ln x)^2\,dx=x\left((\ln x)^2-2\ln x+2\right)+C
[3]
Write your answer here...
C

Find the exact volume generated when RR is rotated through 2π2\pi radians about the xx-axis.

[2]
Write your answer here...
D

vertical line x=kx=k divides the volume in part (c) into two equal volumes. Find kk.

[3]
Write your answer here...

0

Question 49
HL • Paper 3
Hard
Calculator Permitted

The curve

y=x4x2,0x2y=x\sqrt{4-x^2},\quad 0\le x\le 2

forms a single arch above the xx-axis. The region under the arch is denoted by RR.

Arch of y = x sqrt(4 - x^2) on 0<=x<=2.
A
I.

Use the substitution u=4x2u=4-x^2 to find the exact area of RR.

[3]
Write your answer here...
II.

State why the definite integral gives the geometric area without splitting the interval.

[1]
Write your answer here...
B

Find the volume generated when RR is rotated through 2π2\pi radians about the xx-axis.

[3]
Write your answer here...
C

vertical plane perpendicular to the xx-axis cuts the solid into two parts of equal volume at x=cx=c. Find cc.

[3]
Write your answer here...

0

Question 50
HL • Paper 3
Hard
Calculator Permitted

A region RkR_k is bounded by the curve

x=ln(y+1)x=\ln(y+1)

the yy-axis, and the horizontal lines y=0y=0 and y=ky=k, where k>0k>0.

Boundary curve x=ln(y+1) with the coordinate axes.
A

Write down an integral expression for the area of RkR_k.

[1]
Write your answer here...
B
I.

Use the substitution u=y+1u=y+1 and integration by parts to show that

ln2(y+1)dy=(y+1)(ln2(y+1)2ln(y+1)+2)+C\int \ln^2(y+1)\,dy=(y+1)(\ln^2(y+1)-2\ln(y+1)+2)+C
[3]
Write your answer here...
II.

Write down an integral expression for the volume formed when RkR_k is rotated through 2π2\pi radians about the yy-axis.

[1]
Write your answer here...
C

Show that V=π((k+1)(ln2(k+1)2ln(k+1)+2)2)V=\pi\left((k+1)(\ln^2(k+1)-2\ln(k+1)+2)-2\right)

[2]
Write your answer here...
D

Find the value of kk for which V=10V=10.

[3]
Write your answer here...

0

Question 51
HL • Paper 3
Hard
Calculator Permitted

A density profile along a beam is modelled by f(x)=xex22,x0f(x)=x e^{-\frac{x^2}{2}},\quad x\ge 0 For a>0a>0, define A(a)=0axex22dxA(a)=\int_0^a x e^{-\frac{x^2}{2}}\,dx

Graph of f(x)=x e^{-x^2/2} for x≥0.
A
I.

Use the reverse chain rule to show that A(a)=1ea22A(a)=1-e^{-\frac{a^2}{2}}.

[3]
Write your answer here...
II.

Find limaA(a)\lim\limits_{a\to\infty}A(a).

[1]
Write your answer here...
B

Find aa such that A(a)=0.8A(a)=0.8.

[2]
Write your answer here...
C

The region under the curve from x=0x=0 to the value of aa found in part (b) is rotated through 2π2\pi radians about the xx-axis. Write down an integral expression for the volume and find its value. If you did not obtain aa, use a=1.79a=1.79.

[4]
Write your answer here...

0

Question 52
HL • Paper 1
Hard
Non Calculator

Let

I=01x2e2xdxI=\int_0^1 x^2e^{2x}\,dx
A
I.

By using integration by parts, show that

I=e2201xe2xdxI=\frac{e^2}{2}-\int_0^1 xe^{2x}\,dx
[3]
Write your answer here...
II.

Hence evaluate II exactly.

[4]
Write your answer here...
B
I.

Find the exact value of 01x(1x)e2xdx\displaystyle \int_0^1 x(1-x)e^{2x}\,dx. If you did not obtain II in part (a)(ii), use I=e214I=\frac{e^2-1}{4}.

[2]
Write your answer here...
II.

Determine the exact value of kk such that

01(x2k)e2xdx=0\int_0^1 (x^2-k)e^{2x}\,dx=0
[3]
Write your answer here...

0

Question 53
HL • Paper 1
Hard
Non Calculator

Let

I=0π2exsinxdx,J=0π2excosxdxI=\int_0^{\frac{\pi}{2}}e^x\sin x\,dx, \qquad J=\int_0^{\frac{\pi}{2}}e^x\cos x\,dx
A
I.

Show that I=eπ2JI=e^{\frac{\pi}{2}}-J.

[2]
Write your answer here...
II.

Show that J=I1J=I-1.

[2]
Write your answer here...
B

Hence find the exact values of II and JJ.

[3]
Write your answer here...
C

Find the total area between the graph of y=ex(sinxcosx)y=e^x(\sin x-\cos x) and the xx-axis for 0xπ20\le x\le \frac{\pi}{2}.

[4]
Write your answer here...

0

Question 54
SL • Paper 2
Hard
Calculator Permitted

Water enters a tank at a rate

R(t)=201+0.15t2R(t)=\frac{20}{1+0.15t^2}

litres per minute, where tt is the time in minutes after the tap is opened. At the same time, water leaves the tank at a constant rate of 44 litres per minute. Initially the tank contains 6060 litres of water.

Water inflow and outflow rates over the first 8 minutes.
A
I.

Write down an integral expression for the net change in the volume of water in the tank during the first 8 minutes.

[1]
Write your answer here...
II.

Calculate this net change.

[3]
Write your answer here...
B
I.

Find the time at which the accumulated volume predicted by the inflow and outflow rates is greatest during the first 8 minutes, ignoring any capacity limit.

[2]
Write your answer here...
II.

Find this greatest accumulated volume, ignoring any capacity limit.

[2]
Write your answer here...
C

The capacity of the tank is 9090 litres. Determine the time at which the tank first overflows.

[3]
Write your answer here...
D

Explain why the tank is not still overflowing at t=8t=8.

[2]
Write your answer here...

0

Question 55
HL • Paper 2
Hard
Calculator Permitted

Let

f(x)=exsin(2x),0xπf(x)=e^{-x}\sin(2x), \quad 0\leq x\leq \pi

The function is positive on 0<x<π20<x<\dfrac{\pi}{2} and negative on π2<x<π\dfrac{\pi}{2}<x<\pi.

Graph of f(x)=e^{-x}sin(2x) on 0≤x≤π, with zeros marked.
A
I.

Use integration by parts twice to show that

exsin(2x)dx=ex(sin(2x)2cos(2x))5+C\int e^{-x}\sin(2x)\,\mathrm dx=\frac{e^{-x}(-\sin(2x)-2\cos(2x))}{5}+C
[4]
Write your answer here...
II.

Hence find the signed area 0πf(x)dx\displaystyle\int_0^\pi f(x)\,\mathrm dx.

[1]
Write your answer here...
B

Find the total area between the graph of ff and the xx-axis for 0xπ0\leq x\leq \pi.

[3]
Write your answer here...
C
I.

The value kk, where 0<k<π20<k<\dfrac{\pi}{2}, is such that the area between the graph and the xx-axis from 00 to kk is half of the positive lobe area. Write down an equation for kk.

[2]
Write your answer here...
II.

Hence find kk. If you did not obtain the positive lobe area, use 0.4830.483.

[2]
Write your answer here...

0

Question 56
HL • Paper 2
Hard
Calculator Permitted

The region RR is bounded by the curve

y=x1+x,0x4y=\frac{\sqrt{x}}{1+x}, \quad 0\leq x\leq 4

the xx-axis, and the lines x=0x=0 and x=4x=4.

Curve and bounded region under $y=\frac{\sqrt{x}}{1+x}$ for $0\le x\le 4$, with the x-axis and boundary $x=4$ shown.
A
I.

Use the substitution u=xu=\sqrt{x} to show that the area of RR is

42arctan24-2\arctan 2
[4]
Write your answer here...
II.

Find the area of RR to three significant figures.

[1]
Write your answer here...
B

The region RR is rotated through 2π2\pi radians about the xx-axis. Find the volume of the solid formed.

[4]
Write your answer here...
C
I.

Let x=cx=c divide the solid into two parts of equal volume. Write down an equation for cc.

[1]
Write your answer here...
II.

Hence find cc, taking the root in the interval 0<c<40<c<4. If you did not obtain the volume in part (b), use V=2.54V=2.54.

[2]
Write your answer here...

0

Question 57
HL • Paper 3
Hard
Calculator Permitted

For real a>0a>0, define K(a)=0π2eaxcosxdxK(a)=\int_0^{\frac{\pi}{2}} e^{ax}\cos x\,dx This question first considers the case a=2a=2 and then generalizes.

Family of curves for y=e^{ax}cos x on 0≤x≤π/2.
A
I.

Using integration by parts twice, show that e2xcosxdx=e2x(2cosx+sinx)5+C\int e^{2x}\cos x\,dx=\frac{e^{2x}(2\cos x+\sin x)}{5}+C

[4]
Write your answer here...
II.

Hence find K(2)K(2) exactly.

[1]
Write your answer here...
B

By considering eaxcosxdx\int e^{ax}\cos x\,dx, show that K(a)=eaπ2aa2+1K(a)=\frac{e^{\frac{a\pi}{2}}-a}{a^2+1}

[4]
Write your answer here...
C

Find the positive value of aa for which K(a)=10K(a)=10.

[2]
Write your answer here...
D

Explain why no value 0<a<20<a<2 can satisfy K(a)=10K(a)=10.

[1]
Write your answer here...

0

Question 58
HL • Paper 3
Hard
Calculator Permitted

The tangent half-angle substitution is t=tanx2,cosx=1t21+t2,dx=21+t2dtt=\tan\frac{x}{2},\quad \cos x=\frac{1-t^2}{1+t^2},\quad dx=\frac{2}{1+t^2}\,dt. This question investigates I(a)=0π21a+cosxdx,a>1I(a)=\int_0^{\frac{\pi}{2}}\frac{1}{a+\cos x}\,dx,\quad a>1.

A
I.

For a=2a=2, show that the integral becomes 0123+t2dt\int_0^1 \frac{2}{3+t^2}\,dt.

[2]
Write your answer here...
II.

Hence find I(2)I(2) exactly.

[2]
Write your answer here...
B

Show that, for a>1a>1,

I(a)=2a21arctana1a+1I(a)=\frac{2}{\sqrt{a^2-1}}\arctan\sqrt{\frac{a-1}{a+1}}
[4]
Write your answer here...
C
I.

Find lima1+I(a)\lim\limits_{a\to 1^+} I(a) directly from the original integral.

[2]
Write your answer here...
II.

State what this suggests about the formula in part (b) as a1+a\to 1^+.

[1]
Write your answer here...

0

Question 59
HL • Paper 3
Hard
Calculator Permitted

For nZ0n\in\mathbb{Z}_{\ge 0}, define

An=01xnln(1+x)dx,Bn=01xn1+xdxA_n=\int_0^1 x^n\ln(1+x)\,dx,\qquad B_n=\int_0^1\frac{x^n}{1+x}\,dx

n

A_n

nA_n

1

0.250000

0.250000

2

0.184320

0.368641

5

0.102778

0.513889

10

0.059068

0.590682

20

0.031900

0.638004

50

0.013401

0.670039

A

Find the exact value of A0A_0.

[2]
Write your answer here...
B
I.

Use integration by parts to show that An=ln2n+1Bn+1n+1A_n=\frac{\ln2}{n+1}-\frac{B_{n+1}}{n+1}

[3]
Write your answer here...
II.

Show that Bn+Bn1=1nB_n+B_{n-1}=\dfrac1n for n1n\ge 1.

[1]
Write your answer here...
C

Find the exact values of A1A_1 and A2A_2.

[4]
Write your answer here...
D

Using the table or technology, estimate limnnAn\lim\limits_{n\to\infty} nA_n and explain why this value is reasonable.

[2]
Write your answer here...

0

Question 60
HL • Paper 2
Hard
Calculator Permitted

For k>0k>0, a family of profile curves is given by

x=yeky,0y4x=ye^{-ky}, \quad 0\leq y\leq 4

The region RkR_k is bounded by the curve, the yy-axis, and the lines y=0y=0 and y=4y=4.

Representative curve x = y e^(-0.25y) with the boundary y = 4.
A

For this part, take k=0.25k=0.25.

I.

Write down an integral expression for the area of R0.25R_{0.25}.

[1]
Write your answer here...
II.

Use integration by parts to find this area.

[4]
Write your answer here...
B
I.

The region R0.25R_{0.25} is rotated through 2π2\pi radians about the yy-axis. Write down an integral expression for the volume VV of the solid formed.

[2]
Write your answer here...
II.

Calculate VV.

[3]
Write your answer here...
C
I.

Let y=hy=h divide the solid in part (b) into two parts of equal volume. Write down an equation for hh.

[1]
Write your answer here...
II.

Hence find hh. If you did not obtain the volume in part (b), use V=16.3V=16.3.

[2]
Write your answer here...
D

For general k>0k>0, let

V(k)=π04y2e2kydyV(k)=\pi\int_0^4 y^2e^{-2ky}\,\mathrm dy

Justify that V(k)V(k) decreases as kk increases, and determine the value of kk for which V(k)=10V(k)=10.

[2]
Write your answer here...

0


Differential Equations

Kinematics