FS
O
PR
O
PA
G
E
7
7.1
EC
TE
D
Antidifferentiation
Kick off with CAS
R
7.2 Antidifferentiation
7.4 Families of curves and applications
7.5 Review
U
N
C
O
R
7.3 Antiderivatives of exponential and trigonometric
functions
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7.1 Kick off with CAS
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
To come
Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive
step-by-step guide on how to use your CAS technology.
c07Antidifferentiation.indd 301
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7.2 Antidifferentiation
Antidifferentiation, also known as integration, is the reverse process of
differentiation. Antidifferentiation allows us to find f(x) when we are given f ′(x).
d
In Topic 5, you learned that (x2) = 2x. Alternatively, this can be expressed in
dx
function notation: if f(x) = x2, then f ′(x) = 2x. So if you were given f ′(x) = 2x and
asked to find f(x), you might expect that f(x) = x2.
However, this is not quite as simple as it first appears. Consider each of the following
derivatives.
d( 2
x + 7) = 2x
dx
d( 2
x + 2) = 2x
dx
d( 2
x − 1) = 2x
dx
d( 2
x − 5) = 2x
dx
Units 3 & 4
AOS 3
Topic 4
Concept 1
PR
O
O
FS
Antidifferentiation
Concept summary
Practice questions
PA
G
E
If we are asked to find f(x) given that f ′(x) = 2x, how do we know which of the
equations above is the correct answer? To give a totally correct answer, additional
information about the function must be given.
If f ′(x) = 2x, then f(x) = x2 + c, where c is an arbitrary constant. This means we
have a family of curves that fit the criteria for the function f.
y = x2 + 7
y
y = x2 – 1
y = x2 – 5
x
0
y = x2 – 8
O
R
R
EC
TE
D
y = x2 + 2
U
N
C
To know which specific curve matches f, we must know additional information such
as a point through which the curve passes.
Notation
The notation that is commonly used for antidifferentiation was introduced by
the German mathematician Gottfried Leibniz (1646–1716). An example of this
notation is:
2
32xdx = x + c
This equation indicates that the antiderivative of 2x with respect to x is equal to x2
plus an unknown constant, c. 3 f(x)dx is known as the indefinite integral. It is read as
302 MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
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‘the integral of f(x) with respect to x’. The symbol 3 on the left and the dx on the right
Interactivity
Integration of axn
int-6419
can be thought of as the bread of a sandwich, with the f(x) being the sandwich filling.
The 3 tells us to antidifferentiate and the dx tells us that x is the variable.
Remembering that the reverse process of differentiation is integration, consider the
following:
∴ 33x2dx = x3 + c
d 3
(x ) = 3x2
dx
FS
∴ 34x3dx = x4 + c
d 4
(x ) = 4x3
dx
O
d
(2x5) = 10x4 ∴ 310x4dx = 2x5 + c
dx
PR
O
d
(3x6) = 18x5 ∴ 318x5dx = 3x6 + c
dx
From this series of derivatives and integrals, two important observations can be made.
PA
G
E
xn+1
n
3x dx = n + 1 + c, n ≠ −1
axn+1
n
n
3ax dx = a3x dx = n + 1 + c, n ≠ −1
TE
D
This is the general rule for the antiderivative of axn.
Properties of integrals
As differentiation is a linear operation, so too is antidifferentiation.
EC
3 ( f(x) ± g(x)) = 3f(x)dx ± 3g(x)dx
3af(x)dx = a3f(x)dx, where a is a constant.
U
N
C
O
R
R
Interactivity
Properties of integrals
int-6420
That is, we can antidifferentiate the separate components of an expression.
For example,
3
2
3 (4x + 6x − 9x + 7)dx =
4x3+1 6x2+1 9x1+1
+
−
+ 7x + c
4
3
2
= x4 + 2x3 − 9 x2 + 7x + c
2
To check your antiderivative is correct, it is always good to differentiate your answer
to see if the derivative matches the original expression.
Note: If you are asked to find ‘the’ antiderivative of an expression, then the ‘+ c’
component must be part of the answer. However, if you are asked to find ‘an’
antiderivative, then you can choose what the value of c is. The convention when
finding ‘an’ antiderivative is to let c equal 0. So the example above would have an
antiderivative of x4 + 2x3 − 92x2 + 7x.
Topic 7 Antidifferentiation c07Antidifferentiation.indd 303
303
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WORKED
EXAMPLE
1
Find3 (3x4 − x3 + 2)dx.
1
− 7, find the rule for f .
x2
b Find an antiderivative of (2x − 3)(4 − x).
a If f ′(x) = 2 !x +
x4 − 2x3 + 5
dx.
x3
WRITE
by applying the rule.
b 1 First rewrite any surds as fractional powers,
and rewrite any powers in the denominator
as negative powers. That is, write each term
in the form axn.
− + 2x + c
a 3 ( 3x4 − x3 + 2 ) dx =
5
4
3x5
1
− 7bdx
b f(x) = 3 a2 !x +
x2
= 3 ( 2x2 + x−2 − 7 ) dx
1
3
E
2x2 x−1
=
+ −1 − 7x + c
3
2
2 Antidifferentiate each term separately by
PA
G
applying the rule.
x4
O
a Antidifferentiate each term separately
PR
O
THINK
FS
c Determine 3
3
c 1 First expand the expression.
TE
D
3 Simplify.
2 Antidifferentiate each term separately
R
EC
by applying the rule.
Note: The ‘+ c’ is not needed as the
question asked for ‘an’ antiderivative.
R
d 1 Rewrite the expression as separate
C
O
fractions.
4x2 1
− − 7x + c
=
x
3
c (2x − 3)(4 − x) = 8x − 2x2 − 12 + 3x
= −2x2 + 11x − 12
U
N
2 Antidifferentiate each term separately by
applying the rule.
3 Simplify.
Note: Simplest form usually assumes
positive indices only.
3
2
2 + 11x − 12) dx = − 2x + 11x − 12x
(
−2x
3
3
2
d 3
x4 − 2x3 + 5
x4 2x3 5
dx
=
− 3 + 3 bdx
a
3
x3
x3
x
x
= 3 (x − 2 + 5x−3) dx
=
5x−2
x2
− 2x +
+c
2
−2
=
5
x2
− 2x − 2 + c
2
2x
Note: It is extremely useful to differentiate the answer of an antiderivative in order to
check its validity.
304
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
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Integrals of the form 3f(ax + b)dx, n ≠ 1
Consider the function f : R → R, f(x) = (ax + b) n.
Using the chain rule, f ′(x) = an(ax + b) n−1.
Hence,
3an(ax + b)
n−1
dx = (ax + b) n
n−1dx
=
1
(ax + b) n + c
an
O
3 (ax + b)
Antidifferentiate:
E
2
a (2x + 3) 5
tHinK
TE
D
(ax + b) n+1
+ c.
a(n + 1)
R
R
b 1 Take 2 out as a factor.
C
O
2 Apply the rule 3 (ax + b) ndx =
U
N
b 2(3x − 1) −2.
WritE
EC
a Apply the rule 3 (ax + b) ndx =
(ax + b) n+1
+ c.
a(n + 1)
PA
G
WOrKED
EXAMPLE
PR
O
Thus we obtain the general rule:
n
3 (ax + b) dx =
FS
an3 (ax + b) n−1dx = (ax + b) n
(ax + b) n+1
+ c.
a(n + 1)
a 3 (2x + 3) 5dx
(2x + 3) 6
+c
2(6)
(2x + 3) 6
=
+c
12
=
b 32(3x − 1) −2dx
= 23 (3x − 1) −2dx
=
2(3x − 1) −1
+c
3(−1)
2(3x − 1) −1
+c
−3
2
=−
+c
3(3x − 1)
=
Note: The rules described above only apply if the expression inside the brackets is
linear. If the expression is of any other kind, it must be expanded, if possible, before
integrating, or you must use CAS to integrate the expression.
Topic 7 ANTIDIffErENTIATION
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WORKED
EXAMPLE
3
THINK
1
Find 3 ax + b dx.
x
2
WRITE
1
3 ax + x b dx
2
1 As the inner function is not linear,
there is no antidifferentiation rule
we can apply, so the expression
must first be expanded.
1
1
= 3 ax2 + 2xa b + a b bdx
x
x
2 Write all terms in the form axn.
= 3 (x2 + 2 + x−2) dx
3 Apply the rules for
=
E
PA
G
Integration by recognition
PR
O
x3
x−1
+ 2x −
+c
3
1
1
1
= x3 + 2x − + c
x
3
antidifferentiation.
FS
1
bdx
x2
O
= 3 ax2 + 2 +
2
Sometimes you may be required to find an antiderivative of a very complex function.
In order to complete this task, you will first be given a function to differentiate. The
technique is then to recognise the patterns between the derivative you have found and
the function you have been given to antidifferentiate.
4
If y = (3x2 + 4x − 7) 5, find
EC
WORKED
EXAMPLE
TE
D
In general, if f(x) = g′(x), then 3 g′(x)dx = f(x) + c.
dy
dx
. Hence find an antiderivative of
R
20(3x + 2)(3x2 + 4x − 7) 4.
R
THINK
C
given function.
O
1 Use the chain rule to differentiate the
U
N
2 Remove 2 as a factor from
the linear bracket.
3 Rewrite the result as an integral.
4 Adjust the left-hand side so that it
matches the expression to be integrated.
5 Write the answer.
Note: c is not required because ‘an’
antiderivative was determined.
306
WRITE
y = (3x2 + 4x − 7) 5
dy
= 5(6x + 4)(3x2 + 4x − 7) 4
dx
= 10(3x + 2)(3x2 + 4x − 7) 4
2
4
2
5
310(3x + 2)(3x + 4x − 7) dx = (3x + 4x − 7)
2 × 310(3x + 2)(3x2 + 4x − 7) 4dx = 2 × (3x2 + 4x − 7) 5
2
4
2
5
320(3x + 2)(3x + 4x − 7) dx = 2(3x + 4x − 7)
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
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Exercise 7.2 Antidifferentiation
PRactise
1
Work without CAS
Find:
3
a f(x) if f ′(x) = 2x − 4x2 + 2x3
WE1
b an antiderivative of
c 3x(x − 3)(2x + 5)dx
d 3
2 Find:
a 3 a
3
2
1
+ 2 − 3 bdx
2x
!x x
3x3 − x
dx.
2 !x
Antidifferentiate:
a (3x − 5) 5
b
TE
D
b 3 a !x +
3
1
b dx.
!x
dy
If y = (3x2 + 2x − 4) 3, find . Hence find an antiderivative of
dx
(3x + 1)(3x2 + 2x − 4) 2.
WE4
EC
7
PA
G
3
a 3 ( !x − x) 2dx
.
b 3 (1 − 2x) −5dx.
1
Find 3 a2x2 + b dx.
x
6 Find:
5
2
E
a 3 (2x + 3) 4dx
WE3
1
(2x − 3)
4 Find:
5
O
WE2
PR
O
3
FS
b 3 (x + 1)(2x2 − 3x + 4)dx.
3
2
− 4x3 + 3
5x
!x
4
O
R
R
dy
1
8 If y = a7x + !x −
b , find . Hence find an antiderivative of
dx
!x
3
1
1
1
y = a7 +
+ 3 b a7x + !x −
b .
2 !x 2"
!x
x2
1
9 Given that f ′(x) = x2 − 2 , find the rule for f .
x
10 Find:
C
Consolidate
U
N
Apply the most
appropriate
mathematical
processes and tools
a 3x3dx
b 37x2 −
c 3 (4x3 − 7x2 + 2x − 1)dx
d 3 (2 !x) 3dx.
2
dx
5x3
11 Find the indefinite integral of each of the following functions.
a (3x − 1) 3
5
2
c x2 − 3x5
e !x (2x − !x)
1
4x3
x4 − 2x
d
x3
f !4 − x
b
Topic 7 Antidifferentiation c07Antidifferentiation.indd 307
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21/08/15 5:41 PM
12 Find an antiderivative for each of the following functions.
x3 + x2 + 1
x2
a (2x + 3)(3x − 2)
b
4
c 2 !x −
!x
2
d ax − b
x3
3
e 2(1 − 4x) −3
2
2
f
5
(2x − 3) 2
Find the general rule for the function, y.
dy
dx
= x3 − 3 !x.
FS
13 The gradient function for a particular curve is given by
14 Find the general equation of the curve whose gradient at any point is given
O
x3 + 3x2 − 3
.
x2
15 Find the general equation of the curve whose gradient at any point on the curve
1
.
is given by !x +
!x
dy
5x
16 If y = "x2 + 1, find
and hence find the antiderivative of
.
2
dx
"x + 1
dy
17 If y = (5x2 + 2x − 1) 4, find
and hence find an antiderivative of
16(5x + 1)(5x2 + 2x − 1) 3. dx
dy
15x2 + 8x
18 If y = "5x3 + 4x2, find
and hence find an antiderivative of
.
dx
"5x3 + 4x2
x2
19 Find 3
dx.
Master
"x3 + 1
TE
D
PA
G
E
PR
O
by
20 Find 32(3x + 5) 2 (7x2 + 4x − 1)dx.
EC
1
R
R
of exponential and
7.3 Antiderivatives
trigonometric functions
O
Antiderivatives of cos(x) and sin(x).
U
N
C
We know that:
• if y = sin(x), then
dy
dx
= cos(x)
• if y = sin(ax + b), then
• if y = cos(x), then
dy
dx
dy
dx
= −sin(x)
• if y = cos(ax + b), then
308 = a cos(ax + b)
dy
dx
= −a sin(ax + b).
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
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Hence,
1
3cos(x)dx = sin(x) + c and 3cos(ax + b)dx = a sin(ax + b) + c
1
3sin(x)dx = −cos(x) + c and 3sin(ax + b) dx = − a cos(ax + b) + c
5
WritE
x
3 a2 sin(5x) + 3 cosa 3 b bdx
O
tHinK
x
Find an antiderivative of f(x) = 2 sin(5x) + 3 cosa b.
3
FS
WOrKED
EXAMPLE
PR
O
1 Separate the two terms.
x
=32 sin(5x)dx + 33 cosa bdx
3
sin and cos.
PA
G
3 Apply the antidifferentiation rules for
x
= 23sin(5x)dx + 33cosa bdx
3
x
= −25cos(5x) + 9 sina b
3
E
2 Take out 2 and 3 as factors.
TE
D
The antiderivative of e x
R
Find:
b 38e2xdx.
O
a 3 (x7 − e3x)dx
WritE
U
N
tHinK
6
C
WOrKED
EXAMPLE
1 kx
x
x
kx
3e dx = e + c and 3e dx = k e + c
R
EC
d
d
As we have seen in Topic 5, (ex) = ex and (ekx) = kekx.
dx
dx
Therefore, it follows that
a Integrate each term separately.
b 1 Apply the rule 3ekxdx = ekx + c.
k
1
a 3 (x7 − e3x) dx
1 3x
8
= 18x − 3e + c
b 38e2xdx
= 83e2xdx
= 8 × 12e2x + c
2 Simplify.
= 4e2x + c
Topic 7 ANTIDIffErENTIATION
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309
21/08/15 5:41 PM
Algebraic expansion may also be necessary for questions involving ekx.
7
Find y if it is known that
dy
dx
THINK
= (ex + e−x) 3.
WRITE
dy
= ( ex + e−x ) 3
dx
= ( ex ) 3 + 3 ( ex ) 2 ( e−x ) + 3 ( ex ) ( e−x ) 2 + ( e−x ) 3
= e3x + 3e2xe−x + 3exe−2x + e−3x
= e3x + 3ex + 3e−x + e−3x
1 Expand the brackets.
y = 3 ( e3x + 3ex + 3e−x + e−3x ) dx
O
2 Antidifferentiate each term separately.
FS
WORKED
EXAMPLE
PR
O
= 13e3x + 3ex − 3e−x − 13e−3x + c
8
Given that y = ex , find
2
dy
dx
THINK
2
and hence find an antiderivative of xex .
TE
D
WORKED
EXAMPLE
PA
G
E
For particularly difficult antidifferentiation problems, you may first be asked
to differentiate a function so that you can use this result to carry out the
antidifferentiation.
Recall that if f(x) = g′(x), then 3 g′(x)dx = f(x) + c
WRITE
y = ex
dy
2
= 2xex
dx
1 Use the chain rule to differentiate the
EC
given function.
x
x
32xe dx = e
2
R
2 Rewrite the result as an integral.
R
2
O
1
2
U
N
C
2
23xex dx = ex
3 Adjust the left-hand side so that it matches
the expression to be integrated.
2
× 23xex dx = 12 × ex
2
2
1 x
x
3xe dx = 2e
2
4 Write the answer.
2
2
EXERCISE 7.3 Antiderivatives of exponential and
trigonometric functions
PRACTISE
Work without CAS
310
1
x
1
cos(3x + 4) − 4 sina b.
2
2
2x
1
b Find an antiderivative of cosa b − sin(5 − 2x).
3
4
WE5
a Find the indefinite integral of
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
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21/08/15 5:51 PM
2a Find 3 asina b − 3 cosa b bdx.
x
x
2
2
b If f ′(x) = 7 cos(2x) − sin(3x), find a general rule for f.
3 WE6 Find:
a 3 (x4 − e−4x)dx
b 3 a e2x − e
2
3
1
2
− 1x
2
b dx.
b cos(4x) + 3e−3x.
Find 3 (e2x − e−3x) 3dx.
O
WE7
x
3
x
6 Find the indefinite integral of ae2 −
7
2
1 2
b .
ex
Find the derivative of ecos (x) and hence find an antiderivative of
2
sin(x)cos(x)ecos (x).
WE8
E
5
x
3
a e + sina b +
PR
O
x
3
FS
4 Find the indefinite integral of:
8 Find the derivative of e(x+1) and hence find 39(x + 1) 2e(x+1) dx.
Consolidate
9 Find:
a 3 (2e3x − sin (2x))dx
TE
D
Apply the most
appropriate
mathematical
processes and tools
PA
G
3
c 3 (0.5 cos(2x + 5) − e−x)dx
b 3
3
e2x + 3e−5x
dx
2ex
d 3 (ex − e2x) 2dx.
EC
10 If it is known that 3aebxdx = −3e3x + c, find the exact values of the constants
a and b.
3πx
1
+ sina
b.
2
2
4x
dy
12 The gradient of a tangent to a curve is given by
= cos(2x) − e−3x. Find a
dx
possible general rule for the curve y.
U
N
C
O
R
R
11 Find an antiderivative of
13 Heat escapes from a storage tank at a rate of kilojoules per day. This rate can be
modelled by
π2
πt
dH
= 1 + sina b , 0 ≤ t ≤ 100
dt
9
45
where H(t) is the total accumulated heat loss in kilojoules, t days after June 1.
a Determine H(t).
b What is the total accumulated heat loss after 15 days? Give your answer correct
to 3 decimal places.
14 Differentiate y = 2xe3x and hence find an antiderivative of xe3x.
dy
2
15 Given that y = e2x +3x−1, find
and hence find an antiderivative of
2
dx
2(4x + 3)e2x +3x−1.
d
16 Find (x cos(x)) and hence find an antiderivative of x sin(x).
dx
Topic 7 Antidifferentiation c07Antidifferentiation.indd 311
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21/08/15 5:41 PM
17 If x(t) = 20 + cosa
dy
πt
π
= x(t) − π, find a possible rule for y in terms of t.
b and
4
dt 20
18 If f ′(x) = a sin(mx) − benx and f(x) = cos(2x) − 2e−2x + 3, find the exact
constants a, b, c, m and n.
e2x + ex − 1
dx.
ex + 1
Families of curves and applications
Initial conditions
FS
20 Find 3
0 ≤ x ≤ 2π}.
O
7.4
19 Find {x :3ex sin(x)dx = 3ex cos(x)dx,
dy
PR
O
MastEr
= 2e2x. Because y = e2x + c, this is a series
dx
of an infinite number of exponential functions. We call this a family of curves.
The functions with c values of 3, 1, 0, –2 and –4, shown below, are four of the
possible functions for y = e2x + c.
y
A specific function can only be found if we are given
some additional information to allow us to evaluate
the constant, c. For example, we might be told that the
y = e2x + 1
curve passes through the origin. This lets us know that
y = e2x
when x = 0, y = 0.
x
0
2x – 2
2x
y
=
e
Hence,
y=e +c
(0, 0) ⇒ 0 = e2(0) + c
y = e2x – 4
0=1+c
c = −1
Therefore,
y = e2x − 1
This additional information is referred to as an initial condition. The question could
have been given as follows: ‘If f ′(x) = 2e2x, find f given that f(0) = 0.’
Suppose we are asked to investigate
a Sketch a family of curves that have the derivative function f ′(x) = 2 cos(2x)
O
9
for 0 ≤ x ≤ 2π.
U
N
C
WOrKED
EXAMPLE
R
R
EC
TE
D
PA
G
E
Interactivity
Families of curves
int-6421
b Find the specific rule for this function if f(π) = 2.
tHinK
WritE/draW
a 1 Apply the rule 3cos(ax)dx = sin(ax) + c to
a
1
antidifferentiate the function.
312
a f(x) = 32 cos(2x)dx
= sin(2x) + c
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
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f(x) = sin(2x) + c
2 Graph the function, first with c = 0. That is,
sketch f(x) = sin(2x). Then translate this graph
up or down to sketch the graphs with different
c values. Any c values may be used.
Amplitude = 1 and period =
y
y = sin(2x) + 2
0
2π x
O
PR
O
E
3 State the rule for f(x).
Find the equation of the curve that passes through the point (1, 0) if the
dy
gradient is given by
= 3x2 − 2x + 2.
dx
PA
G
10
TE
D
tHinK
1 Write the gradient rule and antidifferentiate
EC
to find y.
R
R
2 Substitute the known point into the equation.
U
N
C
O
3 State the rule for y.
WOrKED
EXAMPLE
f(x) = sin(2x) + c
⇒
f(π) = 2
2 = sin(2π) + c
b
2 =0+c
c=2
f(x) = sin(2x) + 2
2 Simplify and determine the value for c.
WOrKED
EXAMPLE
FS
π
y = sin(2x)
y = sin(2x) – 3
b 1 Substitute the known point into the equation.
2π
=π
2
11
WritE
dy
= 3x2 − 2x + 2
dx
y = 3 (3x2 − 2x + 2) dx
= x3 − x2 + 2x + c
When x = 1, y = 0:
0=1−1+2+c
c = −2
y = x3 − x2 + 2x − 2
Application questions such as those involving rates of change may also be given in
terms of the derivative function. Integrating the equation for the rate of change allows
us to determine the original function.
A young boy bought an ant farm. It is known that the ant population is
dN
= 20e0.2t, 0 ≤ t ≤ 20, where N is the number of
changing at a rate defined by
dt
ants in the colony and t is the time in days since the ant farm has been set up.
a Find a rule relating N to t if initially there were 50 ants.
b How many ants make up the colony after 8 days?
Topic 7 ANTIDIffErENTIATION
c07Antidifferentiation.indd 313
313
21/08/15 5:41 PM
WRITE
THINK
a 1 Write the rate rule and antidifferentiate to
a
find the function for N.
dN
= 20e0.2t
dt
N = 3 (20e0.2t) dt
20 0.2t
e +c
= 0.2
= 100e0.2t + c
When t = 0, N = 50:
50 = 100e0.2(0) + c
50 = 100 + c
c = −50
N = 100e0.2t − 50
b When t = 8:
N = 100e0.2(8) − 50
= 100e1.6 − 50
= 445.3
There are 445 ants after 8 days.
2 Use the initial condition to determine the
FS
value of c.
O
3 State the equation for N.
PR
O
b 1 Substitute t = 8 into the population equation.
E
2 Answer the question.
PA
G
Sketching the graph of a function given the graph of f ’(x)
When f ′(x) represents the equation of a polynomial function, the graph of f(x) can be
drawn by raising the degree of f ′(x) by one.
Note: When we sketch f(x) from the gradient function, the ‘+ c’ component is
unknown, so there is not just one single answer. Often we will choose c to be zero,
but any vertical translation of this general graph is correct.
Units 3 & 4
TE
D
AOS 3
Topic 4
Concept 3
Gradient function, f ′(x)
A line parallel to the x-axis ( y = m)
is degree 0.
y
Example:
R
EC
Graphs of
antiderivative
functions
Concept summary
Practice questions
R
x
0
O
y
f ʹ(x) = 2
A line of the form y = mx + c
is degree 1.
Example: The line shown has an
x-intercept at x = 3.
y
3 f ʹ(x) = –x + 3
0
314 A line of the form y = mx + c is
degree 1 and its gradient is m.
(1, 2)
f(x) = 2x
x
0
C
U
N
Interactivity
Sketching the
antiderivative graph
int-5965
2
Original function, f (x)
3
x
A quadratic of the form y = ax2 + bx + c
is degree 2. The function shown has a
turning point at x = 3.
y
0
f(x) = – –12 x2 + 3x
x
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
c07Antidifferentiation.indd 314
21/08/15 5:41 PM
Gradient function, f ′(x)
A quadratic of the form
y = ax2 + bx + c is degree 2.
Example: The graph shown has
x-intercepts at x = –1 and x = 5.
y
f ʹ(x) = (x + 1)(x – 5)
f(x) = –13 x3 – 2x2 – 5x
x
0
x
FS
y
Original function, f (x)
A cubic of the form
y = ax3 + bx2 + cx + d is degree 3.
The graph has turning points at x = –1
and x = 5.
(2, –9)
12
The graph of the gradient function y = f ′(x)
is shown. Analyse this derivative function
and sketch a possible graph for y = f(x).
TE
D
WOrKED
EXAMPLE
PA
G
E
PR
O
O
The derivative functions will not always be polynomial functions, so it is important to
analyse the graph of the derivative carefully, as it will give you key information about
the antiderivative graph. Features of derivative graphs include the following:
• x-intercepts on the graph of y = f ′(x) give the x-coordinates of stationary points on
the graph of y = f(x), that is, points where the derivative is zero.
• When the graph of y = f ′(x) is above the x-axis, this indicates the graph of y = f(x)
has a positive gradient for these x-values.
• When the graph of y = f ′(x) is below the x-axis, this indicates the graph of y = f(x)
has a negative gradient for these x-values.
y
y = fʹ(x)
x
O
R
R
EC
0
C
tHinK
1 Use the derivative graph to
U
N
determine the key features of
f — where f ′ lies above or
below the x axis (this indicates
where f has positive or
negative gradient) and where
the x-intercepts are located
(this indicates the position
of stationary points on f ).
As y = f ′(x) is a positive
cubic, y = f (x) will be
a positive quadratic.
WritE/draW
y
f ʹ(x) = 0 which
indicates a stationary
point when x = 1.
f ʹ(x) = 0 which
indicates a stationary
point when x = –1.
f ʹ is negative so
f has a negative
gradient for x < –1.
0
f ʹ(x) is positive
so f has a
y = f ʹ(x)
positive gradient
for –1 < x < 1.
fʹ is positive
so f has a
positive gradient
for x > 3.
x
f ʹ(x) = 0 which
indicates a stationary
point when x = 3.
f ʹ is negative so f has
a negative gradient
for 1 < x < 3.
Topic 7 ANTIDIffErENTIATION
c07Antidifferentiation.indd 315
315
21/08/15 5:41 PM
y = f(x) has stationary points when x = –1, 1, 3.
The gradient of y = f(x) is positive when x ∈ (−1, 1) ∪ (3, ∞) .
The gradient of y = f(x) is negative when x ∈ (−∞, −1) ∪ (1, 3) .
y
2 Use this knowledge to sketch
y = f(x)
Local maximum
at x = 1
O
Local minimum
at x = 3
Local minimum
at x = –1
FS
x
0
PR
O
the function y = f(x).
Note: In this graph of y = f(x),
we have chosen c = 0. Your
graph may have a different c
value. However, it should be
of the same basic shape, just
translated vertically.
Linear motion
PA
G
E
From Topic 6 we know that the study of
the motion of a particle in a straight line is
called kinematics. When this motion is only –5 –4 –3 –2 –1
in a straight line, it is referred to as rectilinear motion.
0
1
2
3
4
5
U
N
C
O
R
R
EC
TE
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Displacement–velocity–acceleration relationship
Because velocity is the derivative of position (displacement)
with respect to time, it follows that position is the
Displacement x(t)
antiderivative of velocity.
Consider a particle whose position, x metres, from the
dx
–
∫v(t)dt
dt
origin at time t seconds is defined by
x(t) = t2 − 5t − 6, t ≥ 0.
Velocity v(t)
Initially, at t = 0, the particle is 6 metres to the left of
the origin. The velocity of the particle can be defined as
dv
v = x′(t) = 2t − 5 metres/second.
–
∫a(t)dt
dt
The initial velocity of the particle is –5 metres/second.
This same situation could have been approached in the
Acceleration a(t)
following way. A particle has an instantaneous velocity
defined by
v = x′(t) = 2t − 5 metres/second.
If it is known that the particle is initially 6 metres to the left of the origin,
then the displacement can be given by
x = 3 (2t − 5)dt
= t2 − 5t + c
When t = 0, x = −6: −6 = 0 + c
c = −6
So: x = t2 − 5t − 6
316 MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
c07Antidifferentiation.indd 316
21/08/15 2:46 PM
WOrKED
EXAMPLE
13
In each of the following cases, find the displacement as a function of t if
initially the particle is at the origin.
1
a v = t3 − t
b v = (2t − 3) 3
c v=
(t − 1) 2
tHinK
WritE
a 1 Write the velocity equation
a v=
and antidifferentiate to find the
displacement function, x.
dx
= t3 − t
dt
x = 3 (t3 − t)dt
3 State the rule.
b 1 Write the velocity equation
O
x = 3 (2t − 3) 3dt
PA
G
and antidifferentiate to find the
displacement function, x.
x = 14t4 − 12t2
dx
b v=
dt
= (2t − 3) 3
PR
O
the formula for x and determine c.
When t = 0, x = 0:
0=0+c
c=0
E
2 Substitute the initial condition into
FS
x = 14t4 − 12t2 + c
TE
D
=
2 Substitute the initial condition into
R
3 State the rule.
R
EC
the formula for x and determine c.
O
c 1 Write the velocity equation
U
N
C
and antidifferentiate to find the
displacement function, x.
(2t − 3) 4
+c
2(4)
= 18 (2t − 3) 4 + c
When t = 0, x = 0:
0 = 18 (−3) 4 + c
0 = 81
+c
8
c = −81
8
x = 18 (2t − 3) 4 − 81
8
dx
c v=
dt
1
=
(t − 1) 2
= (t − 1) −2
x = 3 (t − 1) −2dt
(t − 1) −1
+c
−1
= −(t − 1) −1 + c
1
=−
+c
(t − 1)
=
Topic 7 ANTIDIffErENTIATION
c07Antidifferentiation.indd 317
317
21/08/15 5:41 PM
2 Substitute the initial condition into
the formula for x and determine c.
3 State the rule.
14
The velocity of a particle moving in a straight line along the x-axis is given by
dx
v=
= 9 − 9e−3t
dt
FS
WOrKED
EXAMPLE
When t = 0, x = 0:
1
0=−
+c
(−1)
0=1+c
c = −1
1
x=−
−1
(t − 1)
PR
O
a Show that the particle is initially at rest.
O
where t is the time in seconds and x is the displacement in metres.
b Find the equation relating x to t if it is known that initially the particle
was 3 metres to the left of the origin.
WritE
a 1 Substitute t = 0 and evaluate.
v = 9 − 9e−3t
t = 0 ⇒ v = 9 − 9e0
=9−9×1
= 0 m/s
Initially the particle is at rest as its
velocity is 0 m/s.
dx
b v=
dt
= 9 − 9e−3t
PA
G
a
TE
D
2 Answer the question.
b 1 Write the velocity equation and
E
tHinK
R
EC
antidifferentiate to find the position
equation, x.
R
2 Substitute the initial condition to
C
O
determine c. Remember, left of the origin
means the position is negative.
U
N
3 State the equation.
x = 3 (9 − 9e−3t)dt
= 9t + 3e−3t + c
When t = 0, x = −3:
−3 = 9 × 0 + 3e0 + c
−3 = 3 + c
c = −6
x = 9t + 3e−3t − 6
ExErcisE 7.4 Families of curves and applications
PractisE
Work without cas
318
1
WE9
a Sketch a family of curves which have the derivative function f ′(x) = 3x2.
b Find the rule for the function that belongs to this family of curves and passes
through the point (2, 16).
2 a Sketch a family of curves related to the derivative function f ′(x) = −2 cos(2x).
b Find the rule for the function that belongs to this family of curves and passes
π
through the point a , 4b.
2
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
c07Antidifferentiation.indd 318
21/08/15 5:41 PM
Find the equation of the curve that passes through the point (0, 3) if the
dy
gradient is given by
= 2e2x + e−x.
dx
4 The gradient function of a particular curve is given by f ′(x) = cos(2x) − sin(2x)
Find the rule for this function if it is known that the curve passes through the
point (π, 2).
3
WE10
5
A chemical factory has permission from the Environment Protection
Authority to release particular toxic gases into the atmosphere for a period of
20 seconds no more than once every 3 hours. This maintains safe levels of the
gases in the atmosphere. This rate of emission is given by
dV
= 20t2 − t3 cm3/s
dt
where 0 ≤ t ≤ 20 and V cm3 is the total volume of toxic gases released over
t seconds. Find the total volume of toxic gases released during a 20-second
release period.
PR
O
O
FS
WE11
6 The rate of change of volume of a balloon as it is being blown up can be
modelled by
TE
D
PA
G
E
dV
= πr2
dr
where V cm3 is the volume of the balloon and r cm is the radius of the balloon.
a Find the rule for the volume of the balloon.
b What is the volume of the balloon when its radius is 4 cm?
y
7 WE12 The graph of the gradient function y = f ′(x) is
shown. Analyse this derivative function and sketch the
given function y = f(x).
y = fʹ(x)
x
R
EC
0
U
N
C
O
R
8 The graph of the gradient function y = f ′(x)
9
y
is shown. Analyse this derivative function and
sketch the given function y = f(x).
In each of the following cases, find the
displacement as a function of t if initially the
particle is at the origin.
WE13
a v = (3t + 1)
b v =
1
2
x
0
y = fʹ(x)
1
(t + 2) 2
c v = (2t + 1) 3
Topic 7 Antidifferentiation c07Antidifferentiation.indd 319
319
21/08/15 5:41 PM
10 Find the displacement of a particle that starts from the origin and has a velocity
E
Apply the most
appropriate
mathematical
processes and tools
15 Find the equation of the curve defined by
the point (0, 5).
1x
dy
= e2 , given that it passes through
dx
PA
G
Consolidate
PR
O
O
FS
defined by:
a v = e(3t−1)
b v = −sin(2t + 3).
11 WE14 A particle is oscillating so that its velocity, v cm/s, can be defined by
dx
v=
= sin(2t) + cos(2t)
dt
where t is the time in seconds and x centimetres is its displacement.
a Show that initially the particle is moving at 1 cm/s.
b Find the equation relating x to t if it is known that initially the particle was at
the origin.
12 When a bus travels along a straight road in heavy traffic from one stop to another
stop, the velocity at time t seconds is given by v = 0.25t(50 − t), where v is the
velocity in m/s.
a Find the greatest velocity reached by the bus.
b Find the rule for the position of the bus, x metres, in terms of t.
13 a Sketch a family of curves with the derivative function f ′(x) = 3e−3x.
b Find the rule for the function that belongs to this family of curves and passes
through the point (0, 1).
14 Find the antiderivative of cos(2x) + 3e−3x if y = 4 when x = 0.
16 Find f(x) for each of the following.
x
2
b f ′(x) = sina b and f(π) = 3
TE
D
a f ′(x) = 5 − 2x and f(1) = 4
1
and f(0) = 4
(1 − x) 2
17 The graphs of two gradient functions are shown. Sketch the corresponding
antiderivative graphs.
EC
c f ′(x) =
a
b
R
y
y = fʹ(x)
R
y = fʹ(x)
0
O
C
U
N
y
0
x
x
18 A particle moves in a straight line so that its velocity, in metres per second, can
be defined by the rule v = 3t2 + 7t, t ≥ 0. Find the rule relating the position of the
particle, x metres, to t, if it is known that the particle started from the origin.
19 A particle attached to a spring moves up and down in a straight line so that at
time t seconds its velocity, v metres per second, is given by
πt
v = 3π sina b, t ≥ 0.
8
Initially the particle is stationary.
320 MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
c07Antidifferentiation.indd 320
21/08/15 5:42 PM
a Find the rule relating the position of the particle, x centimetres, to t.
b What is the maximum displacement of the particle?
c Where is the particle, relative to the stationary position, after 4 seconds?
20 A particle starting from rest at the origin moves in a straight line with a velocity
of
12
+ 6 metres per second after t seconds.
(t − 1) 2
a Find the rule relating the position of the particle, x metres, to t.
b Find the position of the particle after 3 seconds.
21 A population of sea lions on a
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
distant island is growing according
to the model
dP
= 30e0.3t, 0 ≤ t ≤ 10
dt
where P is the number of sea lions
present after t years.
a If initially there were 50 sea lions
on the island, find the rule for the
number of sea lions present, P,
after t years.
b Find the number of sea lions on the island after 10 years. Give your answer
correct to the nearest whole sea lion.
22 The rate of change of the depth of water in a canal is modelled by the rule
πt
dh π
= cosa b
dt 2
4
where h is the height of the water in metres and t is the number of hours
since 6 am.
a Find an expression for h in terms of t if the water is 3 metres deep at 6 am.
b What are the maximum and minimum depths of the water?
c For how many hours a day is the water level 4 metres or more?
23 A newly established suburban
Master
area of Perth is growing at a rate
modelled by the rule
dN
= 400 + 1000 !t, 0 ≤ t ≤ 10
dt
where N is the number of families
living in the suburb t years after the
suburb was established in 2015.
a Find a rule relating N and t if
initially there were 40 families
living in this suburb.
b How many families will be living in the suburb 5 years after its establishment?
Give your answer correct to the nearest number of families.
24 If v = 2t cos(t) metres per second, find a rule relating the position x metres to t if
it is known the particle starts from rest at the origin.
Topic 7 Antidifferentiation c07Antidifferentiation.indd 321
321
21/08/15 5:42 PM
ONLINE ONLY
7.5 Review
www.jacplus.com.au
the Maths Quest review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
• Extended-response questions — providing you with
the opportunity to practise exam-style questions.
a summary of the key points covered in this topic is
also available as a digital document.
the review contains:
• short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions without the
use of CAS technology
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
FS
REVIEW QUESTIONS
Activities
to access eBookPlUs activities, log on to
TE
D
www.jacplus.com.au
PA
G
ONLINE ONLY
E
PR
O
O
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
Interactivities
Units 3 & 4
Antidifferentiation
Sit topic test
U
N
C
O
R
R
EC
A comprehensive set of relevant interactivities
to bring difficult mathematical concepts to life
can be found in the Resources section of your
eBookPLUS.
studyON is an interactive and highly visual online
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and weaknesses prior to your exams. You can
then confidently target areas of greatest need,
enabling you to achieve your best results.
322
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
c07Antidifferentiation.indd 322
21/08/15 5:42 PM
7 Answers
EXERCISE 7.2
+
3
2
1
b 6!x − x4 −
5x2
+c
x4 x3 15x2
c
− −
+c
2
3
2
3
1
2a 4!x − +
+c
x 4x2
(3x − 5) 6
3a
+c
18
(2x + 3) 5
4a
7
1 4
x
2
1 3
x
3
b
−
1
b −
3
2
+c
1
2"x
+
1
3 2
2"
x
R
1
1
= q7x + "x −
r
4
"x
1
1
9 f(x) = x3 + + c
x
3
1
r
!x
1
"x
3
r dx
+c
c
x4
−
7 3
x
3
+
x2
−x+c
7
1
b x3 +
+c
3
5x2
d
16 2
x !x
5
+c
1
11a
−
+
− x + c b −
+c
8x2
7
7
15 5
2
1
2
c x2 −
x + c d x2 + + c
x
7
7
2
27 4
x
4
5
9x3
9 2
x
2
dx
3
x
=
"x2
5x
2
dy
dx
+1
= 5"x2 + 1 + c
3
3
2(4 − x) 2
4x2 1 2
e − x + c f −
+c
3
2
5
1
1
5
b x2 + x −
12a 2x3 + 2 x2 − 6x
x
2
=
15x2 + 8x
2"5x3 + 4x2
15x2 + 8x
"5x3
2
193"x3
+
4x2
dx = 2"5x3 + 4x2
+1+c
3
4
20405 (3x + 5) 2 (135x2 − 72x + 35) + c
EXERCISE 7.3
1a
O
C
1 4
x
4
U
N
10a
dy
3
PA
G
r q7x + "x −
4
3
+c
x
2
TE
D
3 2
2"
x
r q7x + !x −
EC
3 q7 +
2"x
1
R
dx
+
1
2
14 y = x2 + 3x +
16
2
3(2x − 3) 2
13 y = 4 x4 − 2x!x + c
18
3
1
f −
2
3
2
4
316(5x + 1)(5x + 2x − 1) dx = 2(5x + 2x − 1)
1
2
2
2
3
3 (3x + 1)(3x + 2x − 4) = 6 (3x + 2x − 4)
= 4 q7 +
1
4(1 − 4x) 2
"x + 1
dy
17 = 8(5x + 1)(5x2 + 2x − 1) 3
dx
x2
x3 4x
−
+ +c
3
5
2
2 2
2
b x !x + 2x!x + 6!x −
5
!x
dy
7
= 6(3x + 1)(3x2 + 2x − 4) 2
dx
dy
e
15 y = 3 x!x + 2!x + c
3(2x − 3)
1
b
+c
8(1 − 2x) 4
5
2
6a
8
+ 4x + c
1
+c
2x2
+ 3x4 + 6x −
d
1
1 2
x
2
+
1 7
4
x − 4x −
7
5x5
4 2
x − 8x2
3
E
5
3
2
3x
x
d
− +c
7
3
+c
10
8x7
7
2
1
c
FS
−
4
x4
O
1a
3
4x3
PR
O
3x2
b
x
1
sin(3x + 4) + 8 cosa b + c
6
2
2x
3
1
sina b − cos(5 − 2x)
2
3
8
x
2
x
2
2a −2 cosa b − 6 sina b + c
7
1
b f(x) = 2 sin(2x) + 3 cos(3x) + c
1
x5 1 −4x
1
4 − x
+ e +c
b 4e2x + 3 e 2 + c
5 4
x
x
x2
4a 3e2 − 3 cosa b +
+ c b 14 sin(4x) − e−3x + c
3
6
3a
1
3
1
5 6e6x − 3ex − 4 e−4x + 9 e−9x + c
6 ex + 4e
−x
2
− 12e −2x + c
Topic 7 Antidifferentiation c07Antidifferentiation.indd 323
323
21/08/15 5:42 PM
7
dy
dx
2a f(x) = −sin(2x) + c
= −2sin(x) cos(x)ecos (x)
2
y
cos (x) = − ecos (x)
3sin(x) cos(x)e
2
1
2
8
dy
dx
= 3(x + 1) 2e(x+1)
39(x +
3
1) 2e(x+1) dx
2
2
y = –sin(2x) + 2
3
=
3
3e(x+1)
1
4
0
+c
1
9a 3e3x + 2 cos(2x) + c
c
y = –sin(2x) + 4
b
sin(2x + 5) + e−x + c
d
1 x
e − 14e−6x + c
2
1 2x
e − 23e3x + 14e4x
2
π
π
–
2
y = –sin(2x) – 5
+c
y = –sin(2x) – 7
3πx
1
2
−
cosa
b
4x 3π
2
e−3x
1
12 y = sin(2x) +
+c
2
3
πt
13a H(t) = t − 5π cosa b
45
b H = 7.146 kilojoules
dy
14 = 6xe3x + 2e3x
dx
11−
1
π
3
6 a V = r3 b dx
= (4x + 3)e2x
2x
32(4x + 3)e
16
dy
dx
dx = 2e2x
2 +3x−1
= −x sin(x) + cos(x)
dy
dt
=
π
πt
cosa b
20
4
y=
EC
3x sin(x)dx = sin(x) − x cos(x)
17
πt
1
sina b + c
5
4
π 3π
2 2
20ex − x + loge (ex + 1) + c
Negative
gradient
Local y
maximum
at
x = –2
8
R
y = f(x)
x
0
Local
minimum
at
x=4
y
2
x3 + 3
y=
x3 – 4
x
0
y = x3 + 1
y = x3
y = x3 – 2
b f(x) = x3 + 8
324 x
0
Positive
gradient
O
U
N
1a f(x) = x3 + c
y=
y = f(x)
C
EXERCISE 7.4
Local maximum
at x = 1
R
18a = −2, b = −4, m = 2, n = −2
19 x = ,
64π
cm3 3
E
2 +3x−1
2 +3x−1
y
7
3
2
PR
O
1
5 V = 13 333 3 cm3
PA
G
dy
1
4 2 sin(2x) + 2 cos(2x) +
O
b f(x) = 4 − sin(2x)
3 e2x − e−x + 2
TE
D
15
FS
10a = −9, b = 3
1 3x
1 3x
3x
3xe dx = 3xe − 9e
y = –sin(2x)
x
2π
y = –sin(2x) – 2
3π
–
2
9a x = 9"(3t + 1) 3 −
b x =
2
9
1
1
−
2 t+2
1
c x = 8 (2t + 1) 4 −
1
8
1
3
1
3e
1
1
b x = 2 cos(2t + 3) − 2 cos (3)
10a x = e3t−1 −
MATHS QUEST 12 MATHEMATICAL METHODS VCE Units 3 and 4
c07Antidifferentiation.indd 324
21/08/15 5:42 PM
11a v = sin(2t) + cos(2t)
t = 0 ⇒ v = sin(0) + cos(0)
=0+1
= 1 cm/s
1
1
b x = −2 cos(2t) + 2 sin(2t) +
b
y
1
2
12a 156.25 m/s
1 3
t
12
13a f(x) = −e−3x + c
b x = 6.25t2 −
y
0
–4
y = –e –3x + 2
y = f(x)
x
O
0
FS
y = –e –3x + 1
y = –e –3x
x
2
y = –e –3x – 4
7
PR
O
y = –e –3x – 2
18 x = t3 + 2t2
πt
b
8
b Maximum displacement = 48 metres
c After 4 seconds the particle is 24 metres above the
stationary position.
E
19a x = 24 − 24 cosa
PA
G
b f(x) = 2 − e−3x
1
14 y = 2 sin(2x) − e−3x + 5
15 y = 2e + 3
TE
D
16a f(x) = 5x − x2
x
2
y
πt
b+3
4
b Minimum depth is 1 metre and maximum depth is
5 metres.
c 8 hours a day
23a N = 400t +
2000
"t3
3
+ 40
b 9494 families
R
0
22a h = 2 sina
24 x = 2t sin(t) + 2 cos(t) − 2
x
U
N
C
O
y = f(x)
Local
maximum
at
x=2
R
17a
EC
b f(x) = 3 − 2 cosa b
1
c f(x) =
+3
1−x
12
− 12
(t − 1)
b At the origin
21a P = 100e0.3t − 50
b 1959 seals
20a x = 6t −
x
2
Local
minimum
at
x = –4
Topic 7 Antidifferentiation c07Antidifferentiation.indd 325
325
21/08/15 5:53 PM
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