IB Calculus SL
Chapters 5-7 Test – Form A
[100 marks]
Name___________________________
Period___________________________
Full marks are not necessarily awarded for a correct answer with no working. Answers must be supported by working and/or explanation. In particular, solutions found from a graphic display calculator should be supported by suitable working, e.g., if graphs are used to find a solution, you should sketch these as part of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided this is shown by written working. You are therefore advised to show all working. Working may be continued below the lines, if necessary.
1. The diagram shows part of the graph of y = 12 x
2
(1 – x ). y
0 x
(a) Write down an integral which represents the area of the shaded region.
(b) Find the area of the shaded region. (Total 4 marks)
2. The diagram shows part of the graph of y =
1 . The area of the shaded region is 2 units. x y
0 1 a x
Find the exact value of a . (Total 4 marks)
3. A car starts by moving from a fixed point A. Its velocity, v m s v = 4 t + 5 – 5e
– t
. Let d be the displacement from A when t = 4.
–1
after t seconds is given by
(a) Write down an integral which represents d .
(b) Calculate the value of d . (Total 6 marks)

4. A part of the graph of y = 2 x – x
2
is given in the diagram below.
The shaded region is revolved through 360

about the x -axis.
(a) Write down an expression for this volume of revolution.
(b) Calculate this volume. (Total 6 marks)
5. The shaded region in the diagram below is bounded by f ( x ) = x , x = a , and the x -axis. The shaded region is revolved around the x -axis through 360
0.845

.

. The volume of the solid formed is
Find the value of a . (Total 6 marks)
6. The following diagram shows part of the graph of y = cos x for 0
 x

2

. Regions A and B are shaded.
(a) Write down an expression for the area of A.
(1)

(b) Calculate the area of A.
(c) Find the total area of the shaded regions.
7. It is given that

1
3 f ( x )d x = 5.
(1)
(4)
(a) Write down

1
3
2 f ( x )d x .
(b) Find the value of

1
3
(3 x
2
+ f ( x ))d x . (Total 6 marks)
8. The velocity v in m s
−1
of a moving body at time t seconds is given by v = e
2t−1
. When t = 0.5 the displacement of the body is 10 m. Find the displacement when t =1. (Total 6 marks)
9. Given

3 k 1 x

2 d x = ln 7, find the value of k . (Total 6 marks)
10. The diagram shows part of the curve y = sin x . The shaded region is bounded by the curve
3
π and the lines y = 0 and x = .
4 y
3

4
 x
Given that sin
3 π
4 region.
=
2
2
and cos
3
π
4
= –
2
2
, calculate the exact area of the shaded
(Total 6 marks)

1. (a)

0
1
12 x
2
(1 – x )d x
(b) 12 x x
3
) d x
= 12


 x
3
3
= 12
= 1


1
3

1
4 

 x
4
4 

 1
0
(A1) (C1)
(M1)
(A1)
(A1) (C3)
2.

1 a 1 x d x = 2

[ln x ]
1 a
= 2

ln a = 2
 a = e
2
(M1)
(M1)
(A1)
(A1) (C4)
Note: If 7.39 given instead of e
2
then deduct [1 mark].
3. (a) d =

0
4
( 4 t

5

5 e
– t
) d t (M1)(A1)(A1) (C3)
Note: Award (M1) for
– 5e
–t

, (A1) for both limits, (A1) for 4t + 5
(b) d = [ 2 t
2 
5 t

5 e
-t
]
4
0
Note: Award (A1) for 2t
2
+ 5t, (A1) for 5e
–t
.
= (32 + 20 + 5e
–4
) – (5)
= 47 + 5e
–4
(47.1, 3sf )
(A1)(A1)
(A1) (C3)
4. (a) Attempting to use the formula V = b   a y
2 d x
Volume =
 
0
2

2 x
 x
2
2  d x
(b) Volume =
 
0
2

4 x
2 
4 x
3  x
4
 d x
=




4 x
3
3

4 x
4
4
 x
5
5 

 2
0
=
16

15
or 3.35 (accept 1.07

)
(M1)
A2 N3
(A1)
(A1)
A1 N3
5. Using V =
  y
2 d x (M1)
[4]
[6]

Correctly integrating
 


 x 2
1



 2 d x
 x
2
2
V =



 x
2
2 

 a
0
=
 a
2
2
Setting up their equation
1
2
 a
2 
0 .
845
 a
2
= 1.69 a = 1.3
6. (a)

3
2
2

 cos x d x
(b) Area of A = 1
(c) Evidence of attempting to find the area of B eg

4
2

3

3 y d x ,

0.134
Evidence of recognising that area B is under the curve/integral is negative eg

3


4
2

3 y d x ,

3
3

4

2 cos x d x ,
3


4
2

3 cos x d x
Area of B = 0.134



 accept
2

2
3 



Total Area = 1 + 0.134
= 1.13



 accept
4

2
3 



7. (a) 10
(b)

1
3
3 x
2  f d x
 
1
3
3 x
2 d x
 
1
3 f
  d x

1
3
3 x
2 d x

 
3
1

27

1
A1
A1
(A1)
M1
A1 N2
A1 N1
A1 N1
(M1)
(M1)
(A1)
A1 N4
A1 N1
(A1)
[6]
[6]

= 26 (may be seen later)
Splitting the integral (seen anywhere) e .
g .

3 x
2 d x
  f d x
Using
3  d x
1 f

5 eg

1
3
3 x
2  f d x

26

5

1
3
3 x
2  f
  d x

31
8. s
  v d s

1
2 e
2 t 1  c
Substituting t = 0.5
1
2
 c

10 c = 9.5
Substituting t = 1 s =
1
2 e

9 .
5


10 .
9 to 3 s .
f .

9. Using
 1 x
 ln x (may be implied)
 k
3 x
1

2 d x

[ln ( x

2)] k
3
= ln ( k

2)

ln1 ln (k

2)

ln1 = ln 7 k

2 = 7
k = 9
10. Area =
 a b sin x d x a = 0, b =
3
π
4
A1
M1
(M1)
A1 N3
(M1)
A1A1
(A1)
M1
A1 N3
[6]
(M1)
(A1)
(A1)(A1)
(A1)
(A1) (C6)
(M1)
(A1)
[6]

Area =

0
3
4

3 π sin x d x = [–cos x ]
0
4
=
– cos
3
π
4
= –


–
2
2
– (– cos 0)

 – (–1)
= 1 +
2
2
Note: Award (G3) for a gdc answer of 1.71 or 1.707.
(A1)
(A1)
(A1)
(A1) (C6)