Antidifferentiation is the process of determining a function that has a known derivative. This

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6

Antidifferentiation and

integration

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MathsWorld Mathematical Methods Units 3 & 4

Derivatives and

antiderivatives

Antidifferentiation is the process of determining a function that has a known derivative. This requires using ‘differentiation in reverse’, that is, antidifferentiation.

From chapter 4, we know that the derivative of x2 with respect to x is 2x. So what function has a derivative of 2x? One such function is f(x) =x2. The derivative of x2 is 2x and so an antiderivative of 2x is x2.

The notation for this is 2x dx= x2 and is read as ‘the antiderivative of 2x with respect to

x is x2’. This notation was used by Gottfried Leibniz (1646–1716), who was one of the cofounders of the study of calculus. The notation has become standard because of the link between antidifferentiation and integral calculus. The symbol is written like a stretched upper case S and derives from the abbreviation for the Latin word summa, which means sum.

The dx is written next to the function to be antidifferentiated so as to specify the

independent variable x with respect to which the original differentiation was made and with respect to the function to which we are to antidifferentiate. If the independent variable is t, then t dt, for example, indicates that we need to antidifferentiate with respect to t.

The figure opposite illustrates how differentiation undoes antidifferentiation.

From the example above, do any other functions have a derivative of 2x? The derivative of x2− 5 is 2x and so 2x dx=x2− 5. Thus, x2 − 5 is another antiderivative of 2x.

The derivative of x2+ 2 is also 2x and so

2x dx=x2+ 2. Thus, x2+ 2 is another antiderivative of 2x.

This demonstrates that a function has many antiderivatives, whereas a function can have only one derivative.

In the example given, there is more than one function that has a derivative of 2x: 2x dx=x2+c, where c is an arbitrary real number.

We interpret this to mean: ‘the general antiderivative of 2x with respect to x is x2 +c’. Because of its connection with integral calculus, it is also interpreted as: ‘the indefinite integral of 2x with respect to x is x2+c’. The term ‘indefinite’ indicates that in the absence of other information, there is no ‘definite’ or unique answer as c may be any real number. The number c is also known as the constant of integration.

Consider the graphs of y=x2+ c for c= −5, 0 and 2.

Taking the graph of y =x2 as the basic graph, the graph of y= x2+ c, for any real value of

c, corresponds to a vertical translation c units parallel to the vertical axis. There is one curve from this family of curves for each value of c. Each member of this family of curves has the same gradient at a particular x-coordinate.

⌠ ⌡

Antidifferentiation Differentiation

f(x)

f′(x) ⌠

⌡

⌠ ⌡

6.1

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The figure opposite shows that at x= 2, each member of this family of curves graphed has a gradient of 4. In fact, the gradient at any point on the family of curves y=x2+c

is 2x.

This figure provides a graphical demonstration of why 2x dx=x2+c, where c is a real number.

Basic integration rules

The constant rule:

As , where m is a real constant, it follows that:

m dx= mx+c

One rule which is very useful in determining other antiderivatives is the constant multiple rule:

k f(x)dx= k f(x)dx, where k is a real constant. The power rule can be obtained using this property.

Since , it follows that

(n+ 1)xndx= xn + 1+c

(n+ 1) xndx=xn+ 1+c xndx= +c, n≠−1

Note that the last line could have been written as but, as also

represents a constant, we elect to leave this constant as c.

Also note the restriction n≠−1. If n= −1, the integral would take the form , which cannot be determined using the power rule. , which is explored in greater

depth later in this chapter. Note that putting n=−1 in the rule results in a zero denominator; we cannot divide by zero.

Another very useful rule is the sum rule:

(f(x) ± g(x))dx= f(x)dx ± g(x)dx

y

x

–4 –3 –2 –1 1 2 3 4 2

4 6 8

–4 –2

–8 –6 0

=4

dy — dx

=4

dy — dx

=4

dy — dx

⌠ ⌡

d dx

---(mx) = m

⌠ ⌡

⌠

⌡ ⌠⌡

d dx

---(xn+1) = (n+1)xn

⌠ ⌡

⌠ ⌡ ⌠

⌡ x

n+1 n+1

---⌠

⌡xndx x

n+1 n+1 --- c

n+1 ---+

= c

n+1 ---⌠

⎮ ⌡ 1

x

---dx

⌠ ⎮ ⌡ 1

x

---dx = log_ex+c

⌠

⌡ ⌠⌡ ⌠⌡

Summary of basic integration rules

Constant multiple rule: k f(x)dx=k f(x)dx

Power rule:

Sum rule: (f(x) ± g(x))dx = f(x)dx ± g(x)dx

⌠

⌡ ⌠⌡

⌠

⌡xndx x

n+1

---+c n, ≠–1

= ⌠

⌡ ⌠⌡ ⌠⌡

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E x a m p l e

1

Find:

a 4dx b x dx c d

S o l u t i o n

a 4dx= 4x+c, using the constant rule to antidifferentiate.

b x dx= , using the power rule to antidifferentiate.

c Write in the form xndx.

= x−2dx

Then use the power rule to antidifferentiate.

x−2dx= =

d Write in the form xndx.

= x1/2dx

Then use the power rule to antidifferentiate.

x1/2dx=

=

E x a m p l e

2

Find:

a (3x− 5)dx b 4x3dx c (2x− 1)2dx d

S o l u t i o n

a The sum rule suggests that we can antidifferentiate each term separately. (3x− 5)dx= 3x dx− 5dx

=

The power rule, the constant multiple rule and the constant rule have been used to antidifferentiate.

In this example 3x dx− 5dx was included in the solution to highlight that each term can be antidifferentiated separately. In practice, the solution is more simply written as

(3x− 5)dx=

⌠

⌡ ⌠⌡ ⌠⎮

⌡

1

x2

---dx ⌠_⌡ x dx

⌠ ⌡ ⌠

⌡ x

2

---+c

⌠ ⎮ ⌡

1

x2

---dx ⌠_⌡

⌠ ⎮ ⌡

1

x2

---dx ⌠_⌡

⌠

⌡ x

1

– 1

– ( ) ---+c

1

x

---– +c

⌠

⌡ x dx ⌠⌡

⌠

⌡ x dx ⌠⌡

⌠

⌡ 1₃

2

---⎝ ⎠ ⎛ ⎞

---x3 2⁄ +c

2 3

---x3 2⁄ +c

⌠

⌡ ⌠⌡ ⌠⌡

⌠ ⎮ ⌡

3x2+1

x2

---dx

⌠

⌡ ⌠⌡ ⌠⌡

3x2

2

---–5x+c

⌠

⌡ ⌠⌡

⌠

⌡ 3x

2

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b Using the constant multiple rule, 4x dx= 4 x dx. Now using the power rule,

4 x3dx =

=x4+c

It is not usual to put in the first step. In this example 4x3dx= 4 x3dx was included in the solution to highlight that x3 is antidifferentiated and the result multiplied by the constant 4. The solution is more simply written as

4x3dx=

=

c Expand first and then antidifferentiate each term. (2x− 1)2dx = (4x2− 4x+ 1)dx

=

d Simplify first and then antidifferentiate each term.

=

= (3 +x−2)dx

= 3x−x−1+c

=

E x a m p l e

3

Find an antiderivative of .

S o l u t i o n

= (x3− 3x−3)dx

=

(Because the question states that only an antiderivative is required, it is not necessary to include the constant of integration. The inclusion of c suggests many antiderivatives rather than a single antiderivative. However, it is not incorrect to include the constant.)

⌠

⌡ ⌠⌡

⌠

⌡ 4x

4

---+c

⌠

⌡ ⌠⌡

⌠

⌡ 4x

4

---+c x4+c

⌠

⌡ ⌠⌡

4x3

3

--- 4x 2

2

---– +x+c

4x3

3

---–2x2+x+c

t i p

The correctness of an antiderivative can be checked by differentiating. The result of the differentiation should be the function you started with.

⌠ ⎮ ⌡

3x2+1

x2

---dx ⌠⎮ ⌡

3x2 x2

--- 1

x2

---dx

+ ⌠ ⌡

3x 1

x

---+c

–

x3 3

x3 ---–

⌠ ⎮ ⌡ x

3 3

x3

---–

⎝ ⎠

⎛ ⎞_dx ⌠

⌡

x4

4

--- 3x 2

– 2

– ( ) ---–

x4

4

--- 3

2x2

---+

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exercise

6.1

1 Find:

a x4dx b 5x3dx c (x2+ 5x− 2)dx

d e (4 − 3x)2dx f (2x+ 1)3dx

g (x− 3)(2x+ 5)dx h i

j k (x3/2− x2/3)dx l (3x

+

1)4dx

2 Find an antiderivative of each of the following.

a b c

d e f

The antiderivative of (

ax

+

b

)

n

The chain rule can be used to show that .

It follows that the reverse is true.

a(n+ 1)(ax+ b)ndx= a(n+ 1) (ax+ b)ndx=

(ax+ b)ndx=

E x a m p l e

4

Find:

a (2x− 5)3dx b c d

S o l u t i o n

a (2x− 5)3dx is in the form (ax+b)ndx, with a= 2, b= −5 and n= 3. Using the rule (ax+b)ndx=

(2x− 5)3dx=

=

b Write in the form (ax+b)ndx.

= (1 − 3x)1/2dx

⌠

⌡ ⌠⌡ ⌠⌡

⌠

⌡2 x dx ⌠⌡ ⌠⌡

⌠

⌡ ⌠⎮

⌡ 1 2x2

---dx ⌠_⎮

⌡

x3–5x2 x5

---⎝ ⎠

⎛ ⎞_dx

⌠ ⎮

⌡ 9 x

4 x

---–

⎝ ⎠

⎛ ⎞_dx ⌠

⌡ ⌠⌡

2x3–6x2–3 1

x4

--- 2

x3

---– 3x x 4

x

---+4x 3x x

2

–

x4

--- (3–2x)2

d dx

---((ax+b)n+1) = a n( +1)(ax+b)n

⌠

⌡ (ax+b)n+1+c

⌠

⌡ (ax+b)n+1+c ⌠

⌡ _{a n}---₍ 1₊₁₎(ax+b)n+1+c n( ≠–1)

⌠

⌡ ⌠⌡ 1 3– xdx ⌠⎮

⌡

1 4x–1

( )3

---dx ⌠_⎮

⌡

24 9+16t

---dt

⌠

⌡ ⌠⌡

⌠

⌡ _{a n}---₍ 1₊₁₎(ax+b)n+1+c ⌠

⌡ _{( )}---₂1_{( )}₄ (2x–5)4+c

1 8

---(2x–5)4+c

⌠

⌡ 1 3– xdx ⌠⌡

⌠

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Now a=−3, b= 1 and .

(1 − 3x)1/2dx=

=

c Write in the form (ax+b)ndx

= (4x− 1)−3dx

Now a= 4, b=−1 and n=−3. (4x− 1)−3dx =

= =

d Note that t is the variable in , hence dt is used rather than dx.

can be written as 24 (9 + 16t)−1/2dt

In this case a= 16, b= 9 and .

24 (9 + 16t)−1/2dt=

= =

The antiderivative of

Recall that . It follows that .

The chain rule can be used to show that . It follows that, provided ax+b> 0:

=

a =

=

n 1

2

---= ⌠

⌡ 1

3

– ( )_{⎝ ⎠}⎛ ⎞3₂

---(1 3– x)3 2⁄ +c

2 9

---– (1 3– x)3 2⁄ +c

⌠ ⎮ ⌡

1 4x–1

( )3

---dx ⌠_⌡

⌠ ⎮ ⌡

1 4x–1

( )3

---dx ⌠_⌡

⌠

⌡ _{( )}---₄ 1_{( )}_–₂ (4x–1)–2+c

1 8

---– (4x–1)–2+c

1 8 4( x–1)2

---– +c

⌠ ⎮ ⌡

24 9+16t

---dt

⌠ ⎮ ⌡

24 9+16t

---dt ⌠_⌡ n 1

2

---– =

⌠

⌡ 24 1

16

( )_{⎝ ⎠}⎛ ⎞1₂

---(9+16t)1 2⁄

⎝ ⎠

⎜ ⎟

⎛ ⎞

c

+

3 9( +16t)1 2/ ₊_c 3 9+16t +c

1

ax

+

b

---d dx

---(logex)

1

x

---,x>0

= ⌠⎮

⌡ 1

x

---dx = logex+c x, >0

d dx

---(loge(ax+b))

a ax+b

---,ax+b>0 =

⌠ ⎮ ⌡

a ax+b

---dx loge(ax+b)+c

⌠ ⎮ ⌡

1

ax+b

---dx loge(ax+b)+c

⌠ ⎮ ⌡

1

ax+b

---dx 1 a

---loge(ax+b)+c

(8)

E x a m p l e

5

Find:

a b c d

S o l u t i o n

a ₌

=

c is in the form ,

with a= 2, b=−1.

E x a m p l e

6

Express in the form . Hence, find .

S o l u t i o n

Long division can be used, but it is simpler to re-express the numerator.

=

CAS 10.7

Warning

CAS notation can be edgy!

The antiderivative produces an unexpected

result on the TI-89. Apart from the equivalent form issue, the absolute value function appears in the result. Note that if ax+b> 0, then and the result agrees with that quoted earlier.

⌠ ⎮ ⌡

1

ax+b

---dx

lnax+b = ln(ax+b)

⌠ ⎮ ⌡

3 2x

---dx ⌠⎮

⌡

x3–2x x2

---dx ⌠⎮ ⌡

1 2x–1

---dx ⌠⎮ ⌡

3 1 5– x

---dx

b =

=

d = 3

In this case a=−5, b= 1. 3

⌠ ⎮ ⌡

x3–2x x2

---dx ⌠⎮ ⌡

x3 x2

--- 2x

x2

---–

⎝ ⎠

⎛ ⎞ _dx

⌠ ⎮ ⌡ x 2 x ---– ⎝ ⎠

⎛ ⎞_dx

⌠

⌡x dx–2⌠⎮_⌡1---_xdx

x2

2

---–2 logex+c x, >0

⌠ ⎮ ⌡

3 1 5– x

---dx ⌠⎮ ⌡

1 1 5– x

---dx

⌠ ⎮ ⌡

1 1 5– x

---dx 3

5

---– loge(1 5– x) c x 1

5 ---< , + = ⌠ ⎮ ⌡ 3 2x

---dx 3

2 ---⌠⎮ ⌡ 1 x ---dx 3 2

---logex+c x, >0

⌠ ⎮ ⌡

1 2x–1

---dx ⌠⎮ ⌡

1

ax+b

---dx

⌠ ⎮ ⌡

1 2x–1

---dx 1

2

---loge(2x–1)+c x, >1₂

---=

2x–3

x+2

--- a b x+2

---+ ⌠⎮

⌡

2x–3

x+2

---dx

2x–3

x+2

--- 2x+4 4– –3

x+2

---2x+4

x+2

--- –7

x+2

---+

2 7

x+2

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=

= − 7

=

In examples 5 and 6 above, it is important to include the restriction on the values that x may take. Remember that we can only take the logarithm of a positive number. In the following exercises, make sure that you include similar restrictions where necessary.

exercise

6.1

3 Find:

a (3x− 2)2dx b (1 − 4x)2dx c (2x− 1)5dx

d (x+ 2)10dx e f

g h i

j k ((2x+ 1)2− (1 − 3x)4)dx l

4 Find:

a b c

d e f

g h 4x−1dx i 3(x− 4)−1dx

j k l

⌠ ⎮ ⌡

2x–3

x+2

---dx ⌠⎮ ⌡ 2

7

x+2

---–

⎝ ⎠

⎛ ⎞ _dx

⌠

⌡2dx ⌠⎮_⌡_x---₊1₂dx

2x–7 loge(x+2)+c x, >–2

CAS 10.4

t i p

The propFrac command on the TI-89 will separate to

a sum of two terms. This is similar to converting the improper

fraction to .

2x–3

x+2

---3 2

--- 1 1

2

---+

continued

⌠

⌡ ⌠⌡ ⌠⌡

⌠

⌡ ⌠⌡ 2x–5dx ⌠⌡ 4–3xdx

⌠ ⎮ ⌡

1 x–1

( )4

---dx ⌠_⎮

⌡ 2 3x+2

( )3

---dx ⌠_⎮

⌡ 1 1–t

---dt

⌠ ⎮ ⌡

3 2p+1

---dp ⌠_⌡ ⌠_⎮

⌡ 1 3q–4 ( )3 2⁄

---dq

⌠ ⎮ ⌡

3 x

---dx ⌠⎮ ⌡

1 2x

---dx ⌠⎮ ⌡

5 4x

---dx ⌠

⎮ ⌡

1 x–2

---dx ⌠⎮ ⌡

2 x+5

---dx ⌠⎮ ⌡

1 2x–3

---dx ⌠

⎮ ⌡

3 4–2x

---dx ⌠_⌡ ⌠_⌡ ⌠

⎮ ⌡

2

–

4–3x

---dx ⌠⎮ ⌡

2

–

2x+5

---dx ⌠⎮ ⌡

1 2x+4

---dx

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5 a Express in the form , where a and b are integers. Hence find

.

b Express in the form , where a and b are integers. Hence find

.

c Express in the form , where a and b are integers. Hence find

.

a b c

d e f

g h i

The antiderivative of

e

kx

It follows that:

kekxdx= k ekxdx= ekxdx=

E x a m p l e

7

Find:

a e3xdx b 0.02e−0.1tdt c (ex+e−x)2dx d

S o l u t i o n

a Applying the rule with k= 3,

e3xdx = .

c Expand first.

(ex+e−x)2dx = (e2x+ 2 +e−2x)dx

Now antidifferentiate. x+2 x–3

--- a b x–3

---+

⌠ ⎮ ⌡

x+2 x–3

---dx 2x+1

x+2

--- a b x+2

---+

⌠ ⎮ ⌡

2x+1 x+2

---dx 3x–5

x+4

--- a b x+4

---+

⌠ ⎮ ⌡

3x–5 x+4

---dx 1 x–5

( )2

--- 3

x+1

--- 1

5x

---3x+2

( )–1 ₄_–₃_x _x 4

3–2x

---+

3x–2 x

--- 2x2–1

x3

--- 3 4( x–3)5

d dx

---(ekx) ₌ _kekx₊_c

⌠

⌡ ekx+c

⌠

⌡ ekx+c

⌠

⌡ 1_k---ekx+c

⌠

⌡ ⌠⌡ ⌠⌡

⌠ ⎮ ⌡

3e2x–5

ex

---dx

b = 0.02 e−0.1tdt

= = d Divide first.

=

Now antidifferentiate.

⌠

⌡0.02e–0.1tdt ⌠⌡

0.02 0.1

–

( )

---e–0.1t+c

0.2e–0.1t

– +c

⌠ ⎮ ⌡

3e2x–5

ex

---dx ⌠⎮ ⌡

3e2x ex

--- 5

ex

---–

⎝ ⎠

⎛ ⎞ _dx

⌠

⌡(3ex–5e–x)dx ⌠

⌡(3ex–5e–x)dx = 3ex+5e–x+c ⌠

⌡ 1₃---e3x+c

⌠

⌡ ⌠⌡

⌠

(11)

Antiderivatives of trigonometric functions

Therefore:

=

k coskx dx=

= Similarly:

Therefore:

= −k sinkx dx=

sinkx dx=

E x a m p l e

8

Find:

a b c

S o l u t i o n

a b

c In this case .

=

exercise

6.1

7 Find:

a b c

d e f

g h i

j k l

d dx

---(sinkx) = kcoskx

⌠

⌡kcoskx xd sinkx+c ⌠

⌡ sinkx+c

⌠

⌡coskx xd 1_k---sinkx+c

Warning

Signs!

The signs of the antiderivatives of sinx and cosx are the opposite of the signs of their derivatives.

= cosx

=−cosx+c =−sinx

= sinx+ c

d dx

---(sinx)

⌠

⌡sinx xd

d dx

---(cosx)

⌠

⌡cosx xd

d dx

---(cos( )kx ) = –ksinkx

⌠

⌡–ksinkx xd coskx+c ⌠

⌡ coskx+c

⌠

⌡ –1_k---coskx+c

⌠

⌡sin 3x dx ⌠⌡(3 cos 4x–5 sin 2x)dx

⌠ ⎮ ⌡sin

πt

12

---dt

⌠ ⎮

⌡sin 3x dx

1 3

---– cos 3x+c

= ⌠

⎮

⌡(3 cos 4x–5 sin 2x)dx

3 4

---sin 4x 5

2

---cos 2x c

+ +

=

k π

12

---= ⌠

⎮ ⌡sin

πt

12

---dt 1

π

12

---⎝ ⎠ ⎛ ⎞ ---– cosπt

12

---+c

12

π ---– cosπt

12

---+c

continued

⌠

⌡e4xdx ⌠_⌡2e–xdx ⌠⎮ ⌡

1 2

---e3xdx ⌠

⌡ex⁄5dx ⌠_⌡4e–0.01xdx ⌠_⌡(ex_–₂_e–x)2_d_x

⌠ ⎮ ⌡

ex–e–x 2

---dx ⌠_⌡500e–0.2tdt ⌠_⌡(ex_–_e–x)5_d_x

⌠ ⎮ ⌡ 2 3

---ex⁄3dx ⌠_⌡(1+ex)2dx ⌠_⌡(e2x–x+1)dx

(12)

8 Antidifferentiate:

a sin 4x b cos (−2x) c

d e 3 cos 6x f

g h cosπx i

j 3 sin 2x− 4 cos 3x k l 3 cos 2x+ 2 cos 3x

9 Find:

a b

c d

e f

a b c

d e f

g h i

11 Let g(x) =ef(x). If , find a rule for f.

12 Let g(x) = cos (f(x)). If , find a rule for f. sinx

3

---cosx 5

--- 2 sinx

4

---2 cos3x 2

--- 4 sinπx 3

---2 sinπx 2

--- 2 cosπx 2

---+

⌠

⌡(e3x+sin 3x)dx _⌡⌠(2x–5+cos 2x)dx

⌠ ⎮ ⌡

1 x

---–ex⁄2

⎝ ⎠

⎛ ⎞dx ⌠⎮

⌡ 3 cos

2πx 3

--- 2 sin3πx 2

---–

⎝ ⎠

⎛ ⎞ dx

⌠

⌡(e–x–x+sin( )–x )dx ⌠⎮ ⌡ cos

x 2

---+(2x–1)3

⎝ ⎠

⎛ ⎞dx

2x+3 –e–x e–2x+sin 2x 1

1+2x

( )

--- 1

1+2x

( )2

---+

3 cos 2x+e2x–1 1

x+1

---+ x+1 e

x

2

–

e2x

---x2–2 cos 2x (1–x)

3

2x

--- 2x+1

x

---g′( )x 2xe_–x2

– =

g′( )x = 3e2xsin(f x( ))

6.1_Appr

oximating ar

eas

CD

SAC analysis task

The area enclosed between the curve with equation y=f(x), the x-axis and the lines x=a

and x= b can be approximated using a number of different techniques. For example, a technique involving sketching the graph on a grid and counting squares could be used. In this task, the use of rectangles to approximate areas will be explored.

Consider a curve with equation y=f(x). The area enclosed between the curve, the x-axis and the lines

x=a and x=b is illustrated in the diagram.

This area is to be approximated using, for example, three rectangles of equal width. The rectangles can be constructed to give either an underestimate or an overestimate of the area.

y=f(x)

x

a b

y

analysis task 1—

(13)

In the first case each of the rectangles lies below the curve.

We will denote by U3 the under-approximation for the area formed by using three rectangles, i.e. U3 =A1 +A2 +A3.

In the second case each of the rectangles extends above the curve.

We will denote by O3 the over-approximation for the area formed by using three rectangles, i.e. O3 =B1 + B2 +B3.

Part 1

a Sketch the graph of y =f(x), where f: [0, 3] →R, f(x) =x2, and shade the region bounded by the curve, the x-axis and the lines x= 1 and x= 3.

b If the interval [1, 3] is divided into two sub-intervals of equal width, what is the width of each sub-interval?

c i On another sketch of the graph of y=f(x), where f: [0, 3] →R, f(x) =x2, show two rectangles of equal width that could be used to give an under-approximation to the area of the region bounded by the curve, the x-axis and the lines x= 1 and x= 3.

ii Calculate the area of each of the two rectangles and hence determine the value of

U2, this approximation to the area.

d i On another sketch of the graph of y=f(x), where f: [0, 3] →R, f(x) =x2, show two rectangles of equal width that could be used to give an over-approximation to the area of the region bounded by the curve, the x-axis and the lines x= 1 and x= 3.

ii Calculate the area of each of the two rectangles and hence determine the value of

O2, this approximation to the area.

e Using your results from question c and question d, between what two values does the actual area of the region lie?

Part 2

f If δ is defined as the difference between On and Un, i.e. δ= On −Un, use your results

from part 1 to find the value of δ, when n= 2.

g i If n= 4, find the values of On and Un and, hence, the value of δ.

ii If n= 10, find the values of On and Un and, hence, the value of δ.

h i What happens to the value of δ as n is increased?

ii What does the value of δ tell you about the accuracy of the approximations On and Un?

y=f(x)

x

a b

y

A₂ A₁ A₃

y=f(x)

x

a b

y

B₂ B₁ B₃

(14)

Part 3

Consider the graph of a continuous, increasing function y=f(x) on the interval [a, b]. The area enclosed by the curve, the x-axis and the lines x=a and x=b is to be approximated using six rectangles of equal width, so the interval [a, b] is divided into six sub-intervals of equal width by the x-coordinates x1, x2, … x5.

i i On the diagram shown draw the six rectangles that would be used to find the under-approximation to the area, U6.

ii If the width of one of these rectangles is defined as being the width of a sub-interval, determine the width of one of these rectangles in terms of a and b.

iii Find an expression for U6.

j i On the same diagram above draw the 6 rectangles which would be used to find the over-approximation to the area, O6.

ii If the width of one of these rectangles is defined as being the width of a sub-interval, determine the width of one of these rectangles in terms of a and b.

iii Find an expression for O6.

k Use your results from above to show that .

l Use the result from question k to write down an expression for δ, where δ =On−Un.

Extension

mFor f(x) =x2, find the value of n for which δ≤ 0.3, when approximating the area of the region enclosed by the curve y= f(x), the x-axis and the lines x= 1 and x= 3.

n Consider the area enclosed by the graph of y= g(x), where g: R→R, g(x) = , the x-axis and the lines x= 1 and x= 3.

i How many rectangles are required so that δ < 0.1?

ii Find an approximation to the area for which δ< 0.1.

o If g: R→R, g(x) = and h: R→R, h(x) = , use a suitable method to approximate the area of the region enclosed between the curves y=g(x) and y =h(x) and the lines x= 1 and x= 3 with δ< 0.1. Explain the method used.

y=f(x)

f(b)

x

a b

y

f(x₅)

f(x₄)

f(x₃)

f(x₂)

f(x₁)

f(a)

x₅ x₄ x₃ x₂ x₁

O6–U6

b–a

6

---(f b( )–f a( ))

=

1+x2

(15)

Approximations to area

under a curve and definite

integration

Approximating areas using numerical methods

There are many problems in mathematics in which the area enclosed between a curve and the

x-axis represents some physical quantity; therefore, it is desirable to be able to calculate such areas.

Algebraic techniques exist for finding the exact areas of some such regions, but they cannot always be used. In such cases, mathematicians often use numerical techniques to approximate these areas.

One straightforward method of approximating the area under the curve involves dividing the region of interest into rectangles whose individual areas are calculated then added together to approximate the whole area.

Consider the area enclosed by the curve with equation y=f(x), the x-axis and the lines x= 0 and x= 1. We can divide the interval [0, 1] into five equal sub-intervals, each of width 0.2, and construct five rectangles, the area of which gives an approximation to the area of the region under consideration.

Dividing the interval [0, 1] into 20 sub-intervals of equal width enables us to construct 20 rectangles whose total area can be used to approximate the area of the required region.

It can be seen from the diagram that the use of more (and therefore narrower) rectangles gives a better approximation to the area under the curve.

0.2 0.4 0.6 1

y=f(x)

0.8 1 x

y

0.8 0.6 0.4 0.2

0

Approximating the area with 5 intervals

0.2 0.4 0.6 1

y₌f(x)

0.8 1 x

y

0.8 0.6 0.4 0.2

0

Approximating the area with 20 intervals

6.2

6 . 2

(16)

Using left-endpoint rectangles to estimate area

Consider the region defined by the curve with equation y= 16 − 0.5x2, the x-axis and the lines

x= 1 and x= 4.

The region can be divided into, for example, six rectangles each of width 0.5, as shown in the diagram. The sum of the areas of the individual rectangles gives an approximation to the area of the region.

A≈ 0.5 ×f(1) + 0.5 ×f(1.5) + 0.5 ×f(2) + 0.5 ×f(2.5) + 0.5 ×f(3) + 0.5 ×f(3.5) = 0.5(15.5 + 14.875 + 14 + 12.875 + 11.5 + 9.875)

= 39.3

The area of the region is approximately 39.3 square units.

This is an overestimate of the actual area. In a case like this, the method of approximation may be referred to as the method of upper rectangles.

Using right-endpoint rectangles to approximate area

As before, consider the region defined by the curve with equation y = 16 − 0.5x2, the x-axis and the lines x= 1 and x= 4.

1 2 3

y=f(x)

4 5 6 7 x

y

15

10

5

0

Note that the top left

vertex of each rectangle intersects with the curve.

The heights of the rectangles can easily be generated in a table on a graphics calculator. In this case enter Y1=16–0.5X2 and start the table at 1 and increase in increments of 0.5.

t i p

y=f(x)

1 2 3 4 5 6 7 x

y

15

10

5

0

Note that the top right

(17)

This time we will approximate the area by dividing the region into six rectangles each of width 0.5, as shown in the diagram. Note that the right-hand end of each sub-interval is used to determine the heights of the rectangles in this case. The sum of the areas of the individual rectangles gives an approximation to the area of the region.

A≈ 0.5 ×f(1.5) + 0.5 ×f(2) + 0.5 ×f(2.5) + 0.5 ×f(3) + 0.5 ×f(3.5) + 0.5 × f(4) = 0.5(14.875 + 14 + 12.875 + 11.5 + 9.875 + 8)

= 35.6

The area of the region is approximately 35.6 square units, according to this method of approximation.

This is an underestimate of the actual area. In a case like this, the method of approximation may be referred to as the method of lower rectangles.

The actual area lies between the lower and upper estimates of 35.6 and 38.3 respectively. Different terminology can be used to describe the type of rectangle used to approximate an area. Left-endpoint and right-endpoint rectangles are used here. When the left-endpoint rectangles were used to approximate the area, each rectangle extended above the curve. These can also be referred to as upper rectangles. When right-endpoint rectangles were used to approximate the area, each rectangle lay below the curve. These can also be referred to as lower rectangles.

Warning

Which is upper/lower?

It is not always the case that left-endpoint rectangles are upper rectangles and that right-endpoint rectangles are lower rectangles. Sometimes, as in the curve shown here it will be the reverse (or neither), depending on the curvature of the graph. It is best to always draw a diagram to determine whether an overestimate or an underestimate of the actual area is being found.

Area approximation (left rectangles) Area approximation (lower rectangles)

1 2 3 4 5 6

x y

2 3 4 5 6 7 8

9

1 0

1 2 3 4 5 6

x y

2 3 4 5 6 7 8

9

1 0

(18)

Approximating areas using other geometric shapes

It is often appropriate to approximate an area using one or a combination of geometric shapes.

E x a m p l e

1

Approximate the area enclosed by the curve with equation f(x) = sinx and the x-axis using the triangle shown in the diagram. S o l u t i o n

Area of a triangle = .

Area =

= =

The area is approximately square units.

E x a m p l e

2

Find an approximation to the area enclosed by the curve with equation

, the x-axis and the lines

x= 1 and x= 3 using two trapezia as shown in the diagram.

S o l u t i o n

Area of a trapezium is given by

, where a and b are the lengths of the parallel sides and h is the perpendicular height.

A1=

= = 1.565

A2=

= = 3.215

A1+A2= 4.78

The area of the region is approximately 4.78 square units.

y= sinx

x y

1

0.5

0 π

2 π

bh

2

---π f π

2

---⎝ ⎠ ⎛ ⎞ ×

2

---π×1

2

---π

2

---π

2

---0.5 1 1.5 2 2.5 3 3.5 x

y

4 6

2

0

A₁

A₂

y=ex2/6

y ex2_⁄₆

=

h

2

---(a+b)

1

A₁

f

(1) f (2)

1 2

---(f( )1 +f( )2 ) 1

2

---(1.181+1.948)

1

A₂

f

(2) f

(3)

1 2

---(f( )2 +f( )3 ) 1

2

(19)

The definite integral

Using a method similar to those shown previously, the area enclosed by a curve with equation

y=f(x), the x-axis and the lines x= 1 and x= 6 can be approximated using rectangles each of width 1. (Note in this case that the rectangles have been placed so that the middle of the top of each rectangle intersects with the curve.)

Using the above the area of these rectangles can be calculated as

Area =f(x1) × 1 + f(x2) × 1 + f(x3) × 1 +f(x4) × 1 +f(x5) × 1

=∑f(xn) × 1

A more accurate approximation could be obtained by reducing the width of these rectangles, and hence fit a greater number of rectangles between x= 1 and x= 6.

Now consider the more general case of the area enclosed by a curve with equation, the x-axis and the lines x= a and x=b can be approximated using rectangles each of width δx and height f(x), as shown right. The area of one of these rectangles is given by f(x) ×δx.

So the total area can be approximated by calculating the sum of the areas of n such rectangles between x= a and x=b. The area can be expressed as

Area ≈f(x1) × δx+f(x2) × δx+ … + f(xn) ×δx

≈∑f(xi) × δx

6.2NUMINT

CD

GC pr

ogram

Most graphing software (e.g. Graphmatica) and graphics calculators calculate and display the approximate areas under curves based on rectangles or trapezia. For example the TI-83/84 program NUMINT calculates and displays such areas and may be useful for checking your manual calculations. The CALCTOOLS APP on the TI-89 may also be helpful.

t i p

1

−1 2 3 4 5 6 7

x

(x₅, f (x₅)) (x₄, f (x₄))

(x₃, f (x₃))

(x₂, f (x₂)) (x₁, f (x₁))

y

y = f (x)

4 6 8 10 12 14 16 18 20

2 0

1 5

δx

f(x)

i= 1 n

(20)

As the widths of these rectangles get smaller i.e. as δx→ 0, the closer this approximation becomes to the actual area. Using limit theory, it can be shown that the actual area enclosed by a curve with equation y= f(x), the x-axis and the lines x= a and x=b is given by

∑f(xi) δx. This limiting sum is referred to as the definite integral, and denoted

as follows.

E x a m p l e

3

The area enclosed by a curve with equation y=x2, the x-axis and the lines x= 1 and x= 5 can be approximated using rectangles of width δx.

a Represent this area for δx= 1, and calculate the sum of the areas of these rectangles.

b Represent this area for δx= 0.5, and calculate the sum of the areas of these rectangles.

c Hence find an approximate value for the definite integral x2dx. S o l u t i o n

a

i= 1 n

δlimx→0

The definite integral

The definite integral is the limiting value of a sum

f(x)dx = ∑f(xi)δx

where the interval [a, b] is partitioned into n equal sub intervals.

⌠ ⌡a

b

i=1

n

δlimx→0

Number

sense with the spence

86

The wealthiest one-fifth of the world’s population consume (or throw away, if they don’t really want it) 86% of its gross domestic product.

⌠ ⌡1

5

1 2 3 4 5 6

x

(1.5, f (1.5)) (2.5, f (2.5))

(3.5, f (3.5)) (4.5, f (4.5))

y

y = f (x)

4 6 8 10 12 14 16 18 20 22 24 26 28

2 0

x =1

Area of rectangles =f(1.5) × 1 +f(2.5) × 1 +f(3.5) × 1 +f(4.5) × 1

=f(1.5) +f(2.5) +f(3.5) +f(4.5)

= 1.52+ 2.52+ 3.52+ 4.52

= 2.25 + 6.25 + 12.25 + 20.25

(21)

b

Area of rectangles =f(1.25) × 0.5 +f(1.75) × 0.5 +f(2.25) × 0.5 +f(2.75) × 0.5 +

f(3.25) × 0.5 +f(3.75) × 0.5 +f(4.25) × 0.5 +f(4.75) × 0.5

= 0.5 × (f(1.25) +f(1.75) +f(2.25) +f(2.75) +f(3.25) +f(3.75) +

f(4.25) +f(4.75))

= 41.25 square units

c f(x)dx = x2dx

= ∑f(xi)δx

≈ 41.25 square units (using δx= 0.5)

Better approximations for this integral can be obtained by choosing smaller values for δx.

exercise

6.2

1 Find an approximation to the shaded area, using:

a three left-endpoint rectangles of equal width.

b three right-endpoint rectangles of equal width.

c three trapezia of equal width.

1 2 3 4 5 6

x (2.75,

f

(2.75)) _(3.25,

f

(3.

25)) (3.75,

f

(3.75)) (4. 25,

f

(4.25)

)

(4.75,

f

(4.75))

(2.25,

f(2.25))

(1.75,

f

(1.75))

(1.25,

f

(1.25)) y

y = f (x)

4 6 8 10 12 14 16 18 20 22 24 26 28

2

x =0.5 δ

⌠ ⌡1

5

⌠ ⌡1

5

i=1

n

δlimx→0

(4, 10)

(3, 6.5) (2, 4)

(1, 2.5)

1 2 3 4 5 x

y

8 10

6 4 2 0

6.3_{A better appr}

oximation

CD

TA

I

(22)

2 The region enclosed by the curve with equation y=f(x), the x-axis and the lines x= 0 and x= 4 is shown.

The following table of values is for y=f(x).

Find an approximation to the area of the region shown, using intervals of width 0.5 and

a left-endpoint rectangles. b right-endpoint rectangles. c trapezia.

3 a Estimate the area under the graph for x= 0 to x= 5, using five right-endpoint rectangles of equal width.

b Is this estimate an overestimate or underestimate of the actual area?

4 A region is enclosed between the curve with equation y= cos(x) the x-axis and the lines x= 0 and x= 1.5.

a Use six rectangles of equal width to find:

i an underestimate of the area. ii an overestimate of the area.

b Find the average of these two estimates.

c Find an approximation to this area, using six trapezia of equal width.

d Compare the results of part b and part c.

5 The region enclosed by the curve with equation the x-axis and the lines x= −2 and x= 2 is illustrated.

a Construct eight lower rectangles of equal width and use them to find an underestimate of the shaded area.

b Construct eight upper rectangles of equal width and use them to find an overestimate of the shaded area.

x 0 0.5 1 1.5 2 2.5 3 3.5 4

y 10 6.67 5 4 3.33 2.86 2.5 2.22 2

y=f(x)

1 2 3 4 5 x

y

8 10

6 4 2 0

y = 25–x2

–2 –1.5 0.5 x

y

y₌e–x2

1.5

0.5 1

–1 –0.5 0 1 1.5 2

(23)

area enclosed by the curve y= 6 − 5x− x2 and the x-axis, by finding the area of the triangle shown.

7 a Sketch the graph of the function .

b Estimate the area under the curve, using four rectangles of equal width and left endpoints.

8 The graph shows the velocity of a body that accelerates from rest to a velocity of 10 metres/second in 25 seconds according to

the rule .

The distance travelled by the body is given by the area under the curve. Approximate the distance travelled by the body in the first 25 seconds, using the five trapezia shown.

9 Use the formula f(x)dx ≈∑f(xi)δx to approximate

a x2dx (use δx= 1) b exdx (use δx= 0.5)

Calculating the definite integral

If F(x) is a function such that F′(x) =f(x) and f(x) is continuous on the interval [a, b], then

f(x)dx= F(b) −F(a).

A definite integral, f(x)dx, is different from an indefinite integral,

∫

, which only involves finding an antiderivative. The presence of the terminals a and b indicate a definite integral that requires further calculation.

The process for evaluating a definite integral is shown in the steps below.

f(x)dx=

=F(b) − F(a)

where F(x) is an antiderivative of f(x).

(1, 0) (–6, 0)

–6 2 x

y =6 − 5x − x2

15

5 10

–4 –2 0 4 6

y 6

2+x2

---,–2≤ ≤x 2

=

5 10 15 20 25 t

v

8 10

6 4 2

0

v= 2√t

v = 2 t

i= 1

n ⌠

⌡a b

⌠ ⌡0

3

⌠ ⌡1

4

⌠ ⌡a

b

⌠ ⌡a

b

f x( )dx

⌠ ⌡a

b

F x( )

[ ]a b

(24)

E x a m p l e

4

Evaluate the definite integral x2dx. S o l u t i o n

x2dx= (antidifferentiating x2)

= (finding F(5) −F(1)

=

= 41.3 (correct to 1 decimal place)

E x a m p l e

5

Evaluate:

a (x3− 4)dx b sin 2x dx

S o l u t i o n

a (x3− 4)dx= (antidifferentiating)

= (substituting upper and lower terminals)

=

b sin 2x dx= (antidifferentiating)

= (substituting terminals)

= =

E x a m p l e

6

Use calculus to find the exact value of the following.

a b (4e2x−ex)dx

S o l u t i o n

a = (antidifferentiating)

= (substituting terminals)

= = = loge3

⌠ ⌡1

5

t i p

Note that the constant of integration,

c, is dropped from the antiderivative when evaluating a definite integral. The reason for this is as follows:

f(x)dx=

= (F(b) +c) − (F(a) +c) =F(b) −F(a) +c−c

=F(b) −F(a)

⌠ ⌡a

b

F x( )+c

[ ]a b

⌠ ⌡1

5 _x3

3 ---1 5 53 3 ---⎝ ⎠ ⎛ ⎞ 13

3 ---⎝ ⎠ ⎛ ⎞ – 124 3 ---⌠ ⌡₋2

1

⌠ ⌡0

π/6

⌠ ⌡₋2

1 _x4

4

---–4x

2

–

1

14

4

---–4 1( )

⎝ ⎠

⎛ ⎞ ( )–2 4

4

---–4 2( )–

⎝ ⎠ ⎛ ⎞ – 63 4 ---– ⌠ ⌡0

π/6 ₁

2

---cos 2x

–

0

π⁄6

1 2

---cosπ 3

---–

⎝ ⎠

⎛ ⎞ 1

2

---cos 0

– ⎝ ⎠ ⎛ ⎞ – 1 4 ---– 1 2 ---+ 1 4 ---⌠ ⎮ ⌡₀ 4 1 2x+1

---dx ⌠⎮ ⌡₀ 4 ⌠ ⎮ ⌡₀ 4 1 2x+1

---dx 1

2

---loge (2x+1)

0 4

1 2

---loge (2 4( )+1)

⎝ ⎠

⎛ ⎞ 1

2

---loge (2 0( )+1)

⎝ ⎠

⎛ ⎞

–

1 2

---loge 9–1₂---loge 1

loge( )9

(25)

b (4e −e )dx= (antidifferentiating)

= (2e2(1)−e1) − (2e2(0)−e0) (substituting terminals)

= 2e2−e− 1

E x a m p l e

7

Find k, if (x3− 3x2)dx= 32 and k is an integer. S o l u t i o n

(x3− 3x2)dx =

=

(x3− 3x2)dx= 32

= 32

k4− 4k3+ 3 = 128

k4− 4k3− 125 = 0

Using the factor theorem or otherwise, k− 5 is a factor of the LHS. Hence: (k− 5)(k3+k2+ 5k+ 25) = 0

Using a numerical approach (for example an appropriate graph) it is easy to check that there are no integer solutions to k3+k2+ 5k+ 25 = 0.

Thus, since k was specified to be an integer, then k= 5.

exercise

6.2

10 Use calculus to evaluate each of the following definite integrals.

a (x2 + 1)dx b (x2− 2x+ 2)dx c

d e f

g (x2− 5)2dx h i

j k (x2 + 3)3dx l

⌠

⌡2 2e

2x_–_ex

[ ]₀

Warning

Use calculus means ‘use calculus’!

If the instruction ‘use calculus’ is given, an appropriate antiderivative must be shown in the working to obtain the solution. In examinations, solutions obtained by other methods, including calculators, may not attract marks.

⌠ ⌡1

k

CAS 5.5, 10.2

t i p

This can be solved directly on a TI-89.

⌠ ⌡1

k _x4

4

---_–x3 1

k

k4

4

---–k3

⎝ ⎠

⎛ ⎞ 1

4

---–1

⎝ ⎠

⎛ ⎞

–

k4

4

--- k3 3

4

---+ – ⌠

⌡1

k

k4

4

--- k3 3

4

---+ –

continued

⌠ ⌡0

3

⌠ ⌡1

5

⌠ ⌡1

9

x xd

⌠ ⎮ ⌡2

4

3–x x3

---dx ⌠_⌡

1 2

3x x xd ⌠_⎮

⌡1 2

1 x2

---dx ⌠

⌡₋5 5

⌠ ⌡1

9

2x+7dx ⌠⎮

⌡0 3

1 4–x

---dx ⌠

⎮ ⌡1

3

1 x

---–3

⎝ ⎠

⎛ ⎞ _dx ⌠

⌡₋1

1 ⌠

⎮ ⌡3

5

1 x–2

( )2

---dx

(26)

11 Find the exact value of the following definite integrals.

a 3 cosx dx b exdx

c d 12x2(1 −x)dx

e (sin 2x− 2 sinx+ 1)dx f

g h

i (cos 2x− sin 2x)dx j

k (1 − 2x)3dx l (3 cos 2x− 1)dx

m e−x/2dx n

o p

12 a If (x− 5)dx= 0, find the value of b.

b Find k if dx= 2.

c If 3x2dx= 128, find a.

d If , find the exact value of k.

e Find the integer value of p, if (x3− 3x2)dx= 30.

⌠ ⌡0

π/4

⌠ ⌡₋1

1

⌠ ⎮ ⌡₀

4

1 x+1

---dx ⌠_⌡

1/4 1/2

⌠ ⌡_−π/4

π/4 ⌠

⎮ ⌡0

1

1 3–x

---dx ⌠

⎮ ⌡₁

3

3e2x 1 x

---–

⎝ ⎠

⎛ ⎞_d_x ⌠

⌡₋1 0

ex_–₂_e–x

( )2

x d

⌠ ⌡_−π

π ⌠

⎮ ⌡2

4

1 3x–4

---dx ⌠

⌡₋2 5

⌠ ⌡_−π/6

π/6

⌠ ⌡₋2

2 ⌠

⎮ ⌡4

9

x2–1 x

---dx ⌠

⎮ ⌡₀

π

sinx x 2

---+

⎝ ⎠

⎛ ⎞ _d_x ⌠

⌡_π/8

π

2 cos 2x xd

⌠ ⌡2

b

⌠ ⎮ ⌡1

k

3 x2

---⌠ ⌡₋a

a

⌠ ⎮ ⌡₁

3

2 3x–2

---dx = logek

⌠ ⌡p

(27)

Using technology to evaluate definite integrals

It is not always possible to evaluate a definite integral by hand. In such cases it is important to be able to use available technology effectively. At other times it is simply more convenient to make use of available technology, or the use of technology can provide a useful check.

E x a m p l e

8

Evaluate , giving the answer correct to 2 decimal places.

S o l u t i o n

Using the TI-83 or TI-84:

Press MATH and select 9:fnint(, and complete the command:

fnint(X/ 1 + 2X^2), X, 0, 4)ENTER.

Hence , correct to 2 decimal places.

exercise

6.2

13 Evaluate each of the following, giving answers correct to 2 decimal places.

a sin2x dx b excos 2πx dx

c logex dx d x

2

sinx dx

e f

g 2xsin 2x dx h

i j sin 2xcos22x dx

0 4 ⌠ ⎮ ⌡

x

1+2x2

---dx

Using the TI-89:

2nd [ ](X/⌠_⌡ (1 + 2X^2), X, 0, 4)♦ ENTER

GC 5.5 CAS 5.5

(

0 4 ⌠ ⎮ ⌡

x

1+2x2

---dx = 2.37

Number

sense with the spence

87

Being 13 less than 100, 87 has scared the pants off generations of cricket players who, let’s be honest, must panic very easily.

continued

⌠ ⌡0

π/2

⌠ ⌡0

2

⌠ ⌡2

4

⌠ ⌡₋2

0

⌠ ⌡1

5

25–x2dx ⌠⎮

⌡5 10

e–x⁄10sinπx 5

---dx ⌠

⌡0 3

⌠ ⌡₋2

2

ex2 x d

⌠ ⌡2

4

x 1+2xdx ⌠_⌡

−π/4

π/4

(28)

6.4_{Investigating pr}

operties of def

inite integrals

CD

SAC analysis task

a Evaluate:

i x2dx ii iii sinx dx iv

b If F′(x) =f(x), evaluate f(x)dx

Comment on the result of evaluating a definite integral for which the upper and lower terminals are the same.

c Evaluate:

i 3x2dx and 3x2dx ii cosx dx and cosx dx

iii exdx and exdx iv and

d If F′(x) =f(x), evaluate f(x)dx and f(x)dx.

Comment on the effect of interchanging the terminals on the result when a definite integral is evaluated.

e Evaluate:

i (x2+ 7)dx+ (x2 + 7)dx and (x2 + 7)dx

ii sin 3x dx+ sin 3x dx and sin 3x dx

iii ₊ and

f If F′(x) =f(x), evaluate f(x)dx+ f(x)dx and f(x)dx. Comment on the result.

⌠ ⌡1

1 ⌠

⎮ ⌡₃

3

1

x

---dx ⌠_⌡

π/4 π/4

⌠ ⌡k

k

x xd

⌠ ⌡a

a

⌠ ⌡1

2

⌠ ⌡2

1

⌠ ⌡_π/3

π/6

⌠ ⌡_π/3

π/6

⌠ ⌡0

2

⌠ ⌡2

0 ⌠

⎮ ⌡₂

3

1

x–1

---dx ⌠⎮ ⌡₃

2

1

x–1 ---dx

⌠ ⌡a

b

⌠ ⌡b

a

⌠ ⌡1

2

⌠ ⌡2

4

⌠ ⌡1

4

⌠ ⌡0

π/2

⌠ ⌡_π/2

π

⌠ ⌡0

π

⌠ ⎮ ⌡₁

3

1

x

---dx ⌠⎮ ⌡₃

5

1

x

---dx ⌠⎮ ⌡₁

5

1

x

---dx

⌠ ⌡a

b

⌠ ⌡b

c

⌠ ⌡a

c

analysis task 2—

(29)

Properties of antiderivatives

and definite integrals

Properties of definite integrals

Properties of definite integrals can be very useful in solving problems and are summarised below.

Let f and g be two continuous functions on the interval [a, b] and let . Then:

f(x)dx= 0

f(x)dx= − f(x)dx

kf(x)dx= k f(x)dx, where k is a real constant

(f(x) ± g(x))dx= f(x) ± g(x)dx

f(x)dx= f(x)dx+ f(x)dx

Also, if the graph of f is symmetrical about the y-axis, then f(x)dx= 2 f(x)dx.

E x a m p l e

1

If f(x)dx= 6 find the value of the following.

a f(x)dx b 3f(x)dx c (3f(x) + 2)dx

S o l u t i o n

a Using the property f(x)dx= − f(x)dx:

f(x)dx =− f(x)dx

=−6

b Using the property kf(x)dx =k f(x)dx: 3f(x)dx= 3 f(x)dx

= 3 × 6

= 18

c Using the property (f(x) +g(x))dx = f(x)dx+ g(x)dx: (3f(x) + 2)dx= 3f(x)dx+ 2dx

Now using the property kf(x)dx=k f(x)dx: 3f(x)dx+ 2dx= 3 f(x)dx+ 2dx

c∈[a b, ]

⌠ ⌡a a ⌠ ⌡a b ⌠ ⌡b a ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡a c ⌠ ⌡c b ⌠ ⌡₋a

a ⌠ ⌡0 a ⌠ ⌡1 4 ⌠ ⌡4 1 ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡4 1 ⌠ ⌡1 4 ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡a b ⌠ ⌡a b ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡1 4 6.3

6 . 3

(30)

This can then be evaluated:

3 f(x)dx+ 2dx=

= 18 + (8 − 2)

= 24

E x a m p l e

2

If f(x)dx= 12 and f(x) = 7, find the value of f(x)dx. S o l u t i o n

Using the property (x)dx= f(x)dx+ (x)dx:

f(x)dx = f(x)dx+ f(x)dx

f(x)dx = f(x)dx− f(x)dx

= 12 − 7

= 5

exercise

6.3

1 If g(x)dx= 8, find the value of:

a g(x)dx b 2g(x)dx c −3g(x)dx

d (1 − 2g(x))dx e 2(g(x) + 1)dx

2 If f(x)dx= 2, find the value of:

a (f(x) − 2)dx b (1 −f(x))dx c 3(f(x) − 3)dx d (3f(x) + 2)dx

3 If f(x) = 7 and g(x) =−4, find (f(x) −g(x))dx.

4 If g(x)dx= 4 and 2f(x)dx= 10, find (g(x) +f(x))dx.

5 Determine whether each of the following is true (T) or false (F).

a f(x)dx= 0 b (2f(x) + 5)dx=2 (f(x) + 5)dx

c (3f(x) − 2)dx= 3 f(x)dx− 2x d (1 − 2g(x))dx= 3 − 2 g(x)dx

e 2(f(x) + 1)dx= 2 f(x)dx+ 2

6 If f(x)dx= 3 and f(x)dx= 5, find the value of f(x)dx.

⌠ ⌡1 4 ⌠ ⌡1 4

3×6+[ ]2x ₁4

⌠ ⌡0 8 ⌠ ⌡5 8 ⌠ ⌡0 5 ⌠ ⌡a b ⌠ ⌡a c ⌠ ⌡c b ⌠ ⌡0 8 ⌠ ⌡0 5 ⌠ ⌡5 8 ⌠ ⌡0 5 ⌠ ⌡0 8 ⌠ ⌡5 8 ⌠ ⌡₋1

5

⌠ ⌡5

−1

⌠ ⌡₋1

5

⌠ ⌡5

−1

⌠ ⌡₋1

5

⌠ ⌡₋1

5 ⌠ ⌡0 3 ⌠ ⌡0 3 ⌠ ⌡3 0 ⌠ ⌡0 3 ⌠ ⌡0 3 ⌠ ⌡1 3 ⌠ ⌡1 3 ⌠ ⌡1 3 ⌠ ⌡₋2

5

⌠ ⌡5

−2

⌠ ⌡₋2

5 ⌠ ⌡2 2 ⌠ ⌡1 4 ⌠ ⌡1 4 ⌠ ⌡0 3 ⌠ ⌡0 3 ⌠ ⌡₋1

2

⌠ ⌡₋1

2

⌠ ⌡₋1

1

⌠ ⌡₋1

1

⌠ ⌡1

4

⌠ ⌡₋3

1

⌠ ⌡₋3

(31)

7 If f(x)dx= −10 and f(x)dx= 2, find the value of f(x)dx.

8 For which of the following functions would the property f(x)dx= 2 f(x)dx be true?

a f(x) =x b f(x) = x2 c f(x) = 4 −x2

d f(x) = (x− 4)2 e f(x) = sinx f f(x) = cos2x

g f(x) =ex h f(x) = x3

9 For which of the functions in question 8 would the property f(x)dx= 0 be true?

Integration by recognition

Integration by recognition relies on an understanding that antidifferentiation is the reverse process of differentiation, and ‘recognising’ that the required antiderivative is present in a differentiated expression.

E x a m p l e

3

Differentiate (3x2− 2)4. Hence find an antiderivative of x(3x2− 2)3. S o l u t i o n

((3x2− 2)4) = 4(6x)(3x2− 2)3 (using the chain rule)

= 24x(3x2− 2)3 So:

= (3x2− 2)4

24 = (3x2− 2)4

= (3x2− 2)4

E x a m p l e

4

Differentiate loge(x2+ 1) and hence find an antiderivative of .

S o l u t i o n

It follows that:

dx = loge(x2+ 1)

2 dx = loge(x2+ 1)

dx =

(Since x2+ 1 > 0 for all x, no restriction is required.)

⌠ ⌡1

⌠ ⌡3

⌠ ⌡1

⌠ ⌡₋a

a

⌠ ⌡0

a

⌠ ⌡₋a

a

If , d then .

dx

---(f x( )) = f′( )x _⌡⌠f′( )x dx = f x( )+c

d dx

---t i p

Only an antiderivative is required, so it is not necessary to include a constant of integration.

⌠

⌡24x(3x2–2)3dx ⌠

⌡x(3x2–2)3dx ⌠

⌡x(3x2–2)3dx ₂₄---1

x x2+1

---d dx

---(loge(x2+1)) 2x

x2₊₁ ---=

⌠ ⎮ ⌡

2x x2+1

---⌠ ⎮ ⌡

x x2+1

---⌠ ⎮ ⌡

x x2+1

--- 1

2

---loge x

2

1

+

( )