CHAPTER 10
SOLUTIONS TO REINFORCEMENT EXERCISES IN
APPLICATIONS OF DIFFERENTIATION AND INTEGRATION
10.3.1 The derivative as a gradient and rate of change
10.3.1A.
For the following functions find a) the rate of change and b) the slope of the
graph at the points specified.
i)
y = x3 + 2x – 1,
ii)
y = sin x
x = 0, 2

x = 0, 3
iii)
iv)
y = ex cos x
y = ln(x2 + 1)
x = 0, 1
x = 0, 2
Solution
i)
For y = x3 + 2x – 1 we have y = 3x2 + 2
So
a) Rate of change at x = 0 is 2
Rate of change at x = 2 is 3  4 + 2 = 14
and
b) Slope of graph at x = 0 is 2
Slope of graph at x = 2 is 14
ii)
y = sin x
y = cos x
a) Rate of change at x = 0 is cos 0 = 1
1


Rate of change at x = 3 is cos 3 = 2
b) Slope of graph at x = 0 is 1
1

Slope of graph at x = 3 is 2
iii)
a)
b)
y = ex cos x
y = ex cos x – ex sin x
y(0) = 1
y(1) = e1 (cos 1 – sin 1)
1, e1 (cos 1 – sin 1)
–1–
y = ln(x2 + 1)
2x
y = 2
x +1
iv)
a)
y(0) = 0
4
y(2) = 5
4
0, 5
b)
10.3.1B.
Find the point on the curve
y = x3 + 3x2 – 9x + 1
where the gradient is – 12.
Solution
If y = x3 + 3x2 – 9x + 1 then y(x) = 3x2 + 6x – 9 is the gradient at any
point x. So the gradient is – 12 where
3x2 + 6x – 9 = – 12
ie
3x2 + 6x + 3 = 0
or
x2 + 2x + 1 = (x + 1)2 = 0
So there is only one such point, at x = – 1, where the value of y is
y(– 1) = (– 1)3 + 3(– 1)2 – 9(– 1) + 1 = 12
So the gradient is – 12 at (– 1, 12)
10.3.2 Tangent and normal to a curve
Find the equations of the tangents and the normals to the following curves at
the points indicated.
i)
y = x2 + 2x – 3
x = 1
–2–
ii)
y = x4 + 1
x = 1
iii)
y = lnx
x = 1
iv)
y = ex sin x
x = 0
Solution
i)
For y = x2 + 2x – 3
y(x) = 2x + 2
So the gradient at x = 1 is y(1) = 4.
The equation of the tangent at (1, 0) is therefore
y – 0 = 4(x – 1)
or
y = 4x – 4
From Section 7.2.5 the gradient of the normal, which is perpendicular to
1
the tangent, is – 4 and so the equation of the normal at (1, 0) is
1
y – 0 = – 4 (x – 1)
or
x + 4y – 1 = 0
ii)
For y = x4 + 1
y(x) = 4x3
So the gradient at x = 1 is y(1) = 4.
The equation of the tangent at (1, 2) is therefore
y – 2 = 4(x – 1)
or
y = 4x – 2
–3–
1
The gradient of the normal is – 4 and so the equation of the normal at
(1, 2) is
1
y – 2 = – 4 (x – 1)
or
x + 4y – 9 = 0
iii)
For y = ln x
1
y(x) = x
So the gradient at x = 1 is y(1) = 1.
The equation of the tangent at (1, 0) is therefore
y – 0 = 1(x – 1)
or
y=x–1
The gradient of the normal is – 1 and so the equation of the normal at (1,
0) is
y – 0 = – 1 (x – 1)
or
x+y–1=0
iv)
For y = ex sin x
y(x) = ex (sin x + cos x)
So the gradient at x = 0 is y(0) = e0 (sin 0 + cos 0) = 1
The equation of the tangent at (0, 0) is therefore
y – 0 = 1(x – 0)
or
–4–
y=x
The gradient of the normal is – 1 and so the equation of the normal at (0,
0) is
y – 0 = – 1 (x – 0)
or
y=–x
10.3.3 Stationary points and points of inflection
10.3.3A.
Locate and classify the stationary points and any points of inflection of the
following functions
iii)
x3
ii) x3 – 12x + 2
x
5
iv) 5 + x
v)
2x3 – 15x2 + 36x – 4
vi) 4x3 + 3x2 – 36x + 6
i)
x2 – 4x + 3
Solution
i) For y(x) = x2 – 4x + 3 we have
y(x) = 2x – 4 = 0
at x = 2.
Further, y = 2 > 0, so we have a minimum at x = 2, at which point y = –
1.
So in this case there is a minimum at (2, – 1)
ii) For y(x) = x3 – 12x + 2 we have
y(x) = 3x2 – 12 = 0
at x =  2.
–5–
Further, y = 6x > 0 at x = 2, so we have a minimum at x = 2, at which
point y = – 14. At x = – 2, y = 6x < 0, so we have a maximum at x = – 2,
at which point y = 18.
Now we also notice that y = 6x is zero at x = 0, and yet y = 6  0 at
that point. So there is a point of inflection at x = 0 (x < 0, y < 0; x > 0, y
> 0, so there is a change in concavity at x = 0), where y = 2.
So in this case there is a minimum at (2, – 14), a maximum at (– 2, 18)
and a point of inflection at (0, 2).
iii)
For y(x) = x3 we have
y(x) = 3x2 = 0
at x = 0.
Further, y = 6x > 0 at x = 0 but y = 6  0 at this point. So there is a
point of inflection at x = 0 (x < 0, y < 0; x > 0, y > 0, so there is a
change in concavity at x = 0), where y = 0.
So in this case there is a point of inflection at (0, 0).
x
5
iv) For y = 5 + x we have
1
5
y(x) = 5 – 2 = 0
x
when x =  5.
Further, y =
10
so we see that y > 0 at x = 5, where y = 2, and y < 0
x3
at x = – 5 where y = – 2. y is never zero, so there are no points of
inflection.
So we have a minimum at (5, 2) and a maximum at (– 5, – 2)
v)
For y(x) = 2x3 – 15x2 + 36x – 4 we have
–6–
y(x) = 6x2 – 30x + 36 = 0
where
x2 – 5x + 6 = (x – 2)(x – 3) = 0
ie at x = 2, 3.
Further, y = 12x – 30 < 0 at x = 2, so we have a maximum at x = 2, at
which point y = 24. At x = 3, y > 0, so we have a minimum at x = 3, at
which point y = 23.
5
Also y = 12x – 30 is zero at x = 2 , but y is not zero at that point. So
5
there is a point of inflection at x = 2
So in this case there is a minimum at (3, 23), a maximum at (2, 24) and a
5
point of inflection at x = 2 .
v)
For y(x) = 4x3 + 3x2 – 36x + 6 we have
y(x) = 12x2 + 6x – 36 = 0
where
2x2 + x – 6 = (2x – 3)(x + 2) = 0
3
ie at x = 2 and – 2.
y = 24x + 6 < 0 at x = – 2, so we have a maximum at x = – 2 (The
answer in the book is wrong)
3
3
y = 24x + 6 > 0 at x = 2 , so we have a minimum at x = 2
1
There is also a point of inflection at x = – 4
–7–
10.3.3B.
Find the maximum and minimum values of the curve of the function y = x(x 2
– 4), and also find the gradient of the curve at the point of inflection.
Solution
y = x(x2 – 4) = y = x3 – 4x
y = 3x2 – 4 = 0 at x = 
y = 6x > 0 at x =
2
3
2
2
16
, so there is a minimum at x =
(of –
)
3
3 3
3
y = 6x < 0 at x = –
2
2
16
, so there is a maximum at x = –
(of
)
3
3 3
3
There is also a point of inflection at x = 0 at which the gradient is y = –
4:
y = 0 at x = 0, y = 6  0, and y = 3(0) – 4 = – 4
10.3.4 Curve sketching in Cartesian coordinates
Sketch the graphs of the following functions.
iii)
x+1
x–1
x
ii) x + 1
x–2
iv) 2
x +1
v)
x2 + 4
x2 + x – 2
x
vi) 3 + cos 2
 
i)
vii)
x3 – 2x2 – x + 2
x e– x
Solution
While in individual examples there may be various short cuts and results that
can be used (for example vi) can be done simply from our knowledge of the
cosine graph) to sketch graphs, it is often a good policy to work methodically
though the S(ketch) GRAPH procedure described in Section 10.2.4. The results
for these exercises are shown below, and the explanation follows.
–8–
i) For y = x3 – 2x2 – x + 2 first note that the function is neither odd nor
even and so there is no particular symmetry we can use here.
Gateways, or the points where it crosses the axes, can be found by
equating the function to zero (for the gateways on the x-axis), or putting
x = 0 (for gateways on the y-axis). Thus we need to solve
x3 – 2x2 – x + 2 = 0
Notice that the LHS factorises quite nicely
x3 – 2x2 – x + 2 = x2(x – 2) – (x – 2) = (x2 – 1)(x – 2)
= (x – 1)(x + 1)(x – 2)
so the roots are x = 1, – 1, and 2 and it is at these points that the curve
crosses the x-axis. It crosses the y-axis when x = 0, ie at y = 2.
There are no restrictions, since a polynomial exists for all values of x.
Asymptotic behaviour shows that for very large values of x, y behaves
like x3, while for very small values it gives an approximate straight line
of the form y = – x + 2 (sloping down to the right at 45 to the x-axis,
through the point (0, 2)).
–9–
Apart from the gateways already done the only other points of interest
are any stationary points (humps and hollows) there might be. We
have
y(x) = 3x2 – 4x – 1
and
y = 6x – 4
There are stationary points where
3x2 – 4x – 1 = 0
ie
x=
4±
16 + 12
2± 7
=
6
3
ie x = 1.549 and – 0.215 to 3dp.
2
There is also a point of inflection at y = 6x – 4 = 0, ie x = 3 .
At x = 1.549, y > 0, so here we have a minimum (and y = – 0.631)
At x = – 0.215, y < 0, so here we have a maximum (and y = 1.29)
Putting all this information together gives us the sketch shown in the
figure (i) above.
x
ii) x + 1 is neither odd nor even and so has no particular symmetry
x
Gateways occur at x + 1 = 0, ie x = 0 (y is never zero, except 'at infinity'
- see asymptotes, below).
Restrictions.
The function is discontinuous at x + 1 = 0 and so we have to consider
each side of x = – 1 separately, and omit the point x = – 1.
For asymptotes it is useful to rewrite the function as
– 10 –
x
x+1–1
1
y= x+1 = x+1
=1–x+1
From this we can see directly that as x , y  1, 'from below'.
Similarly, as x – , y  1 'from above'. Also, as x – 1 from above,
y   , while as x – 1 from below, y  + .
There are no particular points of interest, although you may find it
useful to try out a couple of values to get the asymptotes right - for
1
example, x = – 2 gives y = – 1 and points to the curve descending
through negative values as it approaches the asymptote at x = – 1.
The derivative is y =
1
and is clearly never zero (except 'at
(x + 1)2
infinity'), so there are no stationary points, no humps or hollows.
We now have enough to sketch the graph, as shown in the figure (ii)
above.
Actually, there is a short cut with such a graph, using the form
1
y =1–x+1
and a couple of simple transformations, as described in UEM (page
1
299). We start with the graph of f(x) = x and move it one unit to the left
1
1
to get x + 1 . We then reflect this in the x-axis to get – x + 1 . We now
only have to lift this by one unit upwards to get the required graph.
x+1
iii) y = x – 1
Symmetry
None of note
Gateways
Crosses the x-axis at x = – 1
Crosses the y-axis at y = – 1
– 11 –
Restrictions
x1
Aymptotes
x = 1 is a vertical asymptote
As x 1 from below, (x < 1), y  –
As x 1 from above, (x > 1), y  +
As x , y  1
As x – , y  1
Points of interest
None not considered elsewhere
Humps and hollows
1
x+1
2
y = x – 1 –
2 =–
(x – 1)
(x – 1)2
which is never zero, so there are no stationary points.
Combining all this allows us to sketch the graph as in (iii) above. We
could also have done this by a series of transformations as in ii).
iv)
y=
x–2
x2 + 1
Symmetry
None
Gateways
y = 0 when x = 2
x = 0 when y = – 2
Restrictions
None in particular (other than defined by the max and min values – and
note that since x2 + 1  0, the graph is continuous)
Asymptotes
As noted above x2 + 1  0, so there are no vertical asymptotes.
– 12 –
As x , y behaves like
x
1
2 = x (and so y 0 as x )
x
Similarly as x – , y  0.
Points of interest
Nothing special.
Humps and hollows
As an exercise in differentiation you can confirm that
y =
– x2 + 4x + 1
(x2 + 1)2
=0
when x = 2  5 = 4. 236 or – 0.236 (to 3dp). In this case there is no
need to distinguish between maximum and minimum using the second
derivative, as the information we already have tells us that the
maximum value occurs for x > 0 and the minimum for x < 0 (with
values 0.118 and – 2.117 respectively).
All this gives the sketch shown in (iv) above.
v) y =
x2 + 4
x2 + 4
=
(x – 1)(x + 2)
x2 + x – 2
No symmetry.
Gateways
y is never zero, so the curve does not cross the x-axis. When x = 0, y = –
2, so it crosses the y-axis at y = – 2.
Restrictions
x  1, x  – 2
Asymptotes
Vertical asymptotes at x = 1 and x = – 2.
As x 1 from above, y 
As x 1 from below, y – 
As x – 2 from above, y – 
As x – 2 from below, y 
As x  , y 1, so y = 1 is a horizontal asymptote
– 13 –
Points of interest
The point where the curve crosses the asymptote y = 1 would be useful.
We can get this by solving
x2 + 4
y= 2
=1
x +x–2
and you can easily confirm that this gives x = 6.
Humps and hollows
Note that by looking at the asymptotes we suspect a maximum between
x = – 2 and x = 1, and a minimum for x > 6. Practice your differentiation
by confirming that
x2 – 12x – 4
y = 2
(x + x – 2)2
This is zero when
x2 – 12x – 4 = 0
ie
x = 6  2 10
or x = 12.325, or – 0.325 (3dp). Our knowledge of the graph so far tells
us that the first is a minimum and the second a maximum.
Putting it all together gives the graph in (v) above
x
vi) y = 3 + cos 2
 
We have no need to use S-GRAPH here. We only need to perform
transformations (scaling and shifting) on the standard cosine curve. The
average level of the curve is shifted up the y-axis by 3 units and the
wavelength is doubled. From this we obtain the graph below. It helps to
x
x
note that cos 2 is zero, and therefore 3 + cos 2 is 3 when x = , 3,
 
 
4,... etc. The amplitude of the wave is 1. The result is given in (vi) in
the figure above.
– 14 –
vii)
y = x e– x
No symmetry. Gateway only at the origin, y = 0 when x = 0. No
restrictions. As x , y  0, so the x-axis is an asymptote. No special
points, except a hump where y = e– x – x e– x = 0, ie at x = 1. Also y =
(x – 2) e– x = 0 when x = 2, and there is a point of inflection at x = 2. All
this together adds up to the graph sketched in the figure (vii).
10.3.5 Applications of integration – area under a curve
10.3.5A.
Find the area enclosed between the curve, the x–axis, and the limits stated for
each of the following curves.
1
i)
y = 2x2 + x + 1 x = 0, 2
ii) y = x – x
x = 1, 2

iii)
(x – 1)ex x = 1,2
iv) y = cos 2x
x = 0, 4
1
v)
y = x2 + 2
x = 1, 3
vi) y = sin2x,
x = 0,  /2
x
Solution
i)
y = 2x2 + x + 1
x = 0, 2
The function is clearly positive for x > 0 and we can simply integrate to
find the area:
2
2

1
16
28
2

A =  (2x2 + x + 1) dx = 3 x3 + 2 x2 + x = 3 + 4 = 3

0

0
1
y = x–x
ii)
x = 1, 2
For x > 1, y > 0 and so the area required is
2

1
1
3
1 2
2
A =  x – x dx = 2 x – ln x = 2 – ln 2 – 2 + ln 1 = 2 – ln 2


1

1
iii) y = (x – 1)ex
x = 1,2
y > 0 for x > 1, so by direct integration the required area is
– 15 –
2
2

A =  (x – 1)ex dx =

2
[(x – 1)ex ] 1

–  ex dx

1
1
= e 2 – [ ex
2
]1
= e2 – (e2 – e) = e

x = 0, 4
iv) y = cos 2x

y > 0 for 0 < x < 4 so

4


1
1
4

 1
A =  cos 2x dx = 2 [sin 2x] = 2 sin 2 – sin 0 = 2



0
0
v) y = x2 +
1
x2
x = 1, 3
Again, y > 0 for 1 < x < 3, so
3
 2 1
28
1 3 1 3
A =  x + 2 dx = 3 x – x = 3

1
x

1
vi)
x = 0,  /2
y = sin2x,

2
A =

2

1

 sin2xdx = 2 (1 – cos 2x) dx = 4


0
0
10.3.5B.
Calculate the total signed area between the curves and the x–axis and the limits
given by: a) geometry, b) integration
i) y = 1 – x,
x = 2, 4
ii) y = x – 1, x = 0, 2
– 16 –
Solution
i) y = 1 – x,
a)
x = 2, 4
By geometry, from the figure, we have
Area = – (Area ABC – Area ADE)
1
1

9 1
= – 2 3  3 – 2  1  1 = – 2 – 2 = – 4




b) By integration we have
4

1 4
16

Area =  (1 – x) dx = x – 2 x2 = 4 – 2 – 2 + 2 = – 4 as for a)

2

2
ii) y = x – 1, x = 0, 2
a) By geometry,
– 17 –
Area = Area ABC – Area ADE = 0
b) By integration
2

1 2
2
Area =  (x – 1) dx = 2 x – x = 0 as for a)

0

0
10.3.5C.
1

Evaluate  |x|dx

–1
Solution
1
0
1


 |x|dx =  |x|dx


–1

+  |x|dx

–1
0
0
1

=  – x dx

–1

+  x dx

0
(because |x| = – x on x < 0)
– 18 –
0
1
1 2
1 2
=  2 x 
+ 2 x 

 –1

0
 1 1
= – – 2 + 2 = 1
 
This is easy to check by geometry, as shown in the figure:
1 1
Area = Area ABC + Area ADE = 2 + 2 = 1 as above.
10.3.5D.
Find the area enclosed by the curves y = x2 + 2, and y = 1 – x, and the lines x
= 0, x = 1.
Solution
As the sketch shows, this is a straightforward case where x 2 + 2 is
always greater than 1 – x over the range of integration.
– 19 –
The required area is thus:
1
1


A =  ((x2 + 2) – (1 – x)) dx =  (x2 + x + 1) dx


0
0
11
1 3 1 2
1
= 3 x + 2 x + x = 6

0
10.3.6 Volume of a solid of revolution
Determine the volumes obtained by rotating the positive area under each of the
following curves about the x-axis, between the given limits.
i)
ii)
iii)
Solution
i)
y = x(1 – x), y = 0
xy = 1,
y = sin x,
y = 0,
x = 0,
x = 2,
x = 
y = x(1 – x), y = 0
– 20 –
x = 5
As the figure shows the limits on x are x = 0, 1, so the volume is
1
1


V =  y2 dx =  x2(1 – x)2dx


0
0
1
1


=  x2(1 – 2x + x2)dx =  (x2 – 2x3 + x4)dx


0
0
1
1
1 
1

1
1 1
= 3 x3 – 2 x4 + 5 x5 = 3 – 2 + 5 = 30

0


ii) xy = 1,
y = 0,
x = 2,
x = 5
1
The graph of y = x is shown in the figure, with the required area to be
rotated shown.
– 21 –
The volume is
5
5

 1
V =  y2 dx =  2dx

x
2
2
5
3
 1
1 1
=  – x  = 2 – 5 = 10

2


iii) y = sin x,
x = 
x = 0,
The volume is





1
V =  y2 dx =  sin2 x dx =  2 (1 – cos 2x) dx



0
0

=  2
– 22 –
2
=

2
0