MEI Core 2
Further differentiation and integration
Chapter assessment
1. Differentiate the following
3  2x
(i) y 
x
( x  2)(2 x  1)
(ii) y 
x4
[3]
[3]
2. Find the following indefinite integrals
3 

(i)   2 x 
 dx
x

1 
 2
(ii)   2  3  dx
 x 2x 
[3]
[3]
3. Find the following definite integrals
(i)

9
(ii)

2
1
1
(2  x) x dx
[4]
 6 8
 2  3  dx
x 
x
[4]
2
at the point where x = -1.
x2
(ii) Find the equation of the tangent to the graph at this point.
(iii) Find the equation of the normal to the graph at this point.
4. (i)
Find the gradient of the graph y  x 
5. Find the coordinates of the stationary points of the graph y 
passes through the point (1, 2).
[3]
[3]
1 1 1
 
and
x x 2 x3
determine the nature of each.
6. Find the equation of the graph with gradient function
[2]
[5]
dy 1

 x x which
dx x 2
[3]
7. A closed rectangular tank is to have dimensions x m, 2x m and y m, and is to have
a capacity of 9 cubic metres. To keep costs down the surface area of the tank is to
be as small as possible.
(i) Write down expressions for the volume and the surface area of the tank in
terms of x and y, and hence find an expression for A in terms of x only. [3]
(ii) Find the value of x for which A has a stationary point, and show that this is a
minimum point.
[4]
(iii) Find the dimensions of the tank for which A is as small as possible, and the
corresponding value of A.
[2]
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MEI C2 Further calculus assessment solutions
8. The diagram shows part of the graph of y  5  x 2 
4
.
x2
y
x
Find the area enclosed between the graph and the x-axis.
[5]
Total 50 marks
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MEI C2 Further calculus assessment solutions
Chapter assessment Solutions
3  2x
1. (i) y 
dy
dx
dx
2. (i)

1
x
1
  23 x  2  x  2
3
1
( x  2)(2 x  1)
(ii) y 
dy
 3x 2  2 x 2
x4
 2 x 2  3 x 3  2 x 4
 4 x 3  9 x 4  8 x 5
  2
x
3 
dx 
x 

 2x
4
3
1
2

 3 x  2 dx
1
3
2
1
2
x  6x  c
1 
 2
dx    2 x 2  21 x 3  dx
(ii)   2 
3 
2x 
x
 2 x 1  41 x 2  c
3. (i)
 (2  x )(
9
1
x ) dx  
9
1
2 x
1
2
3

 x 2 dx
9
  43 x 2  52 x 2 
1
3


4
3
5
3
5
 
 9 2  52  9 2 
4
3
3
5
 1 2  23  1 3

  43  27  52  243    43  52 
4
 36  486
5  3 
2
5
 62 152
2 6
8 
(ii)   2  3  dx 
1
x 
x
 6 x
2
1
2
 8 x 3  d x
2
 6 x 1  4 x 2 1
  6(2)1  4(2)2    6(1)1  4(1)2 
 ( 3  1)  ( 6  4)
 2   2   0
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MEI C2 Further calculus assessment solutions
yx
4. (i)
dy
dx
2
x
2
 x  2 x 2
 1  4 x 3
When x  1,
dy
dx
 1  4( 1)3  1  4  3
(ii) When x  1, y  1 
2
 1  2  3
( 1)2
y  y 1  m( x  x 1 )
y  ( 3)  3( x  ( 1))
y  3  3 x  3
y  3 x  6
(iii) The tangent has a gradient of -3,
1 1
so the normal has a gradient of

3 3
y  ( 3)  31 ( x  ( 1))
3( y  3)  x  1
3y  9  x  1
3y  x  8
1
5. y 
dy
dx
x

1
x
2

1
x3
 x 1  x 2  x 3
  x 2  2 x 3  3 x 4
Stationary points occur where
dy
dx
0
0   x 2  2 x 3  3 x 4
3 x 4  x 2  2 x 3
3  x2  2x
x2  2x  3  0
( x  3)( x  1)  0
x  1 or x  3
When x = 1, y  1  1  1  1
5
When x = -3, y   31  19  271   27
d2 y
dx 2
 2 x 3  6 x 4  12 x 5
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MEI C2 Further calculus assessment solutions
When x  3 :
d2 y
2
dx
d2 y
 2( 3) 3  6( 3)4  12( 3) 5
2
2
  27
 27
 814  814
dx 2
5
As this is positive,  3,  27
 is a local minimum.
When x  1 :
d2 y
dx 2
d2 y
 2(1)3  6(1)4  12(1)5
 2  6  12  4
dx 2
As this is negative, (1,1) is a local maximum.
6.
dy
dx
dy
dx

1
x
2
 x x  x 2  x( x 2 )
1
 x 2  x 2
3
y    x 2  x
3
2
 dx
y   x 1  52 x  c
5
2
When x = 1, y = 2: 2  (1)1  52 (1)2  c
5
2  1  52  c
c
17
5
y   x 1  52 x  175
5
2
7. (i)
Volume  x(2 x )( y )  2 x 2 y
Surface Area  2( x )(2 x )  2( x )( y )  2(2 x )( y )
 4 x 2  2 xy  4 xy
 4 x 2  6 xy
Volume  9  2 x 2 y
y
9
2x2
9 
2 
 2x 
A  4 x 2  6 x 
54 x
2x2
A  4 x 2  27 x 1
A  4x 2 
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MEI C2 Further calculus assessment solutions
(ii)
dA
 8 x  27 x  2
dx
Stationary point occurs when
dA
0
dx
8 x  27 x 2
8 x 3  27
x3 
x
27
8
3
2
d2 A
 8  54 x 3
dx 2
d2 A
When x  1.5 , 2  8  16  24
dx
As this is positive, the stationary point at x  1.5 is a local minimum.
(iii) At the minimum point, x  1.5 m
Dimensions  1.5 m, 2(1.5 ) m,
 1.5 m, 3 m, 2 m
A  4 x  27 x 1
9
m
2(1.5 )2
2
A  4(1.5 2 )  27(1.5 )1
A  9  18  27 m 2
8. y  5  x 2 
4
x2
y  5  x 2  4 x 2
When y  0 : 0  5  x 2  4 x 2
5 x2  x4 4  0
x4  5 x2  4  0
 x 2  4  x 2  1   0
x2  4
x  2
or x 2  1
x=  1
The area we want is for positive x values, so the boundaries are x = 2 and x = 1.
Area  
2
1
 5  x 2  4 x 2  dx
2
  5 x  31 x 3  4 x 1  1
  5(2)  31 (2)3  4(2)1    5(1)  31 (1)3  4(1)1 
  10  83  2    5  31  4 

28
3

2
3
 26
3
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Further differentiation and integration

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