MEI C2 Study Resources Core2 Integration 1-2

MEI Core 2 Integration Section 1: Introduction to integration Study Plan Background This section is about reversing the process of differentiation. i.e. when you are given the gradient of a function, you use integration to work out what curve or family of curves have that gradient. A direct application could be in finding velocity when given acceleration or displacement when given velocity.

Detailed work plan 1. Read the explanation at the beginning of the chapter (pages 234-236) taking note of why the arbitrary constant is necessary. It is all too easy to forget it at times but that can lead to difficulties when solving practical problems. Make a note of the rule on page 235 and keep it handy when working through the questions. If in doubt, differentiate your answer and check that you get back to the function you started with. There are some additional examples in the Notes and Examples. 2. For extra examples, use the Flash resources Basic indefinite integration: single powers of x, Basic indefinite integration: sums of powers of x and Reverse of differentiation. Some notation is used here (explained in the Notes and Examples) which has not yet been introduced by the textbook. 3. Exercise 9A Attempt questions 1*, 3, 4*, 6*, 8, 9, 10*. You can also try the Indefinite integrals puzzle. 4. For extra practice, try the interactive questions Indefinite integration of polynomials.

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MEI Core 2 Integration Section 1: Introduction to integration Notes and Examples These notes contain subsections on:  Reversing differentiation  The rule for integrating xn  Finding the arbitrary constant  An application: the motion of a particle

Reversing differentiation Integration is the reverse of differentiation. If you are given an expression f or dy , and you want to find an expression for y, you need to use integration. dx This is sometimes called solving a differential equation. Remember that when you integrate, you must always add an arbitrary constant (see the textbook for the explanation of this).

Example 1 shows how you can integrate a function by thinking about what function you would need to differentiate to obtain the given function.

Example 1 Find the general solution of the following differential equations: dy dy dy  x3  x6 t (i) (ii) (iii) dx dx dt df ds du  2x2  3t 4 5 (iv) (v) (vi) dx dt dv Solution dy  x3  y  14 x 4  c (i) dx

4

3

The derivative of x is 4x . 3

4

So integrating 4x gives x . 3

Therefore integrating x gives

(ii)

dy  x 6  y  17 x 7  c dx

7

1 4

x4 .

6

The derivative of x is 7x . 6

7

So integrating 7x gives x . 6

Therefore integrating x gives

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x7 .

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MEI C2 Integration Section 1 Notes and Examples (iii)

dy  t  y  12 t 2  c dt

2

The derivative of t is 2t . 2

So integrating 2t gives t . 1 2 2

Therefore integrating t gives

(iv)

t .

df  2 x 2  f  23 x3  c dx 2

3

The derivative of x is 3x . 2

3

So integrating 3x gives x . 2

1 3 3

Therefore integrating x gives 2

and integrating 2x gives

(v)

t ,

2 3 . 3

t

ds  3t 4  s  53 t 5  c dt 5

4

The derivative of t is 5t . 5

4

So integrating 5t gives t . 4

Therefore integrating t gives 4

and integrating 3t gives

(vi)

du  5  u  5v  c dv

1 5 5

t ,

3 5 5

t .

The derivative of v is 1. So integrating 1 gives v. Therefore integrating 5 gives 5v.

The rule for integrating xn The method for integrating any polynomial function can be summed up as: x n 1  Integrating x n , where n is a positive integer, gives n 1 n  Integrating kx , where n is a positive integer and k is a constant, kx n 1 gives n 1  You can integrate the sum of any number of such functions by simply integrating one term at a time.

Example 2 Integrate each of the following gradient functions. dy  x3  3x  2 (i) dx

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MEI C2 Integration Section 1 Notes and Examples (ii) (iii)

dy  4 x2  5x 1 dx dy  ( x  3)( x  2) dx

Solution dy (i)  x3  3x  2 dx  y  14 x 4  32 x 2  2 x  c dy (ii)  4x2  5x 1 dx  y  43 x3  52 x 2  x  c dy (iii)  ( x  3)( x  2) dx  x2  x  6  y  13 x3  12 x 2  6 x  c

Remember the arbitrary constant

You can see further examples using the Flash resources Basic indefinite integration: single powers of x and Basic indefinite integration: sums of powers of x. Note that these resources use notation for integration which is not used in the textbook until a little later: for now you just need to know that  f ( x) dx means “integrate the function f(x)”. You can also practise integration using the interactive questions Indefinite integration of polynomials. Note that “w.r.t. x” means “with respect to x” and just means that x is the variable that you are using. In these questions, x is not always the variable being used, but you integrate in exactly the same way whatever letter is used. You could also try the Indefinite integrals puzzle. This also uses the notation  f ( x) dx to mean “integrate the function f(x).

Finding the arbitrary constant If you are given additional information, you can find the value of the arbitrary constant by substituting the given information. This is sometimes called finding the particular solution of a differential equation. The next example shows how this is done.

Example 3 The gradient of a curve at any point (x, y) is given by

dy = x2(2x + 1). dx

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MEI C2 Integration Section 1 Notes and Examples The curve passes through the point (1, 5). Find the equation of the curve. Solution dy = x2(2x + 1) = 2x3 + x2 dx Integrating: y  2  14 x 4  13 x3  c

Just as with differentiating, you need to expand the brackets first

 12 x 4  13 x3  c When x = 1, y = 5

5=

 13  c  c = 5  12  13 = 1 2

25 6

So the equation of the curve is y = 12 x 4  13 x3  256

Substitute the given values of x and y

You can see more examples like the one above using the Flash resource Reverse of differentiation.

An application: the motion of a particle One very useful application of integration and differentiation is in the relationships between the displacement, velocity and acceleration of a particle. You are not required to know these relationships for this component, but if you study Mechanics you may have already looked at them. The velocity, v, of a particle is the derivative of its displacement, s, with respect to time, t. ds v dt So you can find the displacement by integrating the velocity. The acceleration, a, of a particle is the derivative of its velocity with respect to time. dv a dt So you can find the velocity by integrating the acceleration.

Example 4 The velocity of a particle t seconds after leaving a point P is given by: v = 3t + 1 Find the displacement of the particle from the point P after 3 seconds. Solution The velocity is the derivative of the displacement with respect to t. So: ds = 3t + 1 dt Integrate to find s in terms of t.

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MEI C2 Integration Section 1 Notes and Examples 

s = 32 t 2  t  c

The particle passes through P when t = 0, so substituting t = 0 and s = 0: 0 = 32  02  0  c  c = 0. 

s = 32 t 2  t

To find the displacement after 3 seconds, substitute t = 3 into this equation: s = 32  32  3 = 16.5 m.

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Core 2 Integration Section 1: Introduction to integration Crucial points 1. Don’t muddle up the formulae for differentiation and integration dy = x 3 , find y Example: Given that dx I’m being asked to differentiate

"

Wrong

! Right

dy = x3 ⇒ y = 3x 2 dx

"

"

dy x4 = x3 ⇒ y = + c 4 dx

3

!

The derivative of y is x , so to find y I need to integrate

!

2. Remember the arbitrary constant c Always remember to put in the arbitrary constant – you will lose marks in an examination if you miss it out!

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MEI Core 2 Integration Section 1: Introduction to integration Exercise 1.

dy = 3x2 – 4. dx (i) Find an expression for y in terms of x. (ii) Find the particular curve that passes through the point (2, -1). (iii) Show that this curve also passes through the point (1, -4). A curve has gradient function

dy = 4x – x2. Find the equation of dx the curve given that it passes through the point (3, 2).

2. The gradient function of a curve is given by

3. The gradient of a curve at the point (x, y) is given by 4(1 – x). Given that the curve has a maximum value of 8, find the equation of the curve. dh = 25 – 10t, where t is the time dt in seconds and h is the height of the stone in metres. Given that when t = 0, h = 30, find the value of t for which h = 0.

4. A stone is thrown vertically upwards such that

5. Find y in terms of x given that

dy = (x + 1)2 and that y = 0 when x = 2. dx

6. The acceleration of a particle is given by

dv = 3t2 – 4t – 7. Given that v = 36 dt

when t = 5, show that v = (t + 1)2(t – 4). 7. The acceleration of a particle is given by

dv = 6t – 3t2. Given that the particle dt

starts from rest: (i) Find an expression for the velocity v in terms of the time t. (ii) For what value(s) of t will the particle be at rest? dy = 4x2 + x. dx (i) Find the equation of the curve given that y = 2 when x = 1. (ii) Find the value of y when x = 3.

8. The gradient function of a curve is given by

9. Find an expression for y in terms of x if

dy = (x – 1)(3x – 5) and y = 2 when x = 1. dx

dy = 4x2 – 1 has a local minimum value of 1. Find dx the equation of the curve and the co-ordinates of the local maximum value.

10. A curve with gradient function

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MEI Core 2 Integration Section 1: Introduction to integration Solutions to Exercise 1.

(i)

dy

= 3x 2 − 4 dx y = 3 × 31 x 3 − 4 x + c = x 3 − 4x + c

(ii) When x = 2, y = -1 −1 = 2 3 − 4 × 2 + c −1 = 8 − 8 + c c = −1 Equation of curve is y = x 3 − 4 x − 1

(iii) When x = 1, y = 1 3 − 4 × 1 − 1 = 1 − 4 − 1 = −4 so the curve passes through the point (1, -4).

2.

dy

= 4x − x 2 dx y = 4 × 21 x 2 − 31 x 3 + c = 2 x 2 − 31 x 3 + c When x = 3, y = 2 2 = 2 × 3 2 − 31 × 3 3 + c

2 = 18 − 9 + c c = 2 − 18 + 9 = −7 y = 2 x 2 − 31 x 3 − 7

3.

dy dx

= 4(1 − x ) dy

=0 ⇒ x =1 dx So the curve passes through the point (1, 8). dy = 4(1 − x ) = 4 − 4 x dx y = 4x − 2 x 2 + c When x = 1, y = 8 At maximum point,

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MEI C2 Integration Section 1 Exercise solutions 8 = 4−2 +c c =6 The equation of the curve is y = 4 x − 2 x 2 + 6 .

4.

dh = 25 − 10t dt h = 25 t − 5 t 2 + c When t = 0, h = 30 ⇒ 30 = c h = 25 t − 5 t 2 + 30 When h = 0, 25 t − 5 t 2 + 30 = 0 5t − t 2 + 6 = 0

t 2 − 5t − 6 = 0 (t − 6)(t + 1) = 0 t = 6 o r t = −1

Since t must be positive, the value of t must be 6.

5.

dy

= ( x + 1)2 = x 2 + 2 x + 1

dx y = 31 x 3 + x 2 + x + c

When x = 2, y = 0 0 = 31 × 2 3 + 2 2 + 2 + c 0 = 83 + 4 + 2 + c

c = − 263 y = 31 x 3 + x 2 + x − 263

6.

dv = 3t 2 − 4t − 7 dt v = t 3 − 2t 2 − 7t + c When t = 5, v = 36 ⇒ 36 = 5 3 − 2 × 5 2 − 7 × 5 + c 36 = 125 − 50 − 35 + c c = −4 3 2 v = t − 2t − 7t − 4 = (t + 1)(t 2 − 3t − 4) = (t + 1)(t + 1)(t − 4) = (t + 1)2(t − 4)

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MEI C2 Integration Section 1 Exercise solutions 7. (i)

dv = 6t − 3t 2 dt v = 3t 2 − t 3 + c When t = 0, v = 0 ⇒ c = 0 v = 3t 2 − t 3

(ii) When v = 0, 3t 2 − t 3 = 0

t 2(3 − t ) = 0 t = 0 or 3

The particle is at rest when t = 0 and when t = 3.

8. (i)

dy

= 4x 2 + x

dx y = 43 x 3 + 21 x 2 + c When x = 1, y = 2 2 = 43 × 1 3 + 21 × 1 2 + c

c = 2 − 43 − 21 = 61 y = 43 x 3 + 21 x 2 + 61 (ii) When x = 3, y = 43 × 3 3 + 21 × 3 2 + 61 = 36 +

9 2

+ 61

= 40 23

9.

dy

= ( x − 1)(3 x − 5 ) = 3 x 2 − 8 x + 5

dx y = 3 × 31 x 3 − 8 × 21 x 2 + 5 x + c = x 3 − 4x 2 + 5 x + c When x = 1, y = 2 2 = 13 − 4 × 12 + 5 × 1 + c

c = 2 −1+4−5 = 0 y = x 3 − 4x 2 + 5 x 10. At turning points, 4 x 2 − 1 = 0 (2 x − 1)(2 x + 1) = 0

x= For x < − 21 ,

dy dx

1 2

or − 21

>0

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MEI C2 Integration Section 1 Exercise solutions For − 21 < x < 21 , For x > 21 ,

dy

dy dx

<0

>0 dx Therefore there is a maximum point where x = − 21 and a minimum point where x = 21 . So the graph passes through the point ( 21 , 1 ) . dy = 4x 2 − 1 dx y = 43 x 3 − x + c When x = 21 , y = 1 ⇒ 1 = 43 ( 21 )3 − 21 + c ⇒ 1 = 61 − 21 + c ⇒ c = 43 The equation of the curve is y =

4 3

x 3 − x + 43 .

The maximum point is when x = − 21 . y = 43 ( − 21 )3 − ( − 21 ) + 43 = − 61 + 21 + 43 =

5 3

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Integration Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must match each integral with its solution. To build up the puzzle, place together the edges on which an integral and its solution are written, so that the triangles are joined along this edge. Begin by just finding pairs of integrals and solutions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

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∫9

x

− 10x + c

8

) +1

dx

x 3 ( 2x

+

2 9 x dx ∫

dx

4

+c

dx

4x − 4 x 2 + c

dx

1 3 x − x2 +c 3

∫

3

10

dx

d 1)

x

∫

1 3 x 3− x+ c

3 dx x − ∫

−4x +c

2

x

∫4x

(x

∫2x

2

x

2x

)(x 1 −

1 x 2

x

∫3

x 2 + x 11 + c

∫

dx

+c

4

dx

2

x−

∫x(x − 1) dx

∫ −x

1 3 x

x

3

+c

5

∫ − 10 dx dx

2 − 4(1

d x)

x

∫ (2x

+1 1

x

)d x

10

1 4 − x +c 4

∫

1

∫

5

dx

c

3 x 3+ c

∫x(x + 2) dx

dx

4x 6

dx

4x

4

∫ −1

−4 x+

2 x ∫ dx

∫1

x9 +c

∫ ( x +

1 ) dx

3

d 1)

c + x

∫20 x

∫

+c

dx

x − x2 +c

+ x ( x

3

∫2 x

x

x− 2)(

2)

dx

3

x7 +c

x+

dx

3

2 11 x +c 11

∫(

+c

3

3x

∫ x ( x

6x −1

∫5x( x

c

−2

∫

)d x 1 3 x + x2 +c 3

5x +

1 3 1 2 x − x +c 3 2

dx

+c

x

5

+x

∫8x

7

1 x − 2

+c

6

7x

∫

2

−5

x

+c

2

dx 1 1 x3 + x2 +c 3 2

2

x

1 x 2

+c

+

4

8

∫2x dx

x

3

x

−

2x

+c

+c

− 2) dx

7

2x

2

+c

MEI Core 2 Integration Section 1: Introduction to integration Multiple Choice Test 1. If

dy = x10 , then dx

(a) y = 11x11 + c

(b) y = 101 x11 + c

(c) y = 111 x11 + c (e) I don’t know

(d) y = 10 x9 + c

2. Given that

dy = 1 + 3x 2 , then dx

(a) y = x3 + c (c) y = x + x3 (e) I don’t know

3. Given that

(b) y = x + x3 + c (d) y = 6x + c

dy = 2 x5 − 4 x , then dx

(a) y = 10 x 4 − 4 + c

(b) y = 16 x 6 − 12 x 2 + c

(c) y = 15 x 6 − 4 x 2 + c (e) I don’t know

(d) y = 13 x 6 − 2 x 2 + c

4. Given that

dy = (1 + 3 x) 2 , then dx

(a) y = 6 + 18 x

(c) y = x + 3x 3 + c (e) I don’t know

(b) y = x + 3x 2 + 3x 3 (d) y = x + 3x 2 + 3x 3 + c

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MEI C2 Integration Section 1 MC test 5. If

dy ( x 2 + 2 x3 ) = , then dx 2x

(a) y = 14 x 2 + 13 x3 + c

2 x3 + 3x 4 +c 6 x2 (d) y = 12 x + 13 x3 + c (b) y =

x3 + 12 x 4 (c) y = x2 (e) I don’t know 1 3

6. Given that

dy = x 2 and y = 2 when x = 1, then dx (b) y = 2x − 3 (d) y = 13 x 3

(a) y = 2x (c) y = 13 x 3 + 53 (e) I don’t know

7. Given that

dy = 2 x 3 − x − 5 and that y = -1 when x = 2, the value of y when x = 1 dx

is (a) -6 (c) -5 (e) I don’t know

(b) -2 (d) 2

dy = 3x 2 − 2 x + 1 and passes through the point dx (2, 5). The equation of the curve is

8. A curve has gradient function

(a) y = 6 x − 2

(b) y = x 3 − x 2 + x + c (d) y = x 3 − x 2 + x + 5

(c) y = x 3 − x 2 + x − 1 (e) I don’t know

dy = 3x − 2 and passes through the point (2, 0). dx The point on the curve where x = −1 has y-coordinate

9. A curve has gradient function

(a) − 32 (c) −5 (e) I don’t know

(b) (d)

3 2 7 2

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MEI C2 Integration Section 1 MC test dy = x 2 − x + 1 and passes through the point dx (-1, 1). The point on the curve where x = 2 has y-coordinate

10. A curve has gradient function

(a) 112 (c) 113 (e) I don’t know

(b) (d)

9 2 17 6

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MEI Core 2 Integration Section 2: Finding the area under a curve Study Plan Background This section looks at how integration can be used to find exact values of the area bounded by a curve. At G.C.S.E. you had to use approximate methods to do this which could be cumbersome and not perhaps as accurate as would be necessary in real life. However, there are some situations where it may not be possible to find an area using integration. In this section you will learn the trapezium rule, which is one method that can be used to obtain an approximation to the area. The trapezium rule relies heavily on being organised and precise so it is important that you be neat and accurate.

Detailed work plan 1. Read pages 239-241 paying attention to the notes on standardising the procedure on page 241. The following pages (242-247) formalise the process and introduce you to the notation. It is important to remember that indefinite integrals need an arbitrary constant in the answers but with definite integrals the arbitrary constant is eliminated. There are examples of both indefinite and definite integrals using formal notation in the Notes and Examples. 2. You can see additional examples using the Flash resources Definite integration and Areas under a curve above the x-axis. 3. Exercise 9B It is worth doing at least half of the parts of each of questions 1 and 2, then 4, 6*, 8, 9*, 11*, 13 You can also try the Definite integrals puzzle. 4. For extra practice try the interactive questions Definite integration of polynomials. 5. Read pages 250-252. As with many areas of mathematics there are some pitfalls which you need to be aware of when finding areas under curves. Example 9.8 illustrates one of these, and there is a further example in the Notes and Examples. 6. You can try further examples using the Flash resource Areas above and below the x-axis. You may also find the Mathcentre video Finding areas using integration useful. Note that this video may contain examples which use some integration techniques that you have

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MEI C2 Integration Section 2 Study plan not yet met. 7. Exercise 9C Attempt questions 1 (odd parts) and 2*. 8. The next two sections (pages 254-260), including the two exercises, are outside the syllabus. It is worth reading through this section, and you may like to try a few of the questions from Exercises 9D and 9E as extension work. You can also look at the Flash resource Area between a line and a curve. 9. The last part of the chapter covers the trapezium rule, which allows you to find the area under a curve when integration is not possible. Read the section (pages 260-264) and make a note of the rule on page 263. Keep it to hand until you are thoroughly familiar with the process. There is a further example in the Notes and Examples. 10. You can also look at further examples using the Flash resource The trapezium rule. You can also use the Geogebra resource The trapezium rule to see the result of applying the trapezium rule with different numbers of strips. 11. Exercise 9F Attempt the odd numbered questions including 3* and 7*

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MEI Core 2 Integration Section 2: Finding the area under a curve Notes and Examples These notes contain subsections on:  Indefinite integration: formal notation  Definite integration  The definite integral as an area  The trapezium rule

Indefinite integration: formal notation This section introduces the formal notation for integration. In Section 1 you dy were given expressions for and asked to find an expression for y. dx So you would write: dy  2x  y  x2  c . dx Using the formal notation, you would write this as: 2  2 x dx  x  c

You would read this as “the integral of 2x with respect to x”

The example below is the same as Example 2 from section 1, but is expressed in formal notation.

Example 1 Integrate each of the following functions. x3  3x  2 (i) 4 x2  5x  1 (ii) ( x  3)( x  2) (iii)

Remember the arbitrary constant

Solution 3 4 2 (i)   x  3x  2 dx  14 x  23 x  2x  c (ii) (iii)

  4x  5x 1dx  x  x  x  c  ( x  3)( x  2) dx    x  x  6 dx  2

4 3

3

5 2

2

2

1 3

x3  12 x 2  6 x  c

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MEI C2 Integration Section 2 Notes and Examples Definite integration The definite integral from a to b of a function f(x), which is written as



b a

f ( x)dx , is found as follows:  

Integrate f(x) – suppose we call the integral g(x) Write the integral in square brackets, with the limits on the right hand b side:  g( x)a



Work out the value of g(x) with x = a and x = b, and subtract: b g( x)a  g(b)  g(a) .

Example 2 Evaluate



2 1

(2 x  x 2 )dx .

Solution

  2 x  x  dx   x 2

1

2

2

 13 x 3 

  2  13  2 2

2

1

3

  1

2

 13  13 

Watch the signs here

 4  83  1  13 

2 3

For additional examples like this one, look at the Flash resource Definite integration.

For practice in examples like the one above, try the interactive resource Definite integration of polynomials.

You could also try the Definite integrals puzzle.

The definite integral as an area f(x)

The definite integral



b a

f ( x)dx calculates the area

between the curve y = f(x) and the x-axis. The formal proof of this result is given on pages 242 – 244 of the textbook.

a

b

Note that you are not expected to be able to reproduce this formal proof – you need only have a general understanding of the ideas involved.

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MEI C2 Integration Section 2 Notes and Examples If the curve is above the x-axis, so that the value of y is positive, the definite integral works out to be positive. However, if the curve is below the x-axis, so that y is negative, the integral works out to be negative. It is important that you have a sketch of the curve before you use integration to work out areas like this. If the curve crosses over the x-axis, then the integral subtracts any negative area from a positive area, and therefore calculates the difference of the two areas above and below the x-axis.

Example 3



2

(3x 2  4 x)dx = 0, and interpret the result.

(i)

Show that

(ii)

Find the total area enclosed by the curve, the x-axis and the line x = 2.

0

Solution: (i)



2

2

(3x 2  4 x)dx   x3  2 x 2  0 0

  2 3  2  2 2    03  2  0 2 

0 Since the integral is zero, there must be equal areas above and below the axis. (ii)

Sketch the graph:

8

-1

0

1

2

-2

The curve crosses the x-axis when 3x2  4x = 0  x(3x  4) = 0  x = 0 or x = 43 Red area =

4 3

4

2 3 2 3 0 (3x  4 x)dx   x  2 x  0







4 3 3



 2  43    03  2  02  2

  32 27

© MEI, 05/01/11

This integral works out to be negative because the curve is below the x-axis.

3/6

MEI C2 Integration Section 2 Notes and Examples Blue area =



2 4 3

2

(3x 2  4 x)dx   x3  2 x 2  4 3

  23  2  2 2   





4 3 3

 2  43 

2



32 27

64 So the total area enclosed = 2 32 27  27 square units.

You can test yourself using the Flash resources Areas under a curve above the x-axis and Areas above and below the x-axis. You can move the curve around to get different functions, and move the limits of the area to be found. Use integration to work out the area and then check your answer.

You may also find the Mathcentre video Finding areas using integration useful. Note that this video may contain examples which use some integration techniques that you have not yet met.

The work in the textbook on finding the area between two curves and the area between the curve and the y-axis, are outside the syllabus. However if you are interested in this extension work you might like to look at the Flash resource Area between a line and a curve.

The trapezium rule Splitting the area into bits gives a method for calculating the approximate area under a curve. Instead of using rectangles, it is more accurate to use trapezia:

O

y0 y1 y2 y3 … a h

yn b

If the trapezia have width h, and the y-coordinates are y1, y2, …, yn, then the total area A is:

© MEI, 05/01/11

4/6

MEI C2 Integration Section 2 Notes and Examples A

h  y0  yn  2  y1  y2  ...  yn1   2

(b  a) . n

where h 

This result is derived on pages 261-263 of the textbook.

Example 4 (i) Use the trapezium rule with: (a) 3 strips and (b) 6 strips, to find the approximate value of (ii)



3 0

Given that the exact value of this integral is 143 , find the relative error in each case, and explain why the trapezium rule is an under-estimate of the true value.

Solution (i) (a)

With 3 strips of width h = 1: x y

0 1

A  12 1  2  2  (b)



1 2

2 3

3 2



2  3   4.646264... 

With 6 strips of width h = 0.5: x y

0 1

A  14 1  2  2  (ii)

1  x dx .

(a)

relative error =

(b)

relative error =

0.5 1.5



1 2

1.5 2.5



2 3

2.5 3.5

3 2

1.5  2  2.5  3  3.5   4.661488...  4.646264  143 14 3

4.661488  143 14 3

= 0.00437 = 0.00111

From the graph of the function, it is clear that the trapezia lie underneath the curve, so the trapezium rule gives an under-estimate.

-1

0

1

2

3

-1

© MEI, 05/01/11

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MEI C2 Integration Section 2 Notes and Examples You can see more examples like this one using the Flash resource The trapezium rule. You can also use the Geogebra resource The trapezium rule to see the result of applying the trapezium rule with different numbers of strips.

© MEI, 05/01/11

6/6

Core 2 Integration Section 2: Finding the area under a curve Crucial points 1. Remember to integrate when calculating a definite integral! Don’t forget to actually carry out the integration. The expression you write in the square brackets must be the integrated function. Example: Evaluate:

"

Wrong

∫

!

Right

∫

1 0

1 0

∫

1 0

( x3 + 1)dx

( x3 + 1)dx =  x3 + 1 = (13 + 1) − ( 0 + 1) = 1

"

( x3 + 1)dx =  14 x 4 + x  = ( 14 + 1) − ( 0 + 0 ) =

5 4

1

0

1

0

!

2. Don’t add ‘+ c’ inside the square brackets Remember, you don’t need an arbitrary constant when evaluating a definite integral.

3. Be careful not to mix up indefinite and definite integrals Definite integrals have limits of integration – numbers on top and bottom of the integral sign, whereas indefinite integrals don’t! Example: 2 3 ∫ x dx = 13 x + c

– Indefinite integral

1

 x3  1 1 x dx =  3  = = 3−0 = 3 ∫0  0 1

2

– Definite integral

4. Make sure that you take account of the graph crossing the x− −axis when calculating an area Example: Find the area enclosed by the curve y = 3x2 − 4x, the x-axis and the line x = 2.

"

Wrong

!

Right

∫

2 0

(3 x 2 − 4 x)dx =  x3 − 2 x 2  = ( 23 − 2 × 22 ) − ( 0 − 0 ) = 0 2

0

"

From the graph, you can see that there are areas above and below the axis.

© MEI, 12/04/05

1/2

Core 2 8

-1

0

1

2

-2

The curve crosses the x-axis where x = 0 or x = Red area =

∫

4 3

0

4

(3x 2 − 4 x)dx =  x 3 − 2 x 2  3 0 =

Blue area =

∫

2 4 3

4 3

((

4 3

)

3

)

32 − 2 ( 43 ) − ( 03 − 2 × 0 2 ) = − 27 2

2

(3x 2 − 4 x)dx =  x 3 − 2 x 2  4 3

= ( 23 − 2 × 2 2 ) −

So the total area enclosed = 2× 32 27 =

64 27

© MEI, 12/04/05

((

4 3

)

3

− 2 ( 43 )

square units.

2

)=

32 27

!

2/2

MEI Core 2 Integration Section 2: Area under a curve Exercise 1. Find the following indefinite integrals. (i) ∫ 4 x3dx

(ii) ∫ ( x3 − 3x 2 )dx

(iii) ∫ (10 x 4 + 3x 2 + 4)dx (iv) ∫ (3x − 1) 2 dx

(v)

∫ x(3x − 4)dx

2. Evaluate the following definite integrals.

∫ (ii) ∫ (iii) ∫ (iv) ∫ (v) ∫ (i)

1

−1 0 −1 4

2

(4 x + 5)dx (6 x 2 − 2 x)dx

( x 2 − x + 3)dx

2

−1 2

−1

(2 + x − x 2 )dx ( x 3 − x + 4) dx

3. Find the areas enclosed by the x axis and the following curves. (i) y = (1 – x)(x + 2) (ii) y = 3x2 – x3 4. Find the total area enclosed by y = x(x – 1), the x axis and the line x = 2. 5. Find the total area enclosed by y = x2 – 2x – 3, the x axis, x = –3 and x = 3. 6. (i) (ii) (iii) (iv)

Sketch the curve y = x3 – x for values of x from –3 to +3. Find the area bounded by the curve, the x axis and the lines x = 1 and x = 2. Find the area bounded by the curve and the lines x = -1, x = 0 and the x axis. From your answers to (ii) and (iii) and your sketch deduce the total area given by the integral

∫

2

0

( x 3 − x)dx . Explain your reasoning.

7. Use the trapezium rule to obtain approximate answers to 3 d.p. for the following integrals. 2 1 (i) ∫ dx with 4 strips 0 1 + x2

© MEI, 11/08/08

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MEI C2 Integration Section 2 Exercise (ii)

8. (i)

∫

3

(1 + x 2 dx with 3 strips

0

Using the trapezium rule with 5 strips, find an approximate value to 4 d.p. for

∫

0.5

0

(1 + x 2 )12 dx .

(ii) Using the binomial expansion expand (1 + x2)12 in ascending powers of x up to and including the term in x8. (iii) Use the expansion in (ii) to obtain an approximation to 4 d.p. for the integral in (i). (iv) Given that the estimate in (i) was an over estimate and that in iii) an underestimate, what do you think you could do to improve the accuracy of both methods and bring the answers closer together?

© MEI, 11/08/08

2/2

MEI Core 2 Integration Section 2: Finding the area under a curve Solutions to Exercise

 4x

1. (i)

( x

(ii)

3

3

dx  x 4  c

 3 x 2 )dx  41 x 4  x 3  c

(iii) (10 x 4  3 x 2  4)dx  2 x 5  x 3  4 x  c

(3 x  1) dx    9 x 2

(iv)

 6 x  1  dx

2

 3x 3  3x 2  x  c

 x(3 x  4) dx    3 x

(v)

2

 4 x  dx

 x3  2x2 c

2. (i)



1

1

(4 x  5 )dx   2 x 2  5 x  1 1  2  5  (2  5 )  7  ( 3)  10 0

(ii)  (6 x 2  2 x )dx   2 x 3  x 2  1 1  0  ( 2  1) 0

3 4

(iii)  ( x 2  x  3)dx   31 x 3  21 x 2  3 x  2 2

4

8   64 3  8  12    3  2  6 



64 3



56 3

 4  83  4

2

(iv)  (2  x  x 2 )dx   2 x  21 x 2  31 x 3  1 1 2

  4  2  83    2  21  31   6  83  2  21  31  8  3  21 

9 2

© MEI, 29/06/10

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MEI C2 Integration Section 2 Exercise solutions (v)



2

2

( x 3  x  4)dx   41 x 4  21 x 2  4 x  1 1

  4  2  8    41  21  4   10  41  4  14.25

3. (i) y  (1  x )( x  2) The graph cuts the x-axis at x = 1 and x = -2. The coefficient of x² is negative, so the graph is “upside down”.

-2

1

1

Area   (1  x )( x  2)dx 2 1

  (2  x  x 2 )dx 2

1

  2 x  21 x 2  31 x 3  2   2  21  31    4  2  83   2  21  31  6  83 

9 2

square units

(ii) y  3 x 2  x 3  x 2(3  x ) The graph cuts the x-axis at (3, 0) and touches the x-axis at the origin. 3

Area   (3 x 2  x 3 )dx 0

3

  x 3  41 x 4  0

3

 27  814  0  6.75 square units

4. y  x( x  1) The graph cuts the x-axis at the origin and the point (1, 0). y

This area is negative as it is below the x-axis. x

1

2

© MEI, 29/06/10

The areas above and below the x-axis must be calculated separately.

2/5

MEI C2 Integration Section 2 Exercise solutions 1

Area below x-axis   x( x  1)dx 0 1

  ( x 2  x )dx 0

1

  31 x 3  21 x 2  0  31  21   61 2

Area above x-axis   31 x 3  21 x 2  1

  83  2    31  21 

 56 Total area  61  56  1 square unit.

5. y  x 2  2 x  3  ( x  3)( x  1) y

x

-3

-1

3

1

Area above x-axis   ( x 2  2 x  3) dx 3

1

  31 x 3  x 2  3 x  3

   31  1  3    9  9  9 

32 3 3

Area below x-axis   31 x 3  x 2  3 x  1   9  9  9    31  1  3 

 9  53   323 Total area  323  323  64 3 square units

6. (i) y  x 3  x  x( x 2  1)  x( x  1)( x  1)

-1

1

© MEI, 29/06/10

3/5

MEI C2 Integration Section 2 Exercise solutions 2

(ii) Area   ( x 3  x )dx 1

2

  41 x 4  21 x 2  1   4  2    41  21   2.25 square units 0

(iii) Area   ( x 3  x )dx 1

0

  41 x 4  21 x 2  1  0   41  21 

 0.25 square units (iv) By symmetry, the area between x = 0 and x = 1 is the same as the area between x = -1 and x = 0, only below the x-axis. So the area between x = 0 and x = 2 is given by 2.25 + 0.25 = 2.5 square units.

7. (i) f( x ) 

h  0.5

1 1  x2

f0  f(0)  1 f1  f(0.5 )  0.8 f2  f(1)  0.5 1 f3  f(1.5 )  3.25 f4  f(2)  0.2 Using the trapezium rule: Area  21 h  f0  f 4  2  f 1  f 2  f 3    21  0.5  1  0.2  2  0.8  0.5 

1 3.25



 1.104 (3 d.p.) (ii) f( x )  1  x 2 h1 f0  f(0)  1 f1  f(1)  2 f2  f(2)  5 f3  f(3)  10 Using the trapezium rule:

© MEI, 29/06/10

4/5

MEI C2 Integration Section 2 Exercise solutions Area  21 h  f0  f 3  2  f 1  f 2  

 21  1  1  10  2  2  5    5.731 (3 d.p.)

8. (i) f( x )   1  x 2 

12

h  0.1

f0  f(0)  1 f1  f(0.1)  1.01 12 f2  f(0.2)  1.0412 f3  f(0.3)  1.0912 f4  f(0.4)  1.1612 f5  f(0.5 )  1.25 12

Using the trapezium rule: Area  21 h  f0  f 5  2  f 1  f 2  f 3  f 4    21  0.1  1  1.25 12  2  1.01 12  1.04 12  1.0912  1.16 12  

 1.9253 (4 d.p.) (ii)  1  x 2 

(iii)

12

12  11 2 2 12  11  10 2 3 (x )  (x ) 12 123 12  11  10  9 2 4  ( x )  ... 12 34  1  12 x 2  66 x 4  220 x 6  495 x 8  ....

 1  12 x 2 

 1  x  0.5

0

2 12

dx  

0.5

0

 1  12 x 2  66 x 4  220 x 6  495 x 8  dx 0.5

5 7   x  4 x 3  66  220  55 x 9  0 5 x 7 x  1.7655 (4 d.p.)

(iv) The estimate in (i) could be improved by using more strips. The estimate in (iii) could be improved by using more terms in the expansion.

© MEI, 29/06/10

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Integration Hexagonal Jigsaw The three diagrams on the following pages will ultimately fit together to form a large hexagon. Before you start, the three diagrams must be cut along the lines to make twenty-four equilateral triangles. For the triangles to be fitted together, you must match each integral with its solution. To build up the puzzle, place together the edges on which an integral and its solution are written, so that the triangles are joined along this edge. Begin by just finding pairs of integrals and solutions and placing them edge-to-edge. As you progress you will find that all of the pieces will eventually link up to form a large hexagon. By all means try this by yourself, but it is really designed to be a group activity. Working with other students will help to highlight any misconceptions you may have, as well as making the task more enjoyable. At first glance this may appear to be an easy undertaking, but you will find it takes quite a lot of thought and errors can be made very easily. Make sure you check each coupling, or you could be left with one or two pieces that appear to fit nowhere.

© Susan Whitehouse & MEI, 19/02/07

1/1

2

−4 x

0

2 3 −1

4 3 ∫ 5 dx

0 20

8x

−3

1

∫(3x − 1)(x + 1) dx

2

5

∫

1

−4

2 ) dx

4

0

∫(x − 1) dx

1

2

12

1

4

dx

2x

40

− x (2 3

∫x d

8 4 − ∫

1 2 x dx −2

0

∫0

dx ) 3x

x

∫(1 + 2x + 3 x

8

7

2

dx 3 −2

0

∫8x

5

0

∫ 6x

2

dx 2

−2

∫3x

1

− 18

2

1

dx

6

7x

dx

∫1

4

0

6

∫ 20

1 dx ∫ 2

x

4

dx

6 1 dx ∫ 3 −2

4 1

0

∫2 x dx

∫4 −

9 2

4

2

−

x 3d

1

2

∫(4 − 2x) dx

6 x 2 + 3x dx x

10

1

0

∫

28

−2

3

1

− 10

∫ 1 dx

− 30

2

10

∫ 3 x dx

−

9 ∫0

x

x+

d 1)

x

2 − 3

8

−1

∫ 100 x

d 2x

128

∫(0

9

dx

4 1

−3

−6

1

4

−

3

11

2

0

∫− x dx

5

2

3

2x

∫0

dx

4

∫2

xd x

0

−6

+ (7

∫ −6

dx

−1

2

) 2x

∫ 2x 5 dx

0

5

3

1

∫ (x −

0

24

1)( x+

1)

dx

MEI Core 2 Integration Section 2: Finding the area under a curve Multiple choice Test 1)



2

1

( x3  1)dx =

(a) 5.75 (c) 4.75 (e) I don’t know

2)



1

1

(b) 7.25 (d) 6.75

( x  2)2 dx =

(a) 0

(b) 8 23

(c) 4

(d) 8 23

(e) I don’t know

3)



1 2

( x 2  4 x) dx 

(a) 3 (c) 7 23 (e) I don’t know

(b) -3 (d) 7 23

4) The area enclosed by the curve y = 4x  x2 and the x axis is (b) 0

(a) 10 23 (c) 10 (e) I don’t know 2 3

(d) 32

5) The area enclosed by the curve y = x2  1 and the x-axis is: (a) 1 13 (c) 0

(b) 1 13 (d)

2 3

(e) I don’t know

© MEI, 26/08/09

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MEI C2 Integration Section 2 MC test 6) The area in square units enclosed by the curve y = x3  x2 – 2x and the positive x-axis is (a)

8 3 37 12

(b)

(c) (e) I don’t know

(d)

9 4 5 12

7) The area in square units enclosed by the curve y = x2 + x  2 and the lines y = 0, x = 0 and x = 2 is: (a) 0

(b) 1 56 (d) 3

(c) 23 (e) I don’t know

8) The diagram below shows the curve y  f ( x) . y = f(x)

a

c

b

d

x b

The trapezium rule is used to find approximations for A1   f ( x) dx and a

d

A2   f ( x) dx . Which of the following statements is correct? c

(a) The trapezium rule gives an underestimate for A1 and an overestimate for A2 (b) The trapezium rule gives an overestimate for A1 and an underestimate for A2 (c) The trapezium rule gives an underestimate for both A1 and A2 (d) The trapezium rule gives an overestimate for both A1 and A2 (e) I don’t know

9) A function f has the following values: x f(x)

10 10.1

11 4.5

12 5.4

13 10.7

14 14.7

15 16.6

© MEI, 26/08/09

16 17.5

2/3

MEI C2 Integration Section 2 MC test Using the trapezium rule, the integral of the function f between x = 10 and x = 16 is approximately (a) 65.7 (c) 79.5 (e) I don’t know

(b) 131.4 (d) 39.75

10) Using the trapezium rule with four strips gives the approximate value of the integral



2

0

1  x3 dx to be

(a) 3.2832 (c) 3.2834 (e) I don’t know

(b) 3.2833 (d) 3.2831

© MEI, 26/08/09

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MEI Core 2 Integration Chapter assessment 1. Integrate with respect to x (i) x5 (ii) 2x + 1 (iii) 3 (iv) x4 + 2x2 − 3

[1] [2] [1] [2]

2. Evaluate: (i)

∫

(ii)

∫ ∫

(iii)

2

x 2 dx

[3]

(1 + 2 x 3 )dx

[3]

( x − 1)(2 x + 1)dx .

[4]

−1 −1

−2 2 1

3. For a given curve,

dy = 2x − 1, and y = 3 when x = 1. Find y in terms of x. dx

[4]

4. The diagram below shows the curve y = x3 − 2x2 . y

x

P

(i) Find the coordinates of the point P. (ii) Find the gradient of the curve at P. (iii) Find the shaded area.

[2] [2] [5]

5. (i) Sketch the curve y = 2 − x − x2. (ii) Find the area enclosed by the curve and the x-axis between x = 0 and x = 2. 6. The diagram shows a cross-section of a tunnel. The height is measured in metres every 0.5 metres along the cross section. Use the trapezium rule to estimate the area of the cross-section.

[2] [5]

y

1.35

Is it an under-estimate or over-estimate?

1.84

1.85

2.12

1.86

x 0.5

[5]

© MEI, 07/04/09

1/2

MEI C2 Integration Assessment 7. Use the trapezium rule to estimate

∫

4

0

1 + x 2 dx , giving your answers to 3 d.p.

(i) using 4 strips (ii) using 8 strips. 8. (i)

Evaluate

∫

2

1

[4] [4]

( x + x 3 ) dx .

[3]

(ii) Estimate the above integral using the trapezium rule with 5 strips of width 0.2. [4] (iii) Use a sketch to explain why the trapezium rule gives an overestimate of the true area. [2] (iv) Calculate the percentage error in the calculation. [2]

Total: 60 marks

© MEI, 07/04/09

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