Collins Cambridge International AS and A Level Mathematics sample

7

7

INTEGRATION

Mathematics in life and work Together, differentiation and integration form a branch of mathematics called calculus.

Integration deals with properties of quantities that vary over time and has applications in a very wide range of careers. These include:

›› If you were an electronic engineer you could use integration to find the voltage across a capacitor or the quantity of change stored in it. Capacitors are found in many electronic circuits.

›› If you were a car designer you could use integration to calculate a number called the severity index to assess the severity of likely head injuries in a car crash. Your aim would be to make this as small as possible.

›› Oil, water and other liquids are stored in barrels with curved sides and a varying circular cross-section. If you were a manufacturer you might use integration to find the dimensions of a barrel with a specified volume.

LEARNING OBJECTIVES You will learn how to:

›› understand integration as the reverse process of differentiation ›› integrate (ax + b)n for rational values of n (except –1) ›› solve problems involving the evaluation of a constant of integration ›› evaluate definite integrals ›› find areas bounded by curves and the coordinate axes ›› use definite integration to find a volume of revolution LANGUAGE OF MATHEMATICS Key words and phrases you will meet in this chapter:

›› constant of integration, definite integral, indefinite integral, integrand, integration, solid of revolution, volume of revolution

M

Modelling

PS

Problem solving

PS

Proof

CM

Communicating mathematically

7 Integration

PRIOR KNOWLEDGE You should already know how to:

›› differentiate axn for any values of a and n, and differentiate expressions formed by adding and subtracting terms like this

›› manipulate algebraic expressions involving brackets and powers, including fractional and negative powers

›› sketch the graphs of quadratic and cubic expressions.

You should be able to complete the following questions correctly: 4 1 Write in index form. x 2 Sketch the graph of f(x) = –(x – 3)(x + 5). dy 3 Find for the following (x > 0): dx x2 + 4 a y = 4x3 – 6x + 10 b y = c y = 63x 2x

7.1 Indefinite integrals

y 5

dy For a particular curve, = 2x. This means that the dx gradient at the point (x, y) is 2x.

What is the equation of the curve?

One possible equation is y = x2 However, this is not the only possible answer.

4

Remember from Chapter 6 Differentiation that if y = xn dy then = nxn–1. dx In this case n = 2.

It could also be y = x2 + 1 or y = x2 + 10.5 or y = x2 – 3. dy For each of those, = 2x because the derivative of the dx final constant is zero.

3

2

1

–3

In fact, y = x2 + c is a solution where c is any constant you like. dy There is a whole family of curves for which = 2x . dx Here are five of them.

–2

–1

0

1

2

3

x

–1

–2

–3

All these curves have the same gradient for a particular value of x.

For example, if x = 2 they all have a gradient of 4.

We call x2 + c the integral of 2x.

The integral of 2x is the function we differentiate to get 2x. There is a special notation to express this.

180

It is sometimes called the indefinite integral because the value of c is not fixed.

Indefinite integrals

7.1

We write ∫ 2x dx = x2 + c, which is read as ‘the integral of 2x is x2 + c’.

More generally, the derivative of xn+1 is (n + 1)xn .

1 n+1 x , where n ≠ −1.

That means that the integral of xn is n+1 To integrate a power of x:

›› increase the index by 1 ›› multiply by the reciprocal of the new index.

KEY INFORMATION 1 n+1 x , n+1 where n ≠ −1

∫ xn dx =

Stop and think

If n = −1, what is the value of n + 1?

Why does the formula above include the restriction n ≠ −1?

Example 1 f’(x) = 6x2 –

2 x2

Find f(x).

Solution

2 dx x2 = ∫ 6x2 – 2x–2 dx 6 2 = x3 – x–1 + c 3 –1 = 2x3 + 2x–1 + c 2 = 2x3 + + c x f(x) = ∫ 6x2 –

As you can see in Example 1, if you have to integrate the sum or difference of two terms, you just integrate each term separately.

Example 2 Find ∫

(x + 2)(x – 2) dx for values of x > 0. x

Solution Multiply out the brackets. (x + 2)(x – 2) x2 – 4 dx = ∫ dx ∫ x x (x + 2)(x – 2) x2 – 4 dx = ∫ dx ∫ x x Divide by x 2

–1

Integral = ∫ x3 – 4x 2 dx 2 5 4 1 = x2 – 1 x2 + c 5 2 1 2 5 = x2 – 8x2 + c 5

181

7 Integration

Exercise 7.1A 1 Find: a ∫ 6xdx

b ∫ 4x – 2dx

c ∫ 7x – 5dx

d ∫ 3 – 4xdx

b ∫ 2x–4 dx

c ∫ 5x3 dx

d ∫

5 dx x3

b ∫ 2x4 – 5x + 10 dx

c ∫ 8x – 10x2 dx

b ∫ 2x3dx

c ∫ 3x–1dx

d ∫

10 dx x

b ∫ x2(x – 8) dx

c ∫

b ∫ 4x dx

4 c ∫ dx x

2 Find: a ∫ 2x4 dx 3 Find: a ∫ 6x2 – 4x dx 4 Find: a ∫ xdx

2

2

5 Find: a ∫ x(x – 4) dx 6 Find: a ∫ 4 x dx 7 f’(x) = 2x3 – 2x

10 dx x3

Find an expression for f(x).

C

8 A student writes this incorrect solution to ∫ (2x + 1)(x – 1) dx. ∫ (2x + 1)(x – 1) dx = ∫ 2x + 1 dx × ∫ x – 1 dx 1 2 a Explain the error the student has made.

= (x2 + x) ( x2 – x) + c

b Find the correct integral.

9 Find: 4x3 – 2 dx x3 PS 10 f’(x) = g’(x)

a ∫

C

b ∫

2x + 8 dx x

c ∫

4+2 x dx x2

What can you say about f(x) and g(x)?

7.2 Integrating (ax + b)n If f(x) = (3x – 5)4 then you can differentiate this using the chain rule. f’(x) = 4(3x – 5)3 × 3 = 12(3x – 5)4 Notice that the index after the bracket has decreased by 1 and there is now a constant in front of the bracket.

Suppose you want to find the indefinite integral ∫ (3x – 5)4 dx You can start with the expression f(x) = (3x – 5)5 where the index is increased by 1.

182

C

Communication

MM

Mathematical modelling

PS

Problem solving

Integrating (ax + b)n

In this case f’(x) = 5(3x – 5)4 × 3 = 15(3x – 5)4 1 This is the integrand multiplied by 15 so ∫ (3x – 5)4 dx = (3x – 5)5 + c 15 1 The will cancel the 15 that results when you differentiate (3x – 5)5 15 If you cannot write down an integral immediately you can often guess the general form and then check by differentiating.

7.2

The integrand is the function being integrated.

Example 3 Find ∫

1 dx 4x – 2

Solution 1 1 –1 = (4x – 2) 2 so start by considering y = (4x – 2)2 4x – 2

where the index has been increased by 1.

–1 –1 dy 1 = (4x – 2) 2 × 4 = 2(4x – 2) 2 dx 2 1 1 1 Hence ∫ dx = (4x – 2)2 + c 4x – 2 2

Then

Exercise 7.2A 1 Find: a ∫ (x + 7)2dx

b ∫ (x + 7)3dx

c ∫ (x + 7)5dx

b ∫ (6x + 1)3dx

c ∫ (0.5x – 4)4dx

2 Find: a ∫ (2x – 3)2dx

1 dy find 10x + 1 dx 3 b Hence find ∫ dx (10x + 1)2 4 Find:

3 a If y =

a ∫ x + 1 dx

b ∫ 2x + 1 dx

c ∫ 3 4x – 2 dx

5 Find: a ∫ 3 0.6x + 5 dx

1 dx 0.6x + 5

b ∫ 3

6 Find ∫ x(x2 + 2)2dx 7 Find ∫ x2 x3 + 4 dx 8 Find ∫ (x – 1) x2 – 2x dx 9 Find: a ∫

6 dx 2x + 5

b ∫

6x dx 2x2 + 5 183

7 Integration

7.3 Finding the equation of a curve dy , for a particular curve. dx To find the equation of the curve, you need more information.

Suppose you know the derivative,

You also need to know a point on the curve.

Example 4 y = f(x) is the equation of a curve and f’(x) = 3x2 + 4x – 3.

The curve goes through the point at (2, 4).

Find the equation of the curve.

Solution y = ∫ 3x2 + 4x – 3 dx y = x3 + 2x2 – 3x + c To find the value of c, put the coordinates (2, 4) into this equation: 4=8+8–6+c 4 = 10 + c c = −6 The equation of the curve is y = x3 + 2x2 – 3x – 6

Exercise 7.3A 1 A curve passes through (0,3) and

dy = 4x – 2. dx

a Find an expression for y. b Find the equation of the curve.

2 For a curve

dy = x. x > 0 dx

a Find an expression for y.

The point (9, 25) is on the curve. b Find the equation of the curve. dy = 0.4x + 3. dx Find the equation of the curve if it passes through:

3 The gradient of a curve, a (0, 0) b (0, 5)

c (5, 0)

4 The equation of a curve is y = f(x) and f’(x) = The curve passes through (5, 2).

Find the equation of the curve.

184

x2 + 10 x2

Finding the equation of a curve

PS

5 The gradient of a curve is given by

7.3

dy 2 = x ≠ 0 dx x2

a Show that the gradient is always positive, The curve passes through the point (2, 4).

5x – 2 . x dy 6 This graph shows a curve with = 3x2 – 3 that passes through (0, 2). dx

b Show that the equation of the curve is =

y 5 4 3 2 1 –3

–2

0 –1 –1

1

2

3

x

–2

a Find the equation of the curve. b Show that (−1, 4) and (1, 0) are the only two stationary points on the curve.

7 A curve has an equation of the form y = f(x) x > 0.

20 and the curve passes through the point (2, 4). x3 a Find the equation of the curve.

f’(x) =

b Sketch the graph of the curve.

MM

8 A straight road passes through points O and A and OA = 100 metres.

Initially, a car is at A and starts driving along the road away from O. O

A

100 m

The distance of the car from O after x seconds is y metres.

After x seconds the speed of the car, dy dx

dy = 0.03x2 0 ≤ x ≤ 10 dx

20 15 10 5 0

2

4

6

8

10

x

a Find a formula for y in terms of x. b Show that after 10 seconds the car is 200 m away from A. PS

dy = 3x2 – 12x + 8 dx Find the coordinates on the points where the curve crosses the x-axis.

9 A curve passes through the origin and

185

7 Integration

10 A spherical balloon is being inflated.

The radius after x seconds is y centimetres.

–2 dy = 4x 3 dx After 8 seconds the radius is 30 centimetres.

a Find an equation for y in terms of x. b Find the radius after 20 seconds.

Mathematics in life and work: Group discussion Capacitors are electronic devices that store electrical charge. They are very commonly used in electronic circuits. Capacitors have a property called capacitance, C, measured in Farads.

A capacitor is charged by applying an electric current, i, and the capacitor will then have a voltage, V. 1 The voltage is given by the formula V = ∫ i dt c 1 The voltage is initially 0 and a constant current of value k is applied. Find a formula for the current after t seconds.

2 The voltage is initially V0 and an increasing current of value 0.02t is applied. Find a formula for the current after t seconds.

3 The voltage is initially and an increasing current of value 100t(0.01 – t) is applied for 0.01 seconds.

Find the current at the end of this time.

7.4 Definite integrals Here is the graph of y = 0.3x2 + 1

y 5 4 3 2 1 –3

–2

–1

0

1

2

3

4

5

x

Suppose you want to find the shaded area. This is the region under the curve between x = 1 and x = 3.

Draw a vertical line at some point x on the x-axis.

y

A(x) 0

186

1

x

x