8
8
Applications of
MathsWorld Mathematical Methods Units 3 & 4
Graph of an antiderivative
function
If f is a function with derivative f′, then the process of antidifferentiation can be used to find f(x) from f′(x), i.e. f(x) = f′(x)dx. It is possible to identify a family of curves with a given gradient function.
Consider a curve with gradient function f′(x) = 2x, then f(x) =x2+c, where c is a real constant. This describes a family of parabolas, some of which are sketched below.
Depending on the value of c, different parabolas are obtained. Each one is a translation of the curve y =x2 parallel to the y-axis.
Similarly, if f′(x) = 1 then f(x) =x+c where c is a real constant. This describes a family of straight lines, some of which are sketched below.
Depending on the value of c, different straight lines are obtained. Each one is a translation of the line y =x parallel to the y-axis.
⌠ ⌡
2
0 4 6 8 10
–10 –8 –6 –4 –2 x
y
5
–5
y=x2 +5
y=x2 – 1
y=x2 – 5
y=x2 – 3
y=x2 – 8
y=x2 +2
y=x2 +1
y=x2
All of these curves have the same gradient function
dy
= 2x
dx
y= x + 5
y= x + 2
y= x – 3
y= x – 6
y= x
2
0 4 6 8 10
–10 –8 –6 –4 –2 x
y
5
–5 All of these lineshave the same
gradient function
dy
= 1
dx
8.1
Applications of integration
chapter
8
E x a m p l e
1
Sketch two graphs, each of which could be the graph of an antiderivative of f′(x) =−2.
S o l u t i o n
f(x) = f′(x)dx
= −2dx
=−2x+c
Choosing two values of c, say c= 0 and c = 1:
c= 0, f(x) =−2x c= 1, f(x) =−2x+ 1
It is possible to determine a unique antiderivative if more information is provided.
E x a m p l e
2
The graphs of y=f′(x) and y=f(x) are as shown.
Find the rule for:
a f′(x) b f(x)
S o l u t i o n
a f′(x) =mx
(1, −1) is on this line).
So −1 =m and hence f′(x) =−x.
0 1 2
–2 –1 x
y
1 2
–1
–2
y=−2x+1
y=−2x
⌠ ⌡ ⌠ ⌡
2
(1, –1)
0 4 6
–6 –4 –2
y
2 4
–2
–4
y=f′(x)
(0, 4)
y=f(x)
2
0 4 6
–6 –4 –2 x
y
2 4
–2
–4
b f(x) = f′(x)dx
= =
(0, 4) is on this curve
So 4 =c and hence f(x) = . ⌠
⌡ ⌠ ⌡–xdx
x2
2
---– +c
x2
2
---– +4
exercise
8.1
1 For each of the following, sketch two graphs, each of which could be the graph of an antiderivative of f′(x).
a f′(x) =−3 b f′(x) = 4x c f′(x) =x2
2 In each of the following, the graphs of y =f′(x) and y=f(x) are as shown. Find a possible rule for:
i f′(x) ii f(x)
a
b
c
d
1
0 2
–4 –2 x
y
y = f′(x) 2
–2 –4
y=f(x) 2
0 3 4
–2 –1 1 x
y
–2
–4
–6
2
0 4
–2 –1 1 3 x
y
–2 –4 –6
y = f′(x)
2
0 4
–2 –1 1 3 x
y
4 6
2
y=f(x)
2
0 4
–2 3
(1, –3)
1
–1 x
y
4 6
2
–2
y = f′(x)
2
0 4
–2 3
(0, 3)
1
–1 x
y
4 6
2
–2
y = f(x)
0 1 2 –1
–2 x
f′(x)
10 15
5
(1, e 2)
y = f′(x)
f(x)
0 1 2 –1
–2
x
10 15
5
y = –1
Sketching antiderivatives
Graphs of antiderivatives can be sketched from graphical information alone. Recall that when interpreting the graph of a derivative:
.x-axis intercepts correspond to stationary points on the original function.
.If the graph is above the x-axis, then the original function has positive gradient and therefore it is increasing as x increases.
.If the graph is below the x-axis, then the original function has negative gradient and therefore it is decreasing as x increases.
If the graph of a function is given and the graph of the antiderivative is required, the graph that is given should be interpreted as the graph of the gradient function of the required curve.
E x a m p l e
3
Sketch the graph of the antiderivative of the function shown, given that it passes through the origin.
S o l u t i o n
Interpreting the graph above as the gradient function of the required curve, it can be seen that the graph of the antiderivative has a minimum turning point at x=−2 and a maximum turning point at x= 1.
The gradient of the antiderivative function is negative for x<−2 and x> 1 and positive for
−2 <x< 1. The antiderivative function has maximum positive gradient at x=−0.5.
The graph of the antiderivative is required to pass through (0, 0).
A suitable graph of the antiderivative is sketched at right.
1
0 2
–2 –1 3 4
–3
–4 x
y
1 0
For the required curve,
at x = –2 the gradient is zero and therefore there is a stationary point. The gradient changes from negative to positive at this point, so the stationary point is a minimum.
For the required curve,
at x = 1 the gradient is zero and therefore there is a stationary point. The gradient changes from positive to negative at this point, so the stationary point is a maximum.
2
–2 –1 3 4
–3
–4 x
y
1 0
–2 x
y
E x a m p l e
4
Sketch the graph of the antiderivative of the function shown, given that it passes through the origin.
S o l u t i o n
Interpreting the graph above as a gradient function, it can be seen that the graph of the antiderivative has a stationary point of inflection at x=−2 and a minimum turning point at x= 0.
The gradient of the antiderivative function is negative for x<−2 and
−2 <x< 0 and positive for x> 0. The graph of the antiderivative is required to pass through (0, 0), so the minimum turning point, in this case, is at the origin.
A suitable graph of the antiderivative is sketched at right.
E x a m p l e
5
The graph of y=f′(x) is shown. Sketch a possible graph of y=f(x).
S o l u t i o n
The graph of y=f(x) will have a maximum turning point at x=−3 and a minimum turning point at x= 2. The gradient of this curve is positive for x<−3 and x> 2 and negative for −3 <x< 2.
1 0
–2 –1 x
y
1
At x = –2 the gradient is zero, so there is
a stationary point. The gradient does not change sign at this point, so the stationary point is
a stationary point of inflection.
At x = 0 the gradient is zero, so there is a
stationary point. The gradient changes from negative to positive at this point, so the stationary point is a minimum.
0
–2 –1 x
y
1 0
–2 –1 x
y
1 0 2
–2 –1 3 4 5 6 –3
–4 –5
–6 x
y
A possible graph of y=f(x) is shown below.
exercise
8.1
3 For each of the graphs of y =f′(x) shown below, sketch the graph of the antiderivative,
y= f(x), given that it passes through the origin.
a b
c d
t i p
.When we find the indefinite integral of a function, it always includes an arbitrary constant, the constant of integration. As a result, there is not a unique antiderivative graph for the graph of a given function. It may be translated any distance parallel to the y-axis.
.When sketching graphs of antiderivatives, the basic shape can be accurately
represented by key features with associated x-coordinates, but its vertical translation is indeterminate unless extra information is provided.
0 2
–3 x
y
y = f(x)
continued
1
0 5
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 –3
–4
–5 x
y
y = f′(x)
1
0 5
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 –3
–4
–5 x
y
y = f′(x)
1
0 5
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 –3
–4
–5 x
y
y = f′(x)
1
0 5
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 –3
–4
–5 x
y
y = f′(x)
e f
g h
4 The graph of y =f′(x) is shown at right. Sketch the graph of y= f(x) if f(0) = 0.
1
0 5
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 –3
–4
–5 x
y
y = f′(x)
1
0 5
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 –3
–4
–5 x
y
y = f′(x)
1
0 5 6
–2 –1 1
–1 –2 –3 –4 –5 2 4 5
3
2 3 4 x
y
y = f′(x)
1
0 5
–2 –1 2
–2 –4 –6 –8 –10 4 8 10
6
2 3 4 –3
–4
–5 x
y
y = f′(x)
1
0 5 6
–2 –1 2
–2 –4 –6 –8 –10 4 8 10
6
2 3 4 –3
–4 –5
–6 x
y
5 The graph of y =f′(x) is shown. Sketch the graph of
y= f(x) if f(0) = 3.
6 Sketch a graph showing the general shape of y =f(x) for each of the graphs of y= f′(x) sketched below.
a b
c d
1 0 –2 –1
–10 –5 10
5
2 3 4 –3
–4 x
y
y = f′(x)
1 0
–2 –1 2 3 4 –3
–4 x
y
y = f′(x)
1 0
–2 –1 2 3 4 5 x
y
y = f′(x)
x 0
2π
π
y
y = f′(x)
1 0
–2 –1 2 3
–3 x
y
y = f′(x)
Areas under and between
curves
The fundamental theorem of calculus
Let A(x) be the area bounded by the curve withequation y= f(x), the x-axis and the lines x=a and x=b.
Let P be the point (x, f(x)),
Q the point (x+ δx, f(x+δx)) and
A the point (a, f(a)).
A(x) is the red shaded area.
A(x+δx) is the sum of both shaded areas.
A(x+δx) −A(x) is the blue shaded area.
It is clear from the diagram that f(x)δx≤A(x+δx) − A(x) ≤f(x+δx)δx.
Therefore, (dividing each expression by δx)
If we let δx→ 0,
So f(x) ≤ ≤ f(x)
Therefore, since it is sandwiched between two equal quantities.
That is, A(x) = F(x) + c, where F′(x) = f(x).
When x= a, A= 0, so 0 =F(a) +c, i.e. c =−F(a).
When x= b, A(b) is the area bounded by the curve with equation y =f(x), the x-axis and the lines x=a and x=b. Call this area A. Then A=F(b) +c=F(b) −F(a), where F′(x) =f(x).
x
y (b, f(b)) y=f(x)
(a, f(a)) (x, f(x)) A
Q
P
(x+ x, f(x+ x))
f x( ) A x( +δx)–A x( ) δx
--- f x( +δx)
≤ ≤
f x( )
δx→0
lim A x---( +δδxx)–A x( )
δx→0
lim f x( +δx)
δx→0
lim
≤ ≤
dA dx
---dA dx
--- = f x( )
Fundamental theorem of calculus
Let f be a continuous function on the interval [a, b]. Let F be an antiderivative of f on [a, b] so that F′(x) =f(x). Then f(x)dx = F′(x)dx =F(b) −F(a).
If the graph of y=f(x) lies above the x-axis, this represents the area between the graph and the x-axis.
⌠ ⌡a
b
⌠ ⌡a
b
8 . 2
E x a m p l e
1
Consider the graph of y= 3x2− 12. Find the area, in square units, enclosed by the curve, the
x-axis and the lines:
a x= 2 and x= 4 b x= 0 and x= 2 c x= 0 and x= 4
S o l u t i o n
a (3x2− 12)dx
= (antidifferentiating)
= (substituting for x)
= 32
The area is 32 square units.
b If (3x2− 12) (dx) is evaluated, the result is negative:
(3x2− 12)dx
= =
=−16
The presence of the minus sign indicates that the area is located below the x-axis. The area is 16 square units.
c The area of the region bounded by the graph of y= 3x2− 12, the x-axis and the lines x= 0 and x= 4 is 48 square units.
As y is positive for some values of x and negative for other values of x on the interval [0, 4], the area below the x-axis and the area above the x-axis must be calculated separately and then added together in order to determine the required area.
Signed area
The calculation in the warning box above gives what is known as the signed area. Regions below the x-axis have a negative signed area. In some applications, the signed area rather than the actual area, is required.
4 0
–4 2 6
–6 –2 x
y y= 3x2− 12
20 30
10
–10
A=32
⌠ ⌡2
4
x3–12x
[ ]24
43–12 4( )
( )–(23–12 2( ))
4 0
–4 2 6
–6 –2 x
y
20 30
10
–10 A=16
y= 3x2− 12
4 0
–4 6
A= 16 + 32 = 48
2
–6 –2 x
y
20 30
10
–10
y= 3x2− 12
⌠ ⌡0
2
⌠ ⌡0
2
x3–12x
[ ]02
23–12 2( )
( )–0
Warning
Look out for x-intercepts
The value in example 1 part c is very different from the evaluation of (3x2− 12)dx: (3x2− 12)dx=
= = 16
⌠ ⌡0
4
⌠ ⌡0
4
x3–12x
[ ]04
43–12 4( )
( )–(03–12 0( ))
In general, when finding the area enclosed between a curve with equation y =f(x), the x-axis and the lines x=a and x=b, areas above and below the x-axis need to be calculated using separate integrals.
If the required area lies entirely above the x-axis, it can be calculated using a single integral, as shown above right.
If the required area lies below the x-axis, it is calculated as shown below. In this case
f(x)dx is negative so the area is given by
− f(x)dx, which will be positive.
If the required area lies partly above and partly below the x-axis, separate calculations are required as shown in example 1.
E x a m p l e
2
a Find the exact area of the region enclosed by the curve with equation y= 2e−3x+ 1, the coordinate axes and the line x= 1.
b Give the answer in part a correct to 3 decimal places, and check using technology.
S o l u t i o n
a The required region lies entirely above thex-axis and can therefore be calculated using a single integral.
Area = (2e−3x+ 1)dx
=
=
= (square units)
b Correct to 3 decimal places, the answer is 1.633. The screenshots below provide a check on the numerical answer. The one on the left uses fnInt at the home screen; the one on the right finds the area interactively on the graph. Note that on the TI-89, the exact as well as the approximate answer can be found using the integral command at the homescreen. x
y
x=a x=b
y=f(x)
f(x)dx ⌠
⌡a
b
x
y
x=a x=b
y=f(x)
−⌠⌡ f(x)dx
a b
⌠ ⌡a
b
⌠ ⌡a
b
2 0 1
–1 x
y
y =2e−3x+ 1 10
15
5
⌠ ⌡0
1
2 3
---– e–3x+x
0 1
2 3
---– e–3+1
⎝ ⎠
⎛ ⎞ 2
3
---e0
–
⎝ ⎠
⎛ ⎞
–
5 3
--- 2
3e3
---–
GC 5.5 CAS 5.5
E x a m p l e
3
Calculate the exact area of the region bounded by the curve y=x3− 4x2 and thex-axis.
S o l u t i o n
Sketch the graph and identify the required region. The required region lies entirely below thex-axis and can therefore be calculated using a single integral.
Area =− (x3− 4x2)dx
=
=
=
The area is square units.
E x a m p l e
4
Find the exact area of the region enclosed by the curve with equation y= cos 2x, thex-axis and the lines x= 0 and .
S o l u t i o n
Sketch the graph and identify the required region.
The area is made up of one region above thex-axis and one below thex-axis. The shaded area can be calculated using two integrals.
Area = cos 2x dx − cos 2x dx
=
=
=
=
=
The area is square units.
–10
4
–4 2 6
–6 –2 x
y
y = x3− 4x2
–2 –4 2 –6 –8 ⌠ ⌡0 4 x4 4
--- 4x
3 3 ---– 0 4 – 4
( )4
4
--- 4 4( )
3 3 ---– ⎝ ⎠ ⎛ ⎞ – 64 3
---t i p
The region in question is enclosed by
only the x-axis and the curve; there are no vertical boundaries.
64 3
---x 2π
3
---=
x
y
y = cos2x
4 1 –1 0 4 3 3 2 π π π π
t i p
This example shows that it is critical to initially identify the x-intercepts of the graph as a first step so that the correct integral bounds are used. ⌠
⌡0
π/4
⌠ ⌡π/4
2π/3
1 2
--- sin 2x
0
π⁄4 1
2
--- sin 2x
π⁄4 2π⁄3
–
1 2
--- sin π 2
---– sin 0
⎝ ⎠
⎛ ⎞ sin 4π
3
--- sin π 2 ---– ⎝ ⎠ ⎛ ⎞ – ⎝ ⎠ ⎛ ⎞ 1 2
--- 1 3 2
---– –1
⎝ ⎠ ⎛ ⎞ – ⎝ ⎠ ⎛ ⎞ 1 2
E x a m p l e
5
A section of the graph with equation is shown below. The shaded area is equal to 16 square units. Find the value of k.
S o l u t i o n
= 16
= 16
−2k (cosπ − cos 0) = 16
−2k(−2) = 16 4k= 16
k= 4
exercise
8.2
1 Find the exact area of the region bounded by the each of the following curves and the x-axis.
a y= 3 −x2 b y=x3+ 5x2−x− 5
c y=x2(2 −x) d y=x4− 4x2
2 Find the area of the shaded region in each of the following.
a b
y k sin x 2
---=
x
y
k
π
0
π
2
t i p
The value of k can be found using the Solver on the TI-83/84 or the solve command on the TI-89. GC 5.5, 10.2
CAS 5.5, 7.1 ⌠ ⌡0
2π
ksinx 2
--- dx
2kcosx 2
---–
0 2π
y= 2x
x
y
8
6
4
2
0
0.5 1 1.5 2 2.5 3 3.5
2 0
–2 –1 1 3
–3 x
y
3 4
2 1
–1 –2 –3 –4
c d
3 Find the exact area of the region enclosed by the curve y= 2ex− 1 and the coordinate axes.
4 Find the exact area of the region enclosed by the curve with equation and the lines x= 1 and x= 4.
5 a Sketch the graph of the curve with equation .
b Find the exact area of the region bounded by the curve, the x-axis and the lines x= 3 and x= 5.
6 Calculate the exact area of the region bounded by the graph of y= (x− 1)3 and the lines
x= 0 and x= 2.
7 Find the exact area of the region enclosed by the curve y = 3 sin 2x, the x-axis and the
lines and .
8 Find the area of the region bounded by , the x-axis and the line x= 1.
9 Find the exact area of the shaded region.
y=x2 +2x – 3 2
0
–2 –1 1 3
–3 x
y
3 4
2 1
–1 –2 –3 –4
x
y
8
6
4
2
0
0.5 1 1.5 2 2.5 3 3.5
y=
2x + 1 9
y= x
x
y
2.5 2 1.5
0.5 1
0
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5
e
x
y
2.5 2 1.5
0.5 1
0
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5
y= x
f
y 1
2x
---=
y 3 1
x–2 ---– =
x π
4
---= x 2π
3 ---=
y 4 sinπx 2 ---=
y =1 – ex
0 1
–1 –0.5 x
y
1
–1
0.5
10 Find the exact area of the shaded region.
11 a Find .
b Hence, find the exact area of the region shaded in the diagram below.
12 The graph of y =f(x), where
f: [0, π] →R, f(x) =ksin(x), is shown. Find the value of k, if the area of the shaded region is 2 square units.
13 The graph with equation y =ax(4 − x) is shown. The area of the shaded region is 40 square units. Find the value of a.
14 The graph of is shown.
a Find the values of a and b.
b Hence find the exact area of the shaded region.
y=x3 – x
2 0
–2 x
y
6
4
2
y=
(2x+1)3
0.4 0.2 0.6
0 0.8 1 x
y
1.5
1
0.5 1
⌠ ⎮ ⌡
1 2x+1 ( )3
---dx
x
y
y = f(x)
π π
2 0
x
y
0
0.5 1 1.5 2 2.5 3 3.5 4
y = ax(4− x)
y
(0, 1)
(2, 7)
y=a−bcos x 2
π
y a bcosπx
2 ---–
15 Differentiate y= loge(x + 1). Hence find the area of the shaded region.
Area between curves
Consider two curves with equations y=f(x) and y= g(x), where f(x) >g(x) for all x∈ [a, b].
The area enclosed between the two curves can be determined by finding the area of the region enclosed by the curve with equation y=f(x), the x-axis and the lines x=a and x=b and subtracting the area of the region enclosed by the curve with equation y=g(x), the x-axis and the lines x=a and x=b.
Area = f(x)dx− g(x)dx
= (f(x) − g(x))dx
E x a m p l e
6
Find the exact area of the region enclosed between the line with equation y= 2x and the
curve with equation .
2 0
–2 –1.5 –1 –0.5 1 1.5 2.5
–2.5 x
y
0.5
x x2+ 1
y =
x b a
y
f(x)
g(x)
y =
y =
⌠ ⌡a
b
⌠ ⌡a
b
⌠ ⌡a
b
Visualising the area between curves
Area between curves = ⌠⌡ (top curve − bottom curve)dx = (f(x) −g(x))dx
a b
⌠ ⌡a
b
b a
y
x x x
y y
b a
f(x)
=
−
g(x)
b a
f(x)
g(x)
y =
y = y =
y = f(x)
g(x)
y = y =
This rule applies even if the curves are partly above and partly below the x-axis. When calculating the area between two curves it is only necessary to use more than one definite integral if the top curve and the bottom curve interchange.
t i p
y 1
2
---x2
=
S o l u t i o n
Sketch the graph and identify the required region.
The required area is enclosed between the upper graph, y= 2x, and the lower graph,
, between their points of intersection.
In this case thex-coordinates of the points of intersection can be easily read from the graph, but it is also possible to determine these algebraically as follows.
When y= 2x and intersect: 2x=
0 =
0 =
x= 0, 4
The required area is given by:
Area =
=
= =
The area of the region enclosed between the two curves is square units.
E x a m p l e
7
Find the exact area of the shaded region.
0 4 6
–4 –2
–6 x
y
6 8
4 2
–2 –4
2
y=2x y= 1x2
2
y 1
2
---x2
=
t i p
The required area can be shown using the shade command. However, to find the area required enter the rule
Y3=Y1–Y2, adjust the window dimensions and turn off
Y1 and Y2. The area can be found interactively as shown below right. Alternatively, the fnInt command on the TI-83/84 or the integration command on the TI-89 can be used on the home screen.
[0, 4] by [−2, 8] [0, 4] by [−1, 3]
y 1
2
---x2
=
1 2
---x2
1 2
---x2–2x
1 2
---x x( –4)
⌠ ⎮ ⌡0
4
2x 1
2
---x2
–
⎝ ⎠
⎛ ⎞dx
x2 1
6
---x3
–
0 4
4
( )2 ( )4 3
6
---–
16 3
---16 3
---x
y
π
2
–1 0
π
π π 2
2 3 1
S o l u t i o n
The x-coordinates of the points of intersection must be determined. sinx= cosx
tanx=1 (dividing both sides by cosx)
There is one solution in the first quadrant and one in the third quadrant.
Solution in first quadrant:
Solution in third quadrant:x=
=
Hence the curves intersect at and .
On the interval , y= cosx is the top curve and y= sinx is the bottom curve.
On the interval , y= sinx is the top curve and y= cosx is the bottom curve.
On the interval , y= cosx is the top curve and y= sinx is the bottom curve. Three integrals are required to calculate the area of the shaded region.
Area = (cosx− sinx)dx+ (sinx− cosx)dx+ (cosx− sinx)dx
= sinx+ cosx + −cosx− sinx + sinx+ cosx
=
=
The area of the shaded region is square units. (Note that each of these definite integrals is positive.)
exercise
8.2
16 Find the total area of the region or regions bounded by the graphs with the equations given.
a y=x2, y= 16 b y = 1 −x2, y=x+ 1 c y=x, y= 2x2 d y=x− 1, y=x2− 4x+ 3 e y =x2, y =x f y=x3, y= x
g h y = 4 −x2, y= 2x2− 5x+ 2
17 Find the area of the region enclosed by the curves with equations and and the lines x= 2 and x= 4.
18 Find the area of the region enclosed by y= x2+ x− 2 and y= 3x+ 1.
19 Find the area of the region enclosed between the graphs of and .
x π 4 ---= π π 4 ---+ 5π 4 ---x π 4
---= x 5π
4 ---= 0 π 4 ---, π 4
--- 5π
4
---,
5π 4
---,2π
⌠ ⌡0
π/4
⌠ ⌡π/4
5π/4
⌠ ⌡5π/4
2π ⎡
⎢⎣ ⎤⎥⎦0
π/4 ⎡
⎢⎣ ⎤⎥⎦π/4 5π/4
⎡
⎢⎣ ⎤⎥⎦5π/4 2π 2 2 --- 2 2 ---+ ⎝ ⎠
⎛ ⎞–(0+1)
⎝ ⎠
⎛ ⎞ 2
2 --- 2 2 ---+ ⎝ ⎠
⎛ ⎞ 2
2 ---– 2 2 ---– ⎝ ⎠ ⎛ ⎞ – ⎝ ⎠
⎛ ⎞ (0+1) 2
2 ---– 2 2 ---– ⎝ ⎠ ⎛ ⎞ – ⎝ ⎠ ⎛ ⎞ + + 4 2 4 2
continued
y 1 x---,y 5
2 ---–x
= =
y = ex⁄2 y 2
x
---=
y = x2 y = x
20 Find the area of the region enclosed by the curves with equations y =x3 and y = 16 −x3
and the straight lines x= −1 and x= 3.
21 The tangent to the curve y=x3− 5x2+ 6x at x= 1 cuts the curve again at x=b. Find the value of b and hence find the area of the region enclosed between this tangent and the curve.
22 a Find the derivative of xsinx and hence find an antiderivative of xcosx.
b Use the answer to part a to find the area of the region enclosed by the curves with equations y =xcosx and y= −cosx and the lines x= 0 and x=π.
23 Find the area of the region enclosed by the graphs of y=ex2 and y= 4, correct to 2 decimal places.
24 Find the exact area of the shaded region.
25 Find the area of the region shaded.
26 Find the value of k such that the area bounded by the line with equation y= 2x, the x-axis and the line x= 4 is divided into two regions of equal area by the line with equation y= k.
27 Find the exact area of the region in the first quadrant bounded by the x-axis and the curves with equations and .
x y
0.5 1 1.5 0
1.5
1
0.5
y= x
y= 1
y=sin
x
x
y
π
2
–1 0
π
π 2
2
π
2 3 1
y=cosx2
28 a Find the area of the region enclosed by the graphs of y=x and y=x in the first quadrant.
b Repeat part a for the graphs of
i y= x4 and y= x1/4 ii y=x5 and y= x1/5
c From the results above, make a conjecture about the area of the region enclosed by the graphs of y=xn and y=x1/n where n is a positive integer. Verify your conjecture using integration.
A parabolic arch is defined as having a base and height as shown.
a Find the area under the parabolic arch shown on the right with equation y = 8 − 2x−x2, −4 ≤ x≤ 2.
b What is the base length of this arch?
c Find the height of this arch.
Archimedes was a very great mathematician who discovered that the area under a parabolic arch is always equal to two-thirds of the base length multiplied by the height.
d Test Archimedes’ law for the arch defined in question a.
e Sketch the graph of the parabola
defined by the equation ,
showing the coordinates of the x-intercepts and the turning point. Assume that
b and h are positive constants.
f Use calculus to find the area enclosed by the curve and the x-axis and hence verify Archimedes’ law.
Height
Base
4 0 2
–4 –2 x
y
6 8 10
4 2
–2
y h 4h b2
---⎝ ⎠
⎛ ⎞x2
– =
analysis task 1—
the parabolic arch
SAC
8.1The parabolic ar
ch
CD
SAC analysis task
Hoa plans to construct a new fish pond in her back garden. Her proposed design is made up of a shallow section bounded by a parabola of constant depth, in which her fish can be easily visible, attached to a rectangular pool of changing depth in which the temperature of the water is cooler on hot days.
The proposed plan for the pond, as viewed from above, is shown. The rectangular section PQRS is 3 metres long and 1 metre wide.
The equation of the parabolic boundary of the shallow section is of the form y= ax2 +c.
a Find the values of a and c.
b Find the exact area of the total pond surface.
A view of the vertical cross-section of the pond along TU is displayed on a different set of axes in the diagram. All measurements are in metres.
c What is the depth of water in the shallow section?
d Determine the volume of water required to fill the shallow section.
e Write down the coordinates of the points C and D.
f The curve DC has equation y=−ekx+c. Find the exact values of k and c.
g Find the area of the region OUCD, correct to 4 decimal places.
h Find the volume of water required to fill the rectangular section, correct to 3 decimal places.
i Find the total volume of water required to fill the pond, correct to 3 decimal places.
x Q
O P
S U R
Shallow section
T
y
–1.0
0.5 –0.5
PLAN
x
D E
ELEVATION Shallow
section
C
T U
y
1
1.5 0.5
3
O U
analysis task 2—
Hoa’s fish pond
SAC
8.2Hoa’
s f
ish pond
CD
Straight-line motion
The study of the motion of a particle is called kinematics. When the motion of a body, usually referred to as a particle, is in a straight line only, it is referred to as rectilinear motion.
Position and displacement
The position, x, of a particle on a straight line is defined as where a particle is relative to some fixed point. This point is usually referred to as the origin, O. The position of the particle is usually described as being positive when it is to the right of the origin and negative when it is to the left of the origin.
Displacement is defined as the change in the position of a particle, while distance is defined as how far a particle has travelled. The distinction between distance and displacement is most easily seen through an example.
Consider a particle which starts at the origin. It moves 15 metres to the left and then 10 metres to the right.
The distance travelled by the particle is 25 metres but the displacement of the particle is −5 metres.
Velocity
In everyday language the terms velocity and speed are used interchangeably, but in the study of motion there is a very important distinction between the two quantities.
The speed of a particle refers to how fast the particle is travelling, while velocity also takes the direction of motion into account. A particle travelling with a constant velocity of 5 metres per second is conventionally considered to be travelling to the right, but if it travels with a velocity of −5 metres per second it is considered to be travelling to the left. In both cases the particle has a speed of 5 metres per second.
x
–15 –14 –13 –12 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 O 1 2 3 4 5
Average speed and velocity
Average speed and average velocity are both examples of average rates of change:
average speed =
average velocity = = .
distance travelled time taken
---change in position change in time
--- displacement
time taken
---8.3
8 . 3
E x a m p l e
1
A particle starts at the origin and travels 5 metres to the right and then 5 metres to the left in 5 seconds. Find:
a the distance travelled by the particle. b the displacement of the particle. c the average speed of the particle. d the average velocity of the particle.
S o l u t i o n
It is often useful to consider the motion of the particle through a diagram.
a The particle travels a distance of 10 m. b The displacement of the particle is 0 m.
c Average speed =
= = 2
The average speed of the particle is 2 m/s.
d Average velocity =
= = 0
The average velocity of the particle is 0 m/s.
E x a m p l e
2
The position of a particle,x metres from the point O at time t seconds, is given by
x(t) =t2− 4t− 5, t≥ 0. Find: a the initial position of the particle.
b the position of the particle after 6 seconds. c the distance travelled by the particle in 6 seconds. d the average speed over the first 6 seconds. e the average velocity over the first 6 seconds. f the particle’s initial velocity.
g when and where the particle is instantaneously at rest. h the velocity of the particle as it passes through O.
x
5 4
3 2
1
O
distance travelled time taken
---10 5
---change in position change in time
---0 5
---Instantaneous velocity
Velocity measures the rate of change of position of a particle. Instantaneous velocity, v, is the rate of change of position with time, i.e. v dx.
dt
---=
When the velocity of a particle is 0, we say that the particle is
at rest, even when this is only for an instant.
S o l u t i o n
It is useful to consider the position–time graph when answering the questions.
a x(0) =−5
The particle starts 5 metres to the left of O. b x(6) = 62− 4(6) − 5
= 7
The particle is 7 metres to the right of O
after 6 seconds.
c Using completing the square gives:
x(t) = (t− 2)2− 9
Hence the turning point is at (2, −9). The particle travels from x=−5 to x=−9, 4 metres to the left, then from x=−9 to
x= 7, 16 metres to the right. The distance travelled is 20 metres.
d Average speed =
=
=
The average speed over the first
6 seconds is m/s.
f v= = 2t− 4
The initial velocity is the velocity when t = 0. Hence the initial velocity is −4 m/s.
h The particle passes through O whenx = 0. 0 =t2− 4t− 5
0 = (t− 5)(t+ 1)
t= 5, −1
As t≥ 0, the particle passes through the origin after 5 seconds. Substituting t= 5 gives v= 6.
The velocity of the particle as it passes through the origin is 6 m/s.
t
x
1 2 3 4 5 6 7 8
O
5 10
–5
–10 (2, –9)
x = t2 – 4t –5
e Average velocity =
= = 2
The average velocity over the first 6 seconds is 2 m/s.
g The particle is at rest when v = 0. 0 = 2t− 4
t= 2 When t= 2,
x= (2)2− 4(2) − 5
=−9
The particle is at rest after 2 seconds when it is 9 metres to the left of the origin. (Note that (2, −9) is the minimum point on the graph above.)
displacement time taken
---7–( )–5 6
---distance travelled time taken
---20 6
---10 3
---10 3
---dx dt
Calculating distance and displacement from
velocity–time graphs
Velocity is the rate of change of position with respect to time, i.e. . Hence x= v dt.
E x a m p l e
3
Consider a particle travelling with velocity v m/s at time t seconds given by , where 0 ≤t≤ 8. Find:
a the distance travelled in the first 8 seconds. b the displacement of the object after 8 seconds.
S o l u t i o n
The distance travelled by the particle in 8 seconds is given by the total area because, although the section below the t-axis represents velocity in the opposite direction to that in the previous section, it nevertheless represents movement. So, although the displacement is decreasing, distance travelled is increasing.
a Distance = −
= 20
The distance travelled is 20 metres.
Note: In this case it is not necessary to use calculus to find the required area. It could be easily found from the areas of two triangles each with base 4 and height 5.
b The displacement of the particle is given by the signed area i.e. the definite integral calculated over the full 8 seconds.
The displacement of the particle is 0 metres. Since the two triangles are congruent, the sum of their signed areas is zero.
E x a m p l e
4
The velocity of an object, v m/s, at time t seconds is given by . Find: a the distance travelled in the first 10 seconds.
b the displacement of the object after 10 seconds.
S o l u t i o n
a This problem involves finding area, so a graph is required.
The graph cuts the t-axis where v(t) = 0, i.e. t= 8 (since t> 0).
v dx dt
---= ⌠⌡
v t( ) 5 5 4
---t
– =
t
v
2 4 6 8
0 2 4
–2
–4
v(t) = 5 −5t
4
⌠ ⎮ ⌡0
4
5 5
4
---t
–
⎝ ⎠
⎛ ⎞ dt ⌠⎮
⌡4 8
5 5
4
---t
–
⎝ ⎠
⎛ ⎞dt
⌠ ⎮ ⌡0
8
5 5
4
---t
–
⎝ ⎠
⎛ ⎞dt = 0
v t( ) 16 1 4
---t2,t≥0
– =
t
v
2 4 6 8 10 0
10 5 15
–5
Area = −
=
=
= 94
The distance travelled by the object in the first 10 seconds is 94 metres.
b The displacement of the object is given by .
=
=
=
The displacement of the particle is metres.
exercise
8.3
1 The position of an object on a straight line is given by x= t2− 2t, t≥ 0. Displacement is measured in metres and time in seconds. Find:
a the initial position of the object.
b the position of the object after 2 seconds.
c the distance travelled by the object in the first 2 seconds.
d the average speed of the object over the first 2 seconds.
e the velocity of the object after 2 seconds.
f when and where the object is instantaneously at rest.
2 The position of an accelerating object is given by . Find the velocity of the body when:
a t= 0 b t= 1 c t= 4 d t= 9
3 A particle moves in a straight line so that its position x centimetres relative to O at time
t seconds is given by x=t2− 7t+ 6, t≥ 0. Find:
a the initial position of the particle.
b the position of the particle after 6 seconds.
c the distance travelled by the particle in 6 seconds.
d the average speed over the first 6 seconds.
e the average velocity over the first 6 seconds. ⌠ ⎮ ⌡0 8 16 t 2 4 ----– ⎝ ⎠
⎛ ⎞ dt ⌠⎮
⌡8 10 16 t 2 4 ----– ⎝ ⎠
⎛ ⎞ dt
16t t
3 12 ---– 0 8
16t t
3 12 ---– 8 10 –
16 8( ) ( )8
3
12
---– –( )0
⎝ ⎠
⎛ ⎞ 16 10( ) ( )10 3
12
---– 16 8( ) ( )8
3 12 ---– ⎝ ⎠ ⎛ ⎞ – ⎝ ⎠ ⎛ ⎞ – ⌠ ⎮ ⌡0 10
16 t2 4
----–
⎝ ⎠
⎛ ⎞dt
⌠ ⎮ ⌡0 10 16 t 2 4 ----– ⎝ ⎠
⎛ ⎞ dt 16t t3
12
---–
0 10
16 10( )–( )---10123
230 3 ---762 3
---x = 10t3 2⁄ ,0≤ ≤t 10
f the particle’s initial velocity.
g when and where the particle is at rest.
h the velocity of the particle as it passes through O.
4 The velocity of a particle moving in a straight line, in metres per second, is given by
v= 4t2 + 2t, t≥ 0. Find the distance travelled by the particle in the first 5 seconds.
5 A body starts at the origin and moves in a straight line. Its velocity, in kilometres per hour, is given by v = 3t2 +t, t≥ 0. How far does it travel in 3 hours?
6 A particle starts at the origin and travels in a straight line with velocity v m/s given by v = 3t2− 4t− 4, t≥ 0. Find:
a the displacement of the particle after 3 seconds.
b the distance travelled by the particle in the first 3 seconds.
c when and where the particle is at rest.
7 A particle attached to a spring moves in a straight line so that at time t seconds its
velocity, v cm/s, is given by . Find:
a the displacement of the particle after 18 seconds.
b the distance travelled by the particle in the first 18 seconds.
8 The velocity of an object is given by v= 3t2− 2t, t≥ 0. If the object starts at the origin, at what time is it 100 units to the right of the origin?
9 At time t seconds, the velocity v metres per second of an object is given by v= 5 sint.
What is the total distance travelled by the object for:
a b
v 4π sin πt 6 ---,t≥0 =
0 t 2π
3
---≤ ---≤ 2---3π≤ ≤t 2π
Number
sense with the spence
88
Finding a function given its
derivative
In section 8.1 we looked at how to sketch the graph of an antiderivative. This is not a unique curve, as f′(x)dx=f(x) + c, where c is any real number.
The value of c can be determined if more information is provided.
E x a m p l e
1
Find the rule for the function f for which f′(x) = 4x− 5 and f(2) = 1.
S o l u t i o n
f′(x) = 4x− 5
Antidifferentiate to find f(x).
f(x) = 2x2− 5x+c
Determine the value of c, using the information f(2) = 1. 1 = 2(2)2− 5(2) +c
1 =−2 +c c= 3
So, f(x) = 2x2− 5x+ 3.
E x a m p l e
2
Find the equation of the curve defined by , given that the curve passes through (0, 5).
S o l u t i o n
Antidifferentiate to findy.
Find c, using the information that when x= 0, y= 5. 5 =
5 =
c=
The equation of the curve is .
⌠ ⌡
t i p
These types of problems can be solved using a TI-89. First find the antiderivative, then solve for c
to find the unique antiderivative. CAS 10.2,
10.7
dy dx
--- = e2x
dy dx
--- = e2x
t i p
The graphs of , for
c ∈{−10, −5, 0, 5, 10} help to show that only the graph of
(shown in thick style) passes through the point (0, 5).
y 1
2
---e2x+c
=
y 1
2
---e2x 9
2
---+ =
[−1, 2] by [−10, 15]
y 1
2
---e2x+c
=
1 2
---e0+c
1 2
---+c
9 2
---y 1
2
---e2x 9
2
---+ =
8.4
8 . 4
exercise
8.4
1 Find f(x) for each of the following.
a f′(x) = 6 − 2x and f(2) = 7
b f′(x) = 3(x+ 2)2 and f(1) = 23
c f′(x) = 9e3x and f(0) = 5
d f′(x) =x(2 −x) and the graph of y =f(x) passes through the point (3, 0)
e and the graph of y =f(x) passes through the point (1, −2)
f and
g and f(1) =−1
2 Find y in terms of x if and y = 0 when x= 2.
3 Find the function f if f′(x) = sin 2x− sinx and f(0) = 2.
Practical applications
E x a m p l e
3
The costs associated with maintaining a car increase as the car gets older. The rate of increase in maintenance costs, C, in dollars per year, for a particular car can be approximated by:
C′(t) = 30t2+ 1000, 0 ≤t≤ 6 where t is the age of the car inyears. Find the total maintenance costs from the third to fifthyear.
S o l u t i o n
C′(t) = 30t2+ 1000, 0 ≤t≤ 6 So, C(t) = (30t2+ 1000)dt
The total maintenance costs from the third to fifthyear is given by:
This can be calculated using a definite integral.
C′(t)dt= (30t2+ 1000)dt
=
= (10(5)3+ 1000(5)) − (10(3)3+ 1000(3))
= 2980
The total maintenance cost is $2980.
f′( )x = x –3x2
f′( )x cos x 2
---= f π
2
---⎝ ⎠
⎛ ⎞ = 1+ 2
f′( )x 1 x 2
+
x2
---=
dy dx
--- 1 3–x
---=
⌠ ⌡
C( )5 –C( )3 [C t( )]35
=
⌠ ⌡3
5
⌠ ⌡3
5
10t3+1000t
exercise
8.4
4 A model for the rate at which a typist types is given by words per minute. Estimate the number of words typed during:
a the first 3 minutes.
b the 3rd minute.
5 A population of organisms grows according to the rule where P million is the number of organisms present after t days.
a If there are initially 500 million organisms, find a rule for the number present after
t days.
b Find the number of days taken, correct to 2 decimal places, for the population to double.
6 A ball is thrown vertically upwards from a height of 20 metres above the ground so that
t seconds after its release it is h metres above the ground. The rate of change of height of
the ball, in metres per second, is given by . Find:
a a rule for h in terms of t.
b after how many seconds, correct to the nearest second, the ball reaches its maximum height above the ground.
c the maximum height above the ground reached by the ball, correct to the nearest centimetre.
7 The rate of change of depth of water in a tidal river is modelled by , where
h is the depth of water in metres and t is the number of hours after midnight on a given day.
a Find an expression for h in terms of t, if the water is 5 metres deep at midnight.
b What are the maximum and minimum depths of the water?
c For how many hours a day is the depth of water 6 metres or more?
continued
dW dt
--- = –6t2+14t+90
dP dt
--- = 300e0.2t
dh dt
--- = 10–9.8t
dh dt
--- π 3 ---cosπt
6 ---=
8 The rate of decay of a radioactive substance is such that N′(t) = −e−0.05t, where N is the number of grams present after t days.
a If there are 20 grams of substance initially, find an expression for N(t).
b Find the time taken for the initial amount of substance to halve. Give your answer correct to 2 decimal places.
9 A body moves with velocity v metres per second at time t seconds, where
v= 6 − 2t, t≥ 0. If the body starts 3 metres to the left of O, find:
a the displacement of the body after 5 seconds.
b the distance travelled by the body in the first 5 seconds.
c when and where the body is instantaneously at rest.
10 The sales of a particular product are predicted to grow at a rate
S′(t) dollars per day, where S′(t) = 100et and t is the time in days since sales of the product commenced.
a How much money is expected to be made from sales over the first 8 days, correct to the nearest thousand dollars.
b On what day are the accumulated sales predicted to first exceed $150 000?
c How much money is expected to be made from sales during the 6th, 7th and 8th days, correct to the nearest thousand dollars.
11 The flow of blood in a blood vessel is slower towards the outside of the vessel and faster toward the middle. The speed of the blood is given by v(r) =k(R2− r2), where k is a constant related to the pressure and the viscosity of the blood and the length of the blood vessel, R is the radius of the blood vessel and r is the distance of the blood from the middle of the vessel. Find the total blood flow in a vessel in terms of k if this is given by
2π (v(r) ×r)dr.
12 The rate, in kilowatt hours per hour, at which electrical energy is used on a particular day
in a particular household, is given by the rule , where K is the
number of kilowatt hours used and t is the time in hours after midnight.
a Differentiate te−t and hence find an antiderivative of 15te−t.
b How many kilowatt hours, correct to 2 decimal places, are used in the household over the first 4 hours of the day?
c Find an expression that could be used to determine the number of kilowatt hours used during the first T hours of the day.
13 A concrete bridge of width 15 metres spans a river. The cross section of the bridge is the region between the graphs of y= 0.4x(10 − x) and y = 12 as shown in the screenshot on the right. Linear measurements are in metres. Find the volume of concrete in the bridge.
⌠
⌡0
R
dK dt
--- = 15te–t,0≤ ≤t 24
Average value of a function
One day, when cleaning the glass of her new fish tank, Natalie notices that waves form at the surface of the water. The water now has peaks and troughs. Eventually the waves subside and the water once again reaches a mean or average level.
If a function could be found to model the depth of the water in the tank when the waves are present at the surface, then the average value of this function would be the mean level to which the water returns.
The area under the curve shown with equation y=f(x) on the interval [a, b] can be
calculated using f(x)dx.
The same area could be calculated using a rectangle with width =b−a and height = average value of f over the interval [a, b].
It follows that:
f(x)dx= (b−a) × average value of f
Average value of f= f(x)dx
The average value of f is often denoted by fave.
Not level
Mean level
⌠ ⌡a
b
x
x=a x=b
y
A1
A2
Rectangle with
average height of f(x) on the interval (a, b)
Note: the average height of f(x) is the height that makes
A1=A2.
f(x)
y =
⌠ ⌡a
b
1 b–a
---⌠⎮ ⌡a
b
Average (or mean) value of a function
If f is a continuous function on the interval [a, b], then the average (or mean) value of f on [a, b] is given by:
f(x)dx fave 1
b a–
---= ⌠⎮
⌡a
b
8.5
8 . 5
8 . 5
E x a m p l e
1
Find the average value of the function on the given interval. a f(x) =x2; [1, 5] b f(x) = sin 2x; [0, π]
S o l u t i o n
a fave= f(x)dx
= x2dx
=
=
=
b fave= f(x)dx
= sin 2x dx
=
= = 0
E x a m p l e
2
Find the value of c such that fave=f(c) for f(x) =x2− 4 on the interval [0, 3].
S o l u t i o n
It is first necessary to determine the average value of f on the interval [0, 3].
fave=
=
=
=−1
Now, fave=f(c), so:
−1 =c2− 4 3 =c2 c=
But as c must be in the interval [0, 3], .
t i p
This answer to part a can also be obtained using a TI-89 by defining a function as shown below.
The graph of y=f(x) and its average value on [1, 5] is shown below.
Note that A1=A2. A1
A2
CAS 5.5, 10.1
1
b a– ---⌠⎮
⌡a
b
1 5 1– ---⌠⎮
⌡1 5
1 4
--- x3
3 ---1 5 1 12
---(125 1– )
31 3
---1
b a– ---⌠⎮
⌡a
b
1
π–0
---⌠⎮ ⌡0 π 1 π --- 1 2
---cos 2x
–
0
π
1 2π
---(cos 2π–cos 0)
–
CAS 10.1, 10.2
t i p
This solution can also be obtained using the aveval function defined above.
.First find the average value of f on the interval [0, 3]
.Now solve f(c) =−1 for c on the interval [0, 3].
1 3
---⌠⎮ ⌡0
3
x2–4
( )dx
1 3
--- x
3
3
---–4x
0 3 1 3 --- 27 3
---–12
⎝ ⎠
⎛ ⎞
± 3
E x a m p l e
3
The number of people living in a country is given by N(t) = 200e0.008t, where N is in millions and t is the number of years since 1985. Find the average number of people living in this country between 1985 and 1990, correct to the nearest million.
S o l u t i o n
Average number of people =
=
=
= 5000(e0.04− 1)
= 204.054
The average number of people living in the country between 1985 and 1990 was 204 million. This result can be confirmed using a TI-89 as shown above.
exercise
8.5
1 Find the average value of the function over the interval indicated by hand. Confirm your answer using CAS.
a x3; [−2, 1] b 5 − 3x2; [0, 4]
c ; 0 ≤x≤ 16 d ; [2, 4]
e 3x+ 2; [−5, 2] f cosx;
g e−2x; [−1, 1] h ; −3 ≤x≤ −2
i x2 − 5x+ 4; [0, 5] j ; 1 ≤ x≤ 4
2 For each of the following:
i Find the value of fave over the interval given by the domain of the function.
ii Find the value(s) of c in the given interval such that f(c) =fave.
iii Sketch the graph of f and construct a rectangle whose area is the same as the area under the graph of f over the interval.
a f: [−2, 4] →R, f(x) = 4x b f: [−1, 4] →R, f(x) =x2
c f: [−3, −1] →R, d f: [0, 1] →R, f(x) = e−x
3 The amount of drug, A milligrams, in the bloodstream at time t hours after it is administered is given by A= 2e−0.1t.
a What is the initial dose of the drug?
b What is the average amount of drug present in the bloodstream over the first 2 hours after it is administered? Give your answer correct to 2 decimal places.
1 5 0– ---⌠⎮
⌡0 5
N t( )dt
1 5
---⌠⎮ ⌡0
5
200e0.008tdt
1 5
--- 200
0.008
---[e0.008t]
0 5 ×
x 1
x
---π 2 ---≤ ≤x π
1 1–x
---1
x2
---f x( ) 1 x2
---=
4 The air temperature in a particular suburb during a period of 12 hours is modelled by
T= 12 + 3t− 0.17t2, 0 ≤t≤ 12, where t is measured in hours and T in degrees Celsius. Find, correct to the nearest tenth of a degree, the average temperature during:
a the first 8 hour period b the entire 12 hour period.
5 The average value of f(x) = −3x2 + 6x+ 4 is equal to 3 on the interval [0, k] and k is a positive real number. Find the exact value of k.
6 The velocity of a falling object is given by v(t) = 10 + 9.8t metres per second at time
t seconds after its release. Find the average velocity for 0 ≤ t≤ 4.
7 The depth in metres of water in a tidal pool at time t hours is . What is the average depth for 0 ≤ t≤ 10?
8 The speed at which a typist can type over a 4 minute interval is given by
S(t) = 70 + 14t− 6t2, 0 ≤t≤ 4, where S(t) is the number of words per minute at time t minutes. Find:
a the speed at the beginning of the interval
b the maximum speed, correct to 2 decimal places
c the average speed over the interval.
9 A function y=f(x) is positive for all values of x. If the average value of the function on the interval [1, 5] is 3, what is the area under the curve on the same interval?
d 3.5 1.2 sinπt 5 ---+
=
In an alternating current electric circuit, the current, I(t) amperes is given by the
expression I(t) =Acosωt, where A is the amplitude and ω is the frequency measured in cycles per second.
a Write down an expression for the current in an average household circuit for which the frequency is 60 cycles per second.
The power in an electric circuit of this type is given by P(t) = (I(t))2.
b Write down an expression for the power in the circuit defined in question a.
c Sketch a graph of the function defined in question b, showing one complete cycle.
d Write down, but do not attempt to evaluate, a suitable integral to find the average power in this circuit over a complete cycle.
e i It can be shown that .
Use this result to express cos260t in the form
, where a is a positive constant.
ii Hence evaluate the integral found in question d.
f It has been found that average power in an electric circuit is given by kA2, where k is a positive real constant and A is the amplitude. Find the value of k.
cos2x 1
2
---(1–cos 2x)
=
1 2
---(1–cosat)
analysis task 3—
average power in an electric circuit
SAC
8.3A
verage power in an electric circuit
CD
In a dry country like Australia, farm dams are part of the lifeblood of rural life. Knowing the volume of water at any time is important for such things as planning the number and distribution of livestock, and estimating when the supply is likely to run out if there are drought conditions.
Methods of estimating the volume of a partly empty dam depend on i
