Reversing derivatives

Contents

11.1 Finding indefinite integrals 11.2 Graphing indefinite integrals 11.3 Applying indefinite integrals Chapter summary

Chapter review

Syllabus subject matter

Introduction to integration ■ Definition of the indefinite integral ■ Indefinite integrals of simple polynomial functions, simple

exponential functions, sin (ax + b), cos(ax + b) and ■ Practical applications of the integral

Quantitative concepts and skills ■ Calculation and estimation with and without instruments ■ Basic algebraic manipulations Syllabus

Guide Chapter 11

1 ax+b

Reversing derivatives

11.1

Finding indefinite integrals

We have previously defined the derivative of the function y=f(x) as

f′(x) = = f(x) = =

provided the limit exists.

The derivative of a function is a new function and is known as the gradient function. The derivative or gradient function gives the rate of change of the original function. For a polynomial function, we can use the derivative to calculate the gradient of the polynomial at any point on the curve. This is also the instantaneous rate of change at that point. Work through the following investigation.

The differential calculus you have already studied is very useful in science, technology, economics, engineering, etc. However, it is also very important to be able to work backwards. Integration is the reverse of differentiation and is equally important in our modern world. In many applications, the rate of change of a quantity is used to find the quantity itself. For example, the acceleration (rate of change of velocity) of a falling object near the Earth is known to be 9.8 m/s2_{. Integration allows us to use this fact to}

find the distance that an object falls and its speed at any time after it begins falling.

P

Q y = f(x)

f(x + h) − f(x)

x y

h

(x + h, f(x + h))

(x, f (x))

df dx

--- d

dx

--- dy

dx

--- f x( +h)– f x( )

h

---h→0

lim

We use the process called differentiation to derive the gradient function or derivative, f′(x), from the original function, f (x). The reverse of differentiation is known as antidifferentiation or

integration. If we know the derivative (gradient function), we can differentiate in reverse or work

backwards from it to find a possible original function from the gradient function.

1 Copy the table below into your workbook.

2 Complete the table by working down the list in order. 3 As you go down the list, look for patterns that emerge.

4 Describe in words how to calculate the original function, using the derivative.

f (x) f′(x) f (x) f′(x) f (x) f′(x)

5x 5x3 _x3

3 x3 _3x2− _4x

x2 _6x2 _5x4+ _7x2

3x2 _x2 _6x + ₁₁

4x 3x4 _9x2− _8x + ₁

x x4

x3 _16x3

1 4

---3 2

---1 2

--- 1

2

You may see that there is a pattern that can be followed to calculate a possible original function if the antiderivative is known.

You can see that: If f′(x) = 2x, f(x) =x2_. If f′(x) = 3x2_{, f(x)} =_x3_. If f′(x) = 4x3_{, f(x)} =_x4_.

The pattern can be said as ‘Add 1 to the power of x and divide by the new power’. We can write this in symbols as follows.

In Example 1, it would have been possible to find many antiderivatives of f(x) other than

x3− _5x2+ _{4x. For example, x}3− _5x2+ _4x+ _{5 and x}3− _5x2+ _4x − _{7 are also antiderivatives of}

f (x) = 3x2− _10x + _4.

The derivative of a constant is zero, so we can make as many antiderivatives as we like by adding constants to one antiderivative. We usually write the antiderivative of a function with a constant term ‘+c’. Any particular antiderivative can then be obtained by substituting the appropriate

value for the constant.

It is very important to keep in mind that the indefinite integral is a function. In fact, it can be many functions that only differ by a constant. The dx in the sign is not multiplied by the function. It is just part of the symbol for the indefinite integral.

If f′(x) = axn_{, then f (x)} = _(where a xn+1 _n ≠−_{1) is a possible original function.}

n+1

---!

Find an antiderivative of f (x) = 3x2− _10x+ _4.

Solution

Write down the function. f (x) = 3x2− _10x + ₄

Let g(x) be an antiderivative of f (x) and

apply the pattern to each term. g(x) = − +

Simplify. = − +

g(x) = x3− _5x2+ _4x 3 x2+1

2+1

--- 10 x1+1

1+1

--- 4 x0+1

0+1

---3 x3

3

--- 10 x2 2 --- 4 x

1

---Example

1

The indefinite integral of the function f (x) is F(x) + c.

The indefinite integral (or just integral) is also known as the antiderivative.

The symbol represents the integral.

So F(x) =

Alternatively, if y = f (x), the antiderivative of is y + c.

∫

f x( )dx

∫

dy dx

---!

It may be necessary to express a function in polynomial form before you can find the antiderivative.

More information is required if the value of the constant is to be calculated. Find the indefinite integral of g(x) = 2x3− _5x2+ _8x.

Solution

Write down the function. g(x) = 2x3− _5x2+ _8x

Use the rule for finding indefinite integrals. G(x) = − + + c

Simplify. = − + + c

= x4− _x3+ _4x2+ _c 2 x3+1

3+1

--- 5 x2+1

2+1

--- 8 x1+1

1+1

---2 x4

4

--- 5 x3 3

--- 8 x2 2

---1 2

--- 5

3

---Example

2

Find y if = (3x − 5)2_.

Solution

Write down the information. = (3x − 5)2

Expand the RHS. = 9x2− _30x + ₂₅

Find the antiderivative. y = − + 25x + c

Simplify. = 3x3− _15x2+ _25x + _c

dy dx

---dy dx

---9 x3

3

--- 30 x2 2

---Example

3

Find x(2x2− _3x + _{5) dx.}

Solution

Write down the information. x(2x2− _3x + _{5) dx}

Expand the brackets. = (2x3− _3x2+ _{5x) dx}

Find the antiderivative. = − + + c

Simplify. = x4− _x3+ _x2+ _c

∫

∫

∫

2 x4

4

--- 3 x3 3

--- 5 x2 2

---1 2

--- 5

2

Exercise 11.1

Finding indefinite integrals

1 Find any two antiderivatives of each of the following functions. a f(x) = 6x2+ _4x + _{1 b m(t)} = _8t − ₇ c g(z) = 4z3+ _3z − _5z4 _{d h(x)} = ₄

e p(y) = y2− _y − _{1 f s(v)} = _3v3− _v4+ _4v2− ₂ g v(t) = 9.8t − 0.3t2− _{6 h j(m)} = _4.5(m + _0.3)2+ _0.8 i p(d) = 0.1d3+ _7d − _{0.38 j h(k)} = _5k2− _2k3+ _2k − ₅

2 Find the indefinite integral of each of the following functions.

a q(i) = 4i3− _6i2− _2i − _{3 b f(x)} = _x3− _10x4− _6x2+ ₁₂ c y(u) = 2u − 5u3− _4u2+ _{8 d t(m)} = _0.8m2− _6.3m3− _0.7m + ₄ e r(t) = 0.07t + t2 _{f v(x)} = _4.9(x − ₅₎2

3 Find the following.

a (x2− _{5)dx b (3t} + _2)dt

c (7 −u3+ _u2_{)du d h}4_dh

e ( p + 4)2_{dp f (z}4− _z3+ _z2− _z + _1)dz

4 Find y in each of the following.

a = 4x3+ _6x2− _6x + _{7 b} = _12x3− _12x2+ _12x − ₁₂

c = 3t + 5 d = z− z3+ ₈

e = 0.5x − 0.04x3+ _0.7x2− _{0.52 f} = _0.7t − _0.4t2+ _0.6

Find the antiderivative of f (x) = 3x − 5, given that F(3) = 6.5.

Solution

Write down the function. f(x) = 3x − 5

Find the antiderivative from the rules. F(x) = − 5x + c

Write the given information. F(3) = 6.5

Substitute x = 3. 6.5 = − 5 × 3 + c

Simplify. 6.5 = 13.5 − 15 + c

Solve for c. c = 8

Write the result. F(x) = − 5x + 8

3 x2

2

---3×32

2

---3 x2

2

---Example

5

Additional Exercise

11.1

∫

∫

∫

∫

∫

∫

dy dx

--- dy

dx

---dy dt

--- dy

dz

---dy dx

--- dy

dt

Modelling and problem solving

5 = 4x − 5 and (3, 2) is a point on the graph of y. Find the function y and its value at x = 5.

Justify your reasoning.

6 When t = 1, the value of x is −2. Find x at t = 4, given that = 0.6t2− _2t + _{4. Show logical} steps in your working.

7 f(x) = (6x − 12x2_{) dx and f (}−₁₎ = _{5. Find f(1). Give clear reasons for your working.}

8 g′(v) = v − v3− _{1 and g(v) has a zero at v} = _{3. Find g(4), explaining your steps.}

9 m′(r) = 4 and (2, 6) is on the graph of m(r). Find m(r), showing your logic clearly.

10 = and y is zero at x = 5. Find an expression for y.

11.2

Graphing indefinite integrals

We saw in the last section that antiderivatives are not unique. This means that a given derivative (gradient function) may have many different antiderivatives. However, they differ only by a constant, since the derivative of the difference must be zero.

dy dx ---dx dt

---∫

r dy dx

--- 3 x+4

Draw graphs of three different antiderivatives of the derivative (gradient function)

f (x) = 4x − 3 on the one set of axes and comment on the result.

Solution

Use any method to graph the chosen antiderivatives. F1(x) has x-intercepts of − and 2 and a y-intercept of −2.

F2(x) has x-intercepts of 0 and 1 and a y-intercept of 0.

F3(x) has x-intercepts of and 1 and a y-intercept of 1.

F₁, F₂ and F₃ all have a minimum at the same x value.

Write the derivative. f (x)= 4x − 3

Calculate the antiderivative. F(x)= 2x2− _3x + _c

Choose any three values of c, say −2, 0 and 1. F1(x)= 2x2− 3x − 2

F2(x)= 2x2− 3x

F₃(x)= 2x2− _3x + ₁

State the result. The graphs of F1(x), F2(x) and F3(x) have the

same shape (quadratic with a minimum at

x = ). Each graph can be translated up or down to exactly coincide with the others.

1 3 4 x −2 −3

−1 1 2

x =

−

c =−2

c = 0

c = 1

F(x) = 2x2− _{3x + c}

Use a graphics calculator to graph the function f (x) = x − 5 and an integral on the same set of axes. Compare the graphs.

Solution

All calculators Enter Y1= X − 5. Enter Y2= 0.5X2− 5X.

Set domain and range values as follows: −1 X 12 and −15 Y 10 Graph the function and the integral.

Use or (or V-Window) to examine the graphs more closely if needed.

Comparing the two graphs shows that:

■ The derivative is a linear function and the integral is a quadratic function. ■ F(x) has a negative slope for f (x) 0.

■ F(x) has a stationary point when f (x) = 0. ■ F(x) has a positive slope for f (x) 0.

Note regarding the Texas Instruments TI-84

The TI-84 is able to calculate the integral of a function using the command fnInt(.

Enter the function as Y1= X − 5 and press .

The fnInt command is number 9 on the menu.

Enter Y₂= fnInt(Y₁, X, 0, X) as shown by pressing:

9 1 1 0

.

the function and the integral.

Sharp EL-9900

See the instructions given on the CD-ROM.

Write down the function. f (x) = x − 5

Calculate the antiderivative. F(x) = x2+ _5x + _c

We will use the antiderivative F(x) = x2+ _5x.

1 2

---1 2

---F(x) f (x)

ZOOM WINDOW

ENTER

MATH

MATH VARS , X,T,θ,n , ,

X,T,θ,n )

GRAPH

Example

7

Calculator Instructions

Sketch a possible antiderivative of the function shown. y

x f (x)

Example

8

Exercise 11.2

Graphing indefinite integrals

1 Draw three different indefinite integrals of each of these functions. a f(x) = 2x − 8 b g(z) = 4

c d(m) = 6 − 4m d q(x) = 5 − 4x e p(x) = 3x2− _{6x f w(y)} = ₄ − _6y g v(t) = t − 7 h g(m) = 6(m + 1) i p(x) = 3(x + 2)2− _{12 j k(h)} = _10h + _5h2

2 Use a graphics calculator to plot an antiderivative of each of the following functions. a f(x) = 6x3+ _3x + _{5 b m(t)} = _4t − ₃

c g(z) = 30z3− ₃ + _10z2 _{d h(x)} = ₇ − _4x2

e p(y) = y3+ _y2+ _y + _{1 f s(x)} = _0.2x3− _0.3x4+ _0.5x2− ₂

Solution

Label the important points on the function as A, B, C, D, E and F.

We know that the shape of f (x) will indicate the slope of F(x).

Use a table to describe the shape of f (x) and the corresponding shape of F(x).

Use information in the table to draw a graph of a possible antiderivative.

Interval/point Behaviour of f (x) Behaviour of F(x)

AB f (x) 0 and constant Gradient positive. F(x) is a straight line.

BC f (x) 0 and

increasing

Gradient positive. F(x) is curved and becoming steeper.

CD f (x) 0 and constant Gradient positive. F(x) is a straight line.

DE f (x) 0 and

decreasing

Gradient positive. F(x) is curved and becoming less steep.

E f (x) = 0 F(x) has a stationary point.

EF f (x) 0 and

decreasing

Gradient negative. F(x) is curved and becoming steeper.

y

x f(x)

A B

C D

E

F

y

x F(x)

A

B C

D

E F

Additional Exercise

Modelling and problem solving

3 Sketch a possible indefinite integral of each the functions shown below.

4 Sketch a possible indefinite integral of each the functions shown below.

11.3

Applying indefinite integrals

The velocity of an object is the rate of change of its displacement. This is given by the derivative. In reverse, the displacement of an object must be an antiderivative of its velocity. The same principle applies more generally with any rate of change.

y

x

a y

x

b y

x

c

y

x

a y

x

b y

x

c

Rate of change

If f (x) is the rate of change of F(x), then:

■ f (x) is the derivative of F(x)

■ F(x) is an antiderivative of f (x).

!

The volume V of a balloon is changing with respect to time at a rate given by

= 6t2− _10t + _{7 cm}3_/s

The volume of the balloon is 35 cm3 _{after 3 s.} a Express V as a function of t.

b What is the volume after 10 s?

dV dt

---Example

9

Exercise 11.3

Applying indefinite integrals

Modelling and problem solving

1 A stone is thrown upwards from a tree house at a speed of 5 m/s. Gravity accelerates the stone downwards with an acceleration of at 10 m/s2_{. The tree house is 6 m off the ground.} a Find the indefinite integral of the acceleration. Use it to find an expression for the velocity. b At what time does the stone reach its greatest height (when v = 0)?

c Find the indefinite integral of the velocity and use it to find an expression for the height. d What is the greatest height reached?

e Write an equation to find the time when the stone returns to the height of the tree house, and find its velocity on its return.

Solution

a Write down the derivative. = 6t2− _10t + ₇

Calculate the antiderivative. V(t) = 2t3− _5t2+ _7t + _c

Write the given information. V(3) = 35

Substitute t = 3. 35 = 2 × 33− ₅ × ₃2+ ₇ × ₃ + _c

Simplify. 35 = 54 − 45 + 21 + c

Solve for c. c = 5

Write the function V(t). V(t) = 2t3− _5t2+ _7t + ₅

b Find V after 10 s. V(10) = 2 × 103− ₅ × ₁₀2+ ₇ × ₁₀ + ₅

Evaluate. = 1575

Write the result. The balloon has a volume of 1575 cm3 _{after 10 s.}

dV dt

---A truck travelling at 5 m/s accelerates for 4 s according to the equation

a(t) = 10 − 2t m/s2 What is its new speed?

Solution

Acceleration is the rate of change of velocity,

so v(t) must be the antiderivative of a(t). a(t) =

Write down the acceleration function. a(t) = 10 − 2t

Find the antiderivative. v(t) = 10t − t2+ _c

Initial velocity is 5 m/s. v(0) = 5, so c = 5

Write the velocity function. v(t) = 10t − t2+ ₅

Calculate velocity after 4 s. v(4) = 10 × 4 − 42+ ₅

Evaluate. = 40 − 16 + 5 = 29

Write the result. The new speed of the truck is 29 m/s.

dv dt

---Example

10

Additional Exercise

11.3

Extra Material

Area under a curve

Extra Material

2 Barry throws a stone from the beach up towards the top of a cliff. After a few attempts where the stone falls back, he just manages to reach the top of the cliff, which is 8 m above the beach. After he throws the stone, it accelerates downwards at about 10 m/s2 _{under gravity.} How fast does he throw the last stone? What is this in km/h?

3 In a test conducted by a motoring organisation, a car starting from rest accelerates to 100 km/h in 12 seconds. What is the acceleration of the car in m/s2_?

4 As water leaks out of a container, the rate of leaking slows because the level of water in the container drops. In the first 10 seconds of a leak from a tank, the rate of water flow is approximately given by

L = 0.03t2− _3.7t + _39.8

where L is the rate of leaking in litres/second and

t is the time in seconds. Find the amount of water

that leaks out in the first 5 seconds and the amount leaked after 10 seconds.

5 The marginal cost of producing plasma TVs in a

factory is given by

= 20n

where n is the number produced in a day and C is the cost of producing this number. It costs $20 000 to keep the factory open for a day, even if no TVs are made. The manufacturer’s price for each TV is $1800.

a Find an expression for the cost of producing n TVs and the profit made on n TVs.

b Use a graphics calculator to find the number of TVs the factory should make each day to

get the biggest profit.

6 The force exerted on a person’s body as they slow down when a car stops suddenly

is given by B = kx2_{, where x is the distance travelled by the car while slowing and k} = _.

E is the energy that the person’s body has before stopping, and D is the distance over

which the person stops. At a speed of 60 km/h, a 50 kg person has an energy of about 7000 J (joules), while a 100 kg person has an energy of

about 14 000 J. At 100 km/h, the energy goes up to about 20 000 and 40 000 J respectively. An unrestrained person in a collision will

stop when they hit the windscreen over a distance of about 1 cm while their face squashes in. A person wearing a seatbelt will stop over a distance of about 10 cm as the seatbelt stretches. A person wearing a seatbelt with crumple zones at the front of the car will stop over a distance of about 30 cm as the seatbelt stretches and the front of the car crumples. a Find the value of k for a collision at 60 km/h for

each scenario.

b Find an expression for the force exerted on a person’s body for each scenario, assuming that it is zero at the beginning of the collision.

c Find the maximum force exerted on a person’s

body in each case, and comment on your results.

dC dn

---3E

D3

■ The reverse of the process of differentiation is integration.

■ If f′(x) = axn_{, then f (x)} = _{(where n} ≠−_{1) is a possible original function.}

■ The indefinite integral or antiderivative of the function f (x) is F(x) + c. The indefinite

integral is represented by the symbol , so F(x) = .

■ Possible original functions may be graphed using the derivative.

■ If f (x) is the rate of change of F(x), then f (x) is the derivative of F(x) and F(x) is an indefinite integral of f (x).

■ The velocity of an object is the rate of change (derivative) of its displacement, so the displacement of an object must be an antiderivative of its velocity. Similarly, acceleration is the derivative of velocity, so the velocity of an object must be an antiderivative of its acceleration.

a xn+1

n+1

Chapter review

Knowledge and procedures

1 Find two antiderivatives of:

a f(x) = 4x2 _{– 2x} + _{5 b h(x)} = _{3 – 4x}5

2 =

A 36x2+ _{c B 6x}2+ _{c C 12x}4+ _{c D 4x}4+ _{c E 3x}4+ _c

3 =

A 5x3 _{– 10x}2+ _4x + _{c B x}3 _{– 10x}2+ _4x + _{c C x}3+ _10x2 _{– 4x} + _c

D x3 _{– 10x}2+ _4x + _{c E x}3+ _10x2+ _4x + _c

4 The function f(x) = 12x2− _10x + _{2 has an antiderivative:}

A 12x3− _10x2+ _{2x B 4x}2− _5x + _{2 C 4x}3− _5x2+ _2x + ₇ D 6x3− _5x2+ _2x + _{7 E 12x}3− _10x2+ _2x + ₇

5 Find y if:

a = 7x – 3x2 _{b y}′= _5x4+ _2x

6 Draw three different antiderivatives of:

a f (x) = 3x – 5 b g(x) = 2 – x2

7 Draw the graphs of two antiderivatives of f (x) = 3x2 _{– 2x – 4 using a graphics} calculator and sketch your result.

8 If F(x) has a zero at −1 and a derivative f (x) = 3x2− _{3, then F(x) is:}

A 3x2− _{3 B 3x}3− _3x + _{2 C x}3− _{3x D 3x}3− _{3x E x}3− _3x − ₂

Modelling and problem solving

9 m′(x) = 3x − 2x2+ _{4 and m(x) has a zero at x} = _{5. Find the y-intercept of m(x), explaining} your steps.

10 Sketch a possible antiderivative of the gradient function shown on the right.

11 The marginal cost of production of fancy chairs is given by

= 400 − 2x

where x is the number of chairs and C (in dollars) is the daily running cost. a Find an antiderivative of the marginal cost.

b It costs $1500 a day just to keep the factory open. Use this to find C(x), the cost of production of x chairs.

c Find the costs of producing 140 chairs.

d The maximum daily production is 180 chairs. Find the cost of production of 180 chairs. e Compare the unit costs of production for 140 and 180 chairs.

f Write an expression for the unit cost of production of x chairs.

g Use your graphics calculator to find the lowest unit cost of production.

Ex 11.1

12 x3_dx

∫

Ex 11.1

5 x–2

( )2_dx

∫

Ex 11.1

5 3

--- 25

3

---25 3

--- 15

3

---Ex 11.1 Ex 11.1

Ex 11.1

dy dx

---Ex 11.2

Ex 11.2

Ex 11.3

Ex 11.2

y

x f (x)

Ex 11.3

dC dx