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Higher Mathematics
Integration
Paper 1 Section A
Each correct answer in this section is worth two marks.
1. Evaluate
Z
A.
−2
B.
7
− 16
C.
1
2
D.
2
Key
D
4
1
x −1/2 dx .
Outcome
2.2
Grade
C
Facility
0.71
Disc.
0.64
Calculator
NC
Content
C13
Source
HSN 086
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2. The diagram shows the area bounded by the curves y = x 3 − 3x2 + 4
and y = x2 − x − 2 between x = −1 and x = 2.
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y
y = x3 − 3x2 + 4
y = x2 − x − 2
−1
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x
2
Represent the shaded area as an integral.
Z 2
A.
x3 − 4x2 + x + 6 dx
−1
B.
Z
C.
Z
D.
Z
Key
A
2
−1
2
−1
2
−1
−x3 + 4x2 − x − 6 dx
x3 − 4x2 − x + 2 dx
x3 − 2x2 − x + 2 dx
Outcome
2.2
Grade
C
Facility
0.7
Disc.
0.46
Calculator
NC
Content
C17
Source
HSN 094
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[END OF PAPER 1 SECTION A]
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Higher Mathematics
Paper 1 Section B
3. Find
[SQA]
Z 2x + 3 dx .
3
Z 3x3 + 4x dx .
3
2
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4. Find
[SQA]
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5. Evaluate
[SQA]
Z
1
2
1
x +
x
2
2
5
dx .
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6. Find
[SQA]
Z
x2 − 5
√ dx .
x x
4
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7. Find the value of
[SQA]
Z
2
1
u2 + 2
du.
2u2
5
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8.
[SQA]
(a) Find the value of
Z 2
1
4 − x2 dx .
3
(b) Sketch a graph and shade the area represented by the integral in (a).
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2
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9. Evaluate
[SQA]
Z
9
1
x+1
√ dx .
x
5
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10. Find the value of
[SQA]
Z
1
4√
4
x dx .
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11. Evaluate
[SQA]
Z 2
1
this integral.
2
3x + 4 dx and draw a sketch to illustrate the area represented by
5
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12.
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13. A function f is defined by the formula f (x) = 4x 2 (x − 3) where x ∈ R.
[SQA]
(a) Write down the coordinates of the points where the curve with equation
y = f (x) meets the x - and y-axes.
2
(b) Find the stationary points of y = f (x) and determine the nature of each.
6
(c) Sketch the curve y = f (x).
2
(d) Find the area completely enclosed by the curve y = f (x) and the x -axis.
4
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14.
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15.
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16.
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17.
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18.
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19. Functions f and g are defined on the set of real numbers by f (x) = x − 1 and
g(x) = x2 .
[SQA]
(a) Find formulae for
(i) f g(x)
(ii) g f (x) .
4
(b) The function h is defined by h(x) = f g(x) + g f (x) .
Show that h(x) = 2x2 − 2x and sketch the graph of h.
3
(c) Find the area enclosed between this graph and the x -axis.
4
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20. The diagram shows a sketch of the graphs of y
y = x3 − 12x + 1.
[SQA]
=
5x 2 − 15x − 8 and
The two curves intersect at A and touch at B, i.e. at B the curves have a common
tangent.
y
y = x3 − 12x + 1
A
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B
y = 5x2 − 15x − 8
(a) (i) Find the x -coordinates of the point of the curves where the gradients are
equal.
4
(ii) By considering the corresponding y-coordinates, or otherwise,
distinguish geometrically between the two cases found in part (i).
1
(b) The point A is (−1, 12) and B is (3, −8).
5
Find the area enclosed between the two curves.
Part
(ai)
(aii)
(b)
•1
•2
•3
•4
Marks
4
1
5
ss:
pd:
pd:
ic:
•5 ic:
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Level
C
C
C
Calc.
NC
NC
NC
Content
C4
0.1
C17
•1
•2
•3
•4
know to diff. and equate
differentiate
form equation
interpret solution
interpret diagram
•6 ss: know how to find area between
curves
7
• ic: interpret limits
•8 pd: form integral
•9 pd: process integration
•10 pd: process limits
hsn.uk.net
Answer
x = 13 and x = 3
parallel and coincident
21 13
Page 14
U2 OC2
2000 P1 Q4
find derivatives and equate
3x2 − 12 and 10x − 15
3x2 − 10x + 3 = 0
x = 3, x = 13
•5 tangents at x = 13 are parallel, at
x = 3 coincident
R
•6 (cubic
− parabola)
R
R
or (cubic) − (parabola)
R3
•7 −1 · · · dx
R
•8 (x3 − 5x2 + 3x + 9)dx or equiv.
3
•9 14 x4 − 53 x3 + 32 x2 + 9x −1 or equiv.
•10 21 13
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21.
[SQA]
(a) Find the coordinates of the points of intersection of the curves with equations
y = 2x2 and y = 4 − 2x2 .
2
(b) Find the area completely enclosed between these two curves.
3
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22.
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23.
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24.
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25.
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26. The graph of y = g(x) passes through the point (1, 2).
[SQA]
If
1
1
dy
= x3 + 2 − , express y in terms of x .
dx
4
x
4
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27. For all points on the curve y = f (x), f 0 (x) = 1 − 2x .
[SQA]
If the curve passes through the point (2, 1), find the equation of the curve.
4
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28. A curve with equation y = f (x) passes through the point (2, −1) and is such that
f 0 (x) = 4x3 − 1.
[SQA]
5
Express f (x) in terms of x .
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dy
= 3x2 + 1 passes through the point (−1, 2).
dx
Express y in terms of x .
4
dy
= 6x2 − 2x passes through the point (−1, 2).
dx
Express y in terms of x .
3
29. A curve for which
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30. A curve for which
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31.
[SQA]
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32.
[SQA]
Z
π
2
cos 2x dx .
3
(b) Draw a sketch and explain your answer.
2
(a) Evaluate
0
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33.
[SQA]
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34. The graph of y = f (x) passes through the point
[SQA]
π
9,1
.
4
If f 0 (x) = sin(3x) express y in terms of x .
Part
•1
•2
•3
•4
Marks
4
ss:
pd:
ic:
pd:
Level
A/B
Calc.
NC
Content
C18, C23
Answer
y = − 13 cos(3x) +
7
6
U3 OC2
2000 P1 Q8
R
•1 y = sin(3x) dx stated or implied by
•2
•2 − 13 cos(3x)
•3 1 = − 13 cos( 3π
9 ) + c or equiv.
7
4
• c=6
know to integrate
integrate
interpret ( π9 , 1)
process
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35. Differentiate sin3 x with respect to x .
Z
Hence find
sin2 x cos x dx .
[SQA]
4
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36. Evaluate
[SQA]
Z
0
−3
2x + 3
2
4
dx .
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37.
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38.
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[END OF PAPER 1 SECTION B]
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Paper 2
1. Find
[SQA]
Z
Part
•1
•2
•3
•4
(x2 − 2)(x2 + 2)
dx , x 6= 0.
x2
Marks
4
ss:
pd:
pd:
pd:
Level
C
Calc.
CN
4
Content
C14, C12, C13
start to write in standard form
complete process
integrate
integrate a −ve index
Answer
1 3
−1 + c
3 x + 4x
U2 OC2
2001 P2 Q6
x4 − 4
x2
2
2
• x − 4x −2
•3 13 x3 + c
−4x −1
•4
−1
•1
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2.
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y
3. The parabola shown crosses the x -axis at
(0, 0) and (4, 0), and has a maximum at
4
(2, 4).
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[SQA]
The shaded area is bounded by the
parabola, the x -axis and the lines x = 2
and x = k.
(a) Find the equation of the parabola.
O
2
k 4
x
2
(b) Hence show that the shaded area, A,
is given by
A = − 13 k3 + 2k2 −
Part
(a)
(b)
Marks
2
3
Level
C
C
16
3 .
Calc.
CN
CN
3
Content
A19
C16
•1 ic: state standard form
•2 pd: process for x 2 coeff.
•3 ss: know to integrate
•4 pd: integrate correctly
•5 pd: process limits and complete
proof
Answer
y = 4x − x 2
proof
U2 OC2
2000 P2 Q4
•1 ax(x − 4)
•2 a = −1
Rk
•3 2 (function from (a))
•4 − 13 x3 + 2x2
•5 − 13 k3 + 2k2 − − 83 + 8
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4. A firm asked for a logo to be
designed involving the letters A
and U. Their initial sketch is
shown in the hexagon.
[SQA]
y
(4, 15)
A mathematical representation
PSfrag
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of the final logo is shown
in the
coordinate diagram.
O
The
curve
has
equation
y = (x + 1)(x − 1)(x − 3) and
the straight line has equation
y = 5x − 5. The point (1, 0) is
the centre of half-turn symmetry.
0
−2 −1
1
2
3
4
7
Calculate the total shaded area.
(−2, −15)
Part
•1
•2
•3
•4
•5
•6
•7
Marks
7
Level
C
Calc.
CN
Content
C17
ss: express in standard form
ss: split area and integrate
ss: subtract functions
pd: process
pd: process
pd: process
ic: use symmetry or otherwise for
total area
Answer
40 12 units2
U2 OC2
2001 P2 Q8
•1 y = x3 − 3x2 − x + 3
R4
R1
•2 1 (. . .)dx or −2 (. . .)dx
R
3
2
•3
−3 5) − 2(x − 3x − x + 3) dx
R(5x
or
(x
−
3x
−
x
+
3)
−
(5x
−
5)
dx
R
4
3
2
• (−x + 3x + 6x − 8)dx
•5 − 14 x4 + x3 + 3x2 − 8x
•6 20 14 or −20 14
depending on chosen
integrals
•7 40 12
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5. Calculate the shaded area
enclosed between the parabolas
with equations y = 1 + 10x − 2x 2
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and y = 1 + 5x − x2 .
[SQA]
y
y = 1 + 10x − 2x2
6
y = 1 + 5x − x2
O
Part
Marks
6
Level
C
Calc.
CN
Content
C17
•1 ss: find intersections
•2 ss: know to find limits
•3 ss: know to integrate (upper −
lower)
•4 pd: simplify
•5 pd: integrate
•6 pd: process limits
Answer
20 56
•1
•2
•3
•4
•5
•6
x
U2 OC2
2002 P2 Q5
1 + 10x − 2x 2R = 1 + 5x − x2
5
Rx = 0, 5 and 0 ()2
− 2x ) − (1 + 5x − x 2 )) dx
R ((1 + 10x
2
(5x − x ) dx
5 2
x
− 13 x3
2
20 56
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6.
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7.
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8. A point moves in a straight line such that its acceleration a is given by
1
a = 2(4 − t) 2 , 0 ≤ t ≤ 4. If it starts at rest, find an expression for the velocity
v where a = dv
dt .
[SQA]
Part
Marks
4
Level
C
Calc.
NC
Content
C18, C22
•1 ss: know to integrate acceleration
•2 pd: integrate
•3 ic: use initial conditions with const.
of int.
•4 pd: process solution
Answer
3
V = − 43 (4 − t) 2 +
U3 OC2
32
3
2002 P2 Q8
R
1
•1 V = (2(4 − t) 2 ) dt stated or implied
by •2
3
•2 2 × −13 (4 − t) 2
2
•3 0 = 2 ×
•4 c = 10 23
1
− 23
3
(4 − 0) 2 + c
√ dy
5π
= 3 sin(2x) passes through the point 12 , 3 .
9. A curve for which
dx
Find y in terms of x .
[SQA]
Part
Marks
4
•1
•2
•3
•4
10. Find
[SQA]
Part
Z
pd:
pd:
ss:
pd:
Level
A/B
Calc.
CN
Content
C18, C23
integrate trig function
integrate composite function
use given point to find “c”
evaluate “c”
Answer
√
y = − 32 cos(2x) + 14 3
4
U3 OC2
2001 P2 Q10
R
•1 3 sin(2x) dx stated or implied by •2
•2 − 32 cos(2x)
√
5
•3 3 = − 32 cos(2 × 12
π) + c
√
1
4
• c = 4 3 (≈ 0·4)
1
dx .
(7 − 3x)2
2
Marks
Level
Calc.
Content
2
A/B
CN
C22, C14
•1 pd: integrate function
•2 pd: deal with function of function
Answer
1
+c
3(7 − 3x)
U3 OC2
2000 P2 Q10
1
•1 −1
(7 − 3x)−1
1
•2 × −3
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11.
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[END OF PAPER 2]
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