MEI C2 Study Resources Core2 Further Differentiation And Integration 1-2

MEI Core 2 Further differentiation and integration Section 1: Differentiation involving negative and fractional indices Study Plan Background You were introduced to the processes and rules on differentiation earlier in the course. This section shows how these can be extended to include indices that are negative or are fractions.

Detailed work plan 1. Remind yourself of the basics of differentiation (chapter 8, pages 191 – 200) before reading pages 339 – 340 which extend the rule for differentiating xn to all real values of n. You can also look at examples 1 and 2 in the Notes and Examples. 2. For further examples look at the Flash resource Differentiating rational powers of x. 3. Remind yourself of the applications of differentiation which you met in chapter 8: gradients of curves, tangents and normals, stationary points and problems involving maxima and minima. Examples 3, 4 and 5 in the Notes and Examples cover all these situations. 4. Exercise 12A Try these questions: 1 (odd parts), 2 (even parts), 3*, 4, 5*, 7*, 9, 13, 15, 17*, 20*.

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MEI Core 2 Further differentiation and integration Section 1: Differentiation involving negative and fractional indices Notes and Examples These notes contain subsections on  Differentiating kxn for negative and fractional n  Applications of differentiation

Differentiating kxn for negative and fractional n In Chapter 8 you learned that the derivative, or gradient of xn , where n is a positive integer, is given by nxn-1 . In fact this formula for the derivative of xn is true not only when n is a positive integer, but for all real values of n, including negative numbers and fractions.

Example 1 Differentiate the following functions. 1 y (i) (ii) y  x2 x x Solution (i)

(ii)

(iii)

y

1 x

Subtracting 1 from -1 gives -2

1 y   x 1 x dy 1  1x 2   2 dx x

Use the laws of indices to express this as a single power of x 5

y  x2 x  x2 x 2  x 2 dy 5 32  x dx 2 1

y

(iii)

Subtracting 1 from

5 2

gives

3 2

1 1 x 2 x

dy 3   12 x 2 dx

Subtracting 1 from  12 gives  32

To see more examples like the ones above, use the Flash resource Differentiating rational powers of x.

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MEI C2 Further calculus Section 1 Notes & Examples As in chapter 8, you can extend this idea to allow you to differentiate all functions of the form kxn, where k is a constant, and sums and differences of such functions.  The derivative of kxn is knxn-1, where k is a constant and n is any real number  The derivative of sum (or difference) of two or more such functions is the sum (or difference) of the derivatives of the functions.

Example 2 Differentiate the following functions (i) y  (3  2 x  x 2 ) x (ii)

y

3x  x 2 x5

Solution y  (3  2 x  x 2 ) x (i)

 3 x  2x x  x2 x 3

5

 3x 2  2 x 2  x 2 3 1 dy 1  3  12 x 2  2  23 x 2  52 x 2 dx 1

1

3

 32 x 2  3x 2  52 x 2 (ii)

1

3x  x 2 x5 3 1  4 3 x x 4  3 x  x 3 dy  3  4 x 5  (3x 4 ) dx  12 x 5  3x 4 y

Applications of differentiation Now that you can differentiate a wider range of functions, you can also make use of various applications of differentiation in many more contexts. In chapter 8 you learned to use differentiation to find gradients of curves, find the equations of tangents and normals to curves, find stationary points on curves and solve problems involving maximum and minimum values. The following examples cover these applications.

Example 3 For the graph y  x  x

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MEI C2 Further calculus Section 1 Notes & Examples (i) (ii) (iii)

find the gradient at the point (4, 2) find the equation of the tangent at this point find the equation of the normal at this point.

Solution (i)

(ii)

y  x  x  x  x2 dy 1 1  1  12 x 2  1  dx 2 x dy 1 When x = 4,  1  1  14  dx 2 4 The gradient at (4, 2) is 34 . 1

Gradient of tangent at (4, 2) =

3 4

Using the equation of a line y  y1  m( x  x1 ) with

m

3 4

and

 x1 , y1    4, 2

3 4

Equation of tangent at (4, 2) is y  2  34 ( x  4)

y  34 x  3  2 y  34 x  1 (iii)

Gradient of normal at (4, 2) =  43 Equation of normal at (4, 2) is y  2   43 ( x  4)

Remember that when two lines with gradients m1 and m2 are perpendicular, m1m2  1

y   43 x  163  2 y   43 x  223

Using the equation of a line y  y1  m( x  x1 ) with

m   43 and  x1 , y1    4, 2 

Example 4 Find the stationary points of the graph y  Solution x2  3 1 3 y   3  x 1  3x 3 3 x x x dy 1 9   x 2  9 x 4   2  4 dx x x At stationary points,

x2  3 and determine their nature. x3 At stationary points, the gradient is zero. Differentiate the function and find the values of x for which

dy 0 dx

dy 0 dx

1 9  4  0   x2  9  0 2 x x  x2  9  x  3 93 6 2   When x = 3, y  27 27 9 

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Substitute the values of x into the original equation to find the y-coordinates of the stationary points.

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MEI C2 Further calculus Section 1 Notes & Examples 93 6 2   27 27 9 2 The stationary points are  3, 9  and  3,  92  . When x = -3, y 

d 2 y 2 36   Find the value of the second derivative dx 2 x3 x5 at each stationary point to determine d 2 y 2 36 the nature of the stationary point.   0 When x = 3, 2 dx 27 243 2 d y 2 36   0 When x = -3, 2 dx 27 243 so  3, 92  is a local maximum and  3,  92  is a local minimum.

Example 5 A farmer wants to make a pen for a goat, using the side of a barn as one side of the pen. He wants the pen to have an area of 50 m², but wants to use as little fencing as possible. BARN

y

50 m²

y

x (i) (ii) (iii)

Write down expressions for the area of the pen and the length of fencing required in terms of x and y. Find an expression for the length of fencing required, L, in terms of x only. dL Find and hence find the minimum length of fencing required, and show dx that this is a minimum.

Solution (i) Area = xy Length of fencing = x + 2y (ii)

50  xy  y 

L  x  2y 100  x x (ii)

50 x

dL 100  1 2 dx x

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MEI C2 Further calculus Section 1 Notes & Examples dL 100  0  1 2  0 dx x 2  x  100  x  10 100 When x = 10, L  10   20 10 d 2 L 200  3 dx 2 x d2 L  0 so this is a minimum. When x = 10, dx 2 The minimum length of fencing required is 20m. At minimum value of L,

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Core 2 Further differentiation and integration Section 1: Differentiation involving negative and fractional indices Crucial points 1. Be careful when differentiating negative and fractional powers Make sure that you subtract 1 from the index correctly, especially when dealing with negative numbers. 2. Make sure that you write products and quotients in an appropriate form when differentiating 3 1 e.g. ( x − 1) x must be written as x 2 − x 2 before differentiating 1− x 1 1 must be written as 2 − before differentiating 2 x x x

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MEI Core 2 Further differentiation and integration Section 1: Differentiation involving negative and fractional indices Exercise 1. Differentiate the following functions 1 (i) (ii) y= 3 x 2 3 (iii) (iv) y= − 2 x x (v) (vii)

y = 3x −5 − 2 x −7 x2 − 2x + 3 y= 2x2

y=3 x y=4 x− 2

3 x − 23

(vi)

y = 2 x 3 − 5x

(viii)

y = ( x2 − 2) x

2. Find the gradient of each of the following graphs at the given point 1 (i) y = 2 x − at the point (1, 1) x y = 3 − x at the point (4, 1) (ii)

(iii)

y = x 2 x at the point (1, 1)

3. Find any stationary points on the following curves and determine their nature. 4 (i) y = x− 2 x 1 (ii) y= x+ x 4. Find the equation of the tangent to the graph y =

5. Find the equation of the normal to the graph y =

1 at the point where x = 1. x 1 2 at the point where − x x2

x = 2.

6. A cylindrical can with height h metres and radius r metres has a capacity of 2 litres. (i) Find an expression for h in terms of r. (ii) Hence find an expression for the surface area of the can in terms of r only. (iii) Find the value of r which minimises the surface area of the can.

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MEI Core 2 Further differentiation and integration Section 1: Differentiation involving negative and fractional indices Exercise Solutions 1.

(i)

y= dy dx

(ii)

1

x

3

= x −3

= −3 x −4 = −

y=3x =x dy 1 − =3x dx

3

x4

1 3

2 3

(iii) y = dy dx

2

x

−

3

x

2

= −2 x −2 + 6 x −3 = −

(iv) y = 4 x −

dy dx (v)

= 2 x −1 − 3 x −2

3

2

x

2

6

+

x3

= 4x 2 − 3 x −2 1

x

1

= 2 x − 2 + 23 x − 2 1

3

y = 3 x −5 − 2 x −7 dy 15 14 = −15 x −6 + 14 x −8 = − 6 + 8 dx x x

(vi) y = 2 x 3 − 5 x − 3 2

dy dx (vii) y =

dy dx

=

4 3

2

x − + 103 x − 1 3

5 3

x2 − 2x + 3 1 = 2 − x −1 + 23 x −2 2x2 = x −2 − 3 x −3 =

1

x

2

− 5

3

x3 1

(viii) y = ( x 2 − 2)( x ) = x 2 − 2 x 2 dy dx

=

5 2

x − x− 3 2

1 2

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MEI C2 Further calculus Section 1 Exercise solutions 2. (i)

1

y = 2x − dy

x

= 2 + x −2

dx

At (1, 1) :

(ii)

= 2 x − x −1

dy

= 2 + (1)−2 = 3

dx

1 2

y = 3− x = 3− x dy 1 = − 21 x − = − dx 2 x dy 1 1 At (4, 1) : =− =− dx 2 4 4 1 2

5

(iii) y = x 2 x = x 2 dy

=

dx

5 2

x

3 2

At (1, 1) :

3. (i)

y=x− dy

dy

3

= 52 (1)2 =

dx

4

5 2

= x − 4 x −2

x2

= 1 + 8 x −3

dx

Stationary points occur when 0 = 1 + 8 x −3

dy dx

= 0.

−1 = 8 x −3

x 3 = −8 x = −2 When x = −2, y = −3 d2 y dx

2

= −24 x −4

At x = −2,

d2 y 2

= −24( −2)− 4 = −1.5

dx As this is negative, ( −2, −3) is a local maximum.

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MEI C2 Further calculus Section 1 Exercise solutions (ii)

1

y= x+ dy

=

dx

1 2

= x 2 + x −2 1

x

x − − 21 x − 1 2

1

3 2

dy

Stationary points occur when 0=

x − − 21 x − 1 2

1 2

x − = 21 x − 1 2

1 2

x− = x− 1 2

dx

= 0.

3 2

3 2

3 2

1 = x −1 x =1 When x = 1, y = 2 d2 y dx

2

= − 41 x − 2 + 43 x − 2 3

At x = 1,

5

d2 y 2

= − 41 +

3 4

=

1 2

dx As this is positive, (1, 2) is a local minimum.

4.

y= dy dx

1

x

= x −2 1

= − 21 x − 2

3

When x = 1, y = 1 and

y − y 1 = m( x − x 1 )

dy

−

= − (1) 1 2

dx

3 2

= − 21

y − 1 = − 21 ( x − 1) y − 1 = − 21 x + 21 y = − 21 x + 23

5.

y= dy dx

1

x

−

2

x

2

= x −1 − 2 x −2

= − x −2 + 4 x −3

When x = 2, y = 21 − 21 = 0, As normal, m =

dy dx

= −(2)−2 + 4(2)−3 = 0.25

−1 = −4 0.25

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MEI C2 Further calculus Section 1 Exercise solutions y − y 1 = m( x − x 1 ) y − 0 = −4( x − 2) y = −4 x + 8

6. (i)

Volume of a cylindrical can = π r 2 × h 2 litres = 0.002m 3 (1m 3 = 1000litres ) 0.002 = π r 2h 0.002 h= 2

πr

(ii)

Surface area = 2π rh+2π r 2 = 2π r ( h + r ) = 2π r ( = 2π (

0.002

πr 2

0.002

πr

(iii) s = 2π( =

r

+r 2)

0.002

πr

0.004

+r )

+r 2)

+ 2π r 2 = 0.004r −1 + 2π r 2

ds = −0.004r −2 + 4π r dr ds = 0 ⇒ −0.004r −2 + 4π r = 0 dr ⇒ 4π r = 0.004r −2 ⇒ 4π r 3 = 0.004 0.004 ⇒r3 = 4π ⇒ r = 0.0683 m (3 s.f.) d2 s = 0.008r −3 + 4π , which is positive for all values of r, so the 2 dr stationary point must be a minimum point.

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MEI Core 2 Further differentiation and integration Section 1: Differentiation involving negative and fractional indices Multiple Choice Test 1) The derivative of

4

x is

(a) −4x −5

(b)

(a) −

x

− 14

(d) −4x −3

−3

(c) 14 x 4 (e) I don’t know

2) The derivative of

1 4

1 is x5

5 x4

5 x6 1 (d) 4 5x

(b) −

5 x4 (e) I don’t know

(c)

3) The derivative of (3 x − 2) x is

(a) 3 x − (c)

2 x

(b)

3x − 2 2 x

(d)

3 2 x

9 x 1 − 2 x

(e) I don’t know

4) The derivative of (a) −1 +

x4 − 2x2 + 3 is x3 4 x3 − 4 x 3x 2 2 9 (d) 1 + 2 − 4 x x

3 x2

(b)

2 3 (c) x − + 3 x x (e) I don’t know

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MEI C2 Further calculus Section 1 MC test 5) The gradient of the curve y =

2 at the point (4, 1) is x

(a) − 18

(b)

(c) 12 (e) I don’t know

(d) − 12

6) The gradient of the curve y = (a)

1 8

2 3 at the point (2, − x x2

1 4

) is

(b) − 14

1 2

(c) 14 (e) I don’t know

(d)

5 4

7) The equation of the tangent to the curve y = 2 x − (a) y = 3x (c) y = x (e) I don’t know

1 at the point where x = 1 is x

(b) y = 3x – 2 (d) y = x – 1

8) The equation of the normal to the curve y = 2 x − x at the point where x = 4 is (a) 7y = 4x + 26 (c) 7y + 4x = 58 (e) I don’t know

(b) 4y = 7x + 8 (d) 4y + 7x = 52

9) The stationary point of the curve y = x 2 −

2 is x

(a) Maximum (-1, 3) (c) Maximum (1, -1) (e) I don’t know

(b) Minimum (-1, 3) (d) Minimum (1, -1)

10) The stationary point of the curve y = x − 4 x is (a) Minimum (4, -4) (c) Minimum ( 14 , − 74 ) (e) I don’t know

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(b) Maximum (4, -4) (d) Maximum ( 14 , − 74 )

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MEI Core 2 Further differentiation and integration Section 2: Integration involving negative and fractional indices Study Plan Background In earlier work you saw how integration could be used to reverse differentiation or to find areas under or between curves. This section extends these ideas to include integrating functions of the form xn where n is any real number other than -1.

Detailed work plan 1. Revise the basic rules governing integration from chapter 9 and then read Examples 12.3 and 12.4 (pages 347 – 348) and Examples 1 and 2 in the Notes and Examples. 2. You can see further examples using the Flash resource Integrating rational powers of x. 3. Exercise 12B Try the odd parts of questions 1 and 2. 4. For further practice, try the Calculus Advanced puzzle which includes both differentiation and integration, using fractional and negative powers of x. 5. Read examples 12.5 and 12.6 on pages 348 – 349, and Examples 3, 4 and 5 in the Notes and Examples. The warning note on page 349 is very important. You may be able to obtain what looks like a perfectly sensible numerical answer for an area but if the integral contains a point at which the curve is discontinuous, the area is not valid. This is one of the reasons why it is important to sketch a curve before trying to evaluate the integral. 6. Exercise 12B Try questions 3, 4*, 6*, 7, 9, 11*, 12.

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MEI Core 2 Further differentiation and integration Section 2: Integration involving negative and fractional indices Notes and Examples These notes contain subsections on  Integrating kxn for negative and fractional n  Applications of integration

Integrating kxn for negative and fractional n In Chapter 9 you learned that the integral of xn, where n is a positive integer, is given by

1

 x dx  n  1 x n

n 1

c

where c is an arbitrary constant

In fact this formula is true not only when n is a positive integer, but for all real values of n, including negative numbers and fractions, except for n = -1. The formula does not work for n = -1, since this would give a denominator of 0. 1 There is a different way to integrate , which is covered in C3. x Example 1 Find the following indefinite integrals (i)  xdx

1

(ii)

x

(iii)

  x

3

dx

 2 2



3   dx x

Solution 1 x d x  x (i)   2 dx

n  12 , so n  1  32 , so divide by

3 2

, i.e. multiply by

2 3

.

3

 23 x 2  c (ii)

1 3  x3 dx   x dx   12 x 2  c

n  3 , so n  1  2 , so divide by –2.

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MEI C2 Further diff. and int. Sec. 2 Notes & Examples (iii)



n  2 , so n  1  1 , so divide by –1.



3   2 1 2   x 2  x  dx   2 x  3x 2 dx  2 x 1  3  2 x 2  c 2   6 x c x 1

n   12 , so n  1  12 , so divide by

1 2

, i.e. multiply by 2.

For further examples like the ones above, look at the Flash resource Integrating rational powers of x.

You can also try the Calculus Advanced puzzle which includes both differentiation and integration, using fractional and negative powers of x.

Example 2 Find the following definite integrals. 2  4x 1  (i) 1  x4  dx (ii)



4

1

(3  x) x dx

Solution (i)



2

1

2  4x 1  3 4  4  dx  1  4 x  x dx  x 

  4   12 x 2  13 x 3 

2

1

2

1   2   2  3   x 3 x 1    12  241    2  13   (ii)



4

1

Substitute x = 2 in first bracket and x = 1 in second bracket

29 24

4





(3  x) x dx   3x 2  x 2 dx 1

3

1

4

 3  23 x 2  52 x 2   1 3

5

4

Substitute x = 4 in first bracket and x = 1 in second bracket

  2 x x  52 x 2 x  1 2   2  4  2  5  4  2    2 11  52 11  11.2

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MEI C2 Further diff. and int. Sec. 2 Notes & Examples Applications of integration Now that you can integrate a wider range of functions, you can also solve problems which involve integrating these functions, such as finding functions given their gradient function, and finding the area under a curve.

Example 3 The gradient function of a curve is given by dy 1 3 x  dx x and the curve passes through the point (4, 9) Find the equation of the curve. Integrate to find an expression for y in terms of x

Solution 1  dy 1  3 x  y  3 x  dx dx x x 



  3x

1 2

x

 12

3

 dx

 3  23 x 2  2 x 2  c 1

3

Substitute the coordinates of the given point to find the value of the constant c.

 2x 2  2x 2  c 1

When x = 4, y = 9  9  2  8  2  2  c

 c  9  16  4  c  3 3

The equation of the curve is y  2 x 2  2 x 2  3 1

The next two examples are about finding the area under a curve. Make sure that you have read the exclamation point warning on page 349 of the textbook, about discontinuous graphs.

Example 4 Find the area under the graph y  1  x between x = 0 and x = 4. Solution 4





Area under graph   1  x dx 0

4





  1  x 2 dx 0

1

4

  x  23 x 2   0 2   4  3  8  0 3



28 3

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MEI C2 Further diff. and int. Sec. 2 Notes & Examples Example 5 The diagram shows the graph of y 

1  1 and the line y = 3. x2 Q

P

(i) (ii)

Find the coordinates of points P and Q. Find the area bounded by the curve, the line y = 3 and the x axis.

Solution (i)

1 1 1 1  1  3  2  4  x2   x   . 2 x x 4 2 1 The coordinates of P are   2 ,3 and the coordinates of Q are At P and Q,

 12 ,3 .

(ii) You cannot integrate across a discontinuity. Instead, find the areas of A, B and C separately.

Q

P

B A



C 1 2

1 2

1

 1   1  Area C is given by 1  2  1 dx     x  2  x   x  12 1

  1  1   2  12  

By symmetry area A is also Area B = 3  1 = 3 Total area =

1 2

1 2

1 2

.

 12  3  4 .

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MEI Core 2 Further differentiation and integration Section 2: Integration involving negative and fractional indices Crucial points 1. Be careful when integrating negative and fractional powers Make sure that you add 1 to the index correctly, and remember that you must divide by the new index. 2. Remember that you cannot yet integrate xn for n = -1 The rule for integrating xn is valid for all values of n except for n = -1. 1 You will learn to integrate in C3. x 3. Look out for discontinuities when finding the area under a graph Remember that you cannot integrate across a discontinuity. See Example 5 in the Notes and Examples.

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MEI Core 2 Further differentiation and integration Section 2: Integration involving negative and fractional indices Exercise 1. Find the following indefinite integrals 1 (i) (ii) ∫ x 2 dx (iii)

∫ ( 3x

(v)

∫ (3

−3

− 4 x −4 ) dx

)

3

x dx

(iv)

∫ ( 2x

(vi)

( x − 1) 2 ∫ x 5 dx

2

x − 2 dx

∫

3 4

2

)

− 3x 3 dx

2. Evaluate the following definite integrals 3 1 (i) ∫1 x3 dx 9 1 (ii) ∫1 x dx 2 4 2 − x + 3x (iii) dx ∫1 x 2 2 x −1 (iv) ∫1 x 4 dx dy x − 3 = 3 and passes through the point (1, 1). x dx

3. A curve has gradient function Find the equation of the curve. 4. A curve has gradient function

dy = 2 x − 3 x and passes through the point dx

(1, -1). Find the equation of the curve. 5. Find the area under the graph y = 6. (i) (ii)

1 + x between x = 1 and x = 4. x2

Find the points where the graph of y = x − x meets the x-axis. Find the area enclosed by the graph and the x-axis.

© MEI, 13/08/08

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MEI Core 2 Further differentiation and integration Section 2: Integration involving negative and fractional indices Exercise Solutions 1. (i)

1

∫x

dx = ∫ x −2 dx

2

= − x −1 + c =−

(ii)

∫

3

1

+c

x 1 3

x d x = ∫ x dx =

3 4

4 3

x +c

(iii)

∫ (3x

(iv)

∫ (2 x

3 4

(v)

∫ (3

x − 2 ) dx = ∫ ( 3 x − 2 ) dx

−3

− 4 x −4 ) dx = − 23 x −2 + 43 x −3 + c

)

2

− 3 x 3 dx =

8 7

2

7 4

5 3

x − 59 x + c 2

1 2

(

)

= ∫ 9 x − 12 x 2 + 4 dx =

(vi)

2. (i)

1

3 2

x 2 − 8 x + 4x + c

9 2

⎛ ( x − 1)2 ⎞ −3 −4 −5 ∫ ⎜⎝ x 5 ⎟⎠ dx = ∫ ( x − 2 x + x ) dx = − 21 x −2 + 23 x −3 − 41 x −4 + c

∫

3

1

1

x

3

3

dx = ∫ x −3 dx 1

3

= ⎡⎣ − 21 x −2 ⎤⎦ 1 = − 21 (3)− 2 − ( − 21 (1)−2 ) = − 181 + =

1 2

4 9

© MEI, 13/08/08

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MEI C2 Further calculus Section 2 Exercise solutions 1

9

∫

(ii)

9

x

1

dx = ∫ x − 2 dx 1

1

9

= ⎡⎣ 2 x 2 ⎤⎦ 1

(

1

1 2

) (

1

= 2( 9) − 2(1)2

)

=6−2 =4

∫

(iii)

4

1

4 1 1 3 ⎛ 2 − x + 3x 2 ⎞ dx = ∫ 2 x − 2 − x 2 + 3 x 2 d x ⎜ ⎟ 1 x ⎝ ⎠

(

)

= ⎡⎣ 4 x 2 − 23 x 2 + 1

(

3

1 2

6 5

x ⎤⎦

3 2

5 2

4 1 5

) (

1

3

5

= 4(4) − (4) + (4)2 − 4(1)2 − 23 (1)2 + 56 (1)2 = ( 8 − 163 +

∫

(iv)

2

1

=

616 15

=

548 15

2 3

192 5

6 5

) − (4 −

2 3

)

+ 56 )

− 68 15

2 ⎛ x2 − 1 ⎞ −4 2 ⎜ x 4 ⎟ dx = ∫1 (( x − 1)( x )) dx ⎝ ⎠

=∫

2

1

( x −2 − x −4 ) dx 2

= ⎡⎣ − x −1 + 31 x −3 ⎤⎦ 1 = ( −(2)−1 + 31 (2)−3 ) − ( −(1)−1 + 31 (1)−3 )

= ( − 21 +

1 24

) − ( −1 + 31 )

11 = − 24 − ( − 23 )

=

3.

dy

5 24

x −3 = x −2 − 3 x −3 3 x dx y = ∫ ( x −2 − 3 x −3 ) dx =

y = − x −1 + 23 x −2 + c When x = 1, y = 1 1 = −(1)− 1 + 23 (1)− 2 + c 1 = −1 + 23 + c

c=

1 2

y =−

1

x

+

3 2x

2

+

1 2

© MEI, 13/08/08

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MEI C2 Further calculus Section 2 Exercise solutions 4.

dy dx

1

= 2 x − 3x = 2 x 2 − 3x

y = ∫ ( 2 x − 3 x ) dx 1 2

3 2

y = 43 x − 23 x 2 + c When x = 1, y = −1 3

−1 = 43 (1)2 − 23 (1)2 + c −1 = 43 − 23 + c −1 = − 61 + c

c = − 56 3 2

y = 43 x − 23 x 2 − 56

5.

∫

4

1

4 ⎛ 1 ⎞ −2 ⎜ 2 + x ⎟ dx = ∫1 ( x + x ) dx ⎝x ⎠ 4 = ⎣⎡ − x −1 + 21 x 2 ⎦⎤ 1

= ( −(4)−1 + 21 (4)2 ) − ( −(1)−1 + 21 (1)2 ) = ( − 41 + 8 ) − ( −1 + 21 )

6. (i)

=

31 4

=

33 4

+

1 2

y=x− x The graph meets the x-axis when y = o .

y = x(1 − x − ) 1 2

1 − x −2 = 0 1

1=x

x =0

or

− 21

x =1 (ii)

∫ ( x − x ) dx = ⎡⎣ 1

1 2

0

=

(

1 2 1 2

1

x 2 − 23 x ⎤⎦ 0 3 2

3

) (

(1)2 − 23 (1)2 −

1 2

3

(0)2 − 23 (0)2

)

= ( 21 − 23 ) − 0 = − 61 Area cannot be negative so area =

1 6

© MEI, 13/08/08

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Calculus Hexagonal Jigsaw - Advanced 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 find two expressions that are equivalent to one another. To build up the puzzle, place the edges on which equivalent expressions are written together, so that the triangles are joined along this edge. Begin by just finding pairs of matching expressions 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. Note that in this puzzle, there are no blank edges – there are some expressions which have no matching expression, and these will be on the perimeter of the final 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 Wall, MEI 2006

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Calculus Hexagonal Jigsaw - Advanced

© Susan Wall, MEI 2006

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Calculus Hexagonal Jigsaw - Advanced

© Susan Wall, MEI 2006

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Calculus Hexagonal Jigsaw - Advanced

© Susan Wall, MEI 2006

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MEI Core 2 Further differentiation and integration Section 2: Integration involving negative and fractional indices Multiple Choice Test 1)

1

∫x

7

dx =

1 +c 6 x6 1 (c) − 6 + c 7x (e) I don’t know

(a) −

∫x

2) (a)

4 5

1 4

dx =

5

x4 + c

(b)

5

(c) 14 x 4 + c (e) I don’t know

3)

6 +c x6 7 (d) − 6 + c x

(b) −

5 4

5

(d) 4x 4 + c

2x −1 dx = x

∫

3

(a) 3x 2 − 12 x 2 + c 1

x2 − x

(b) 2 x ( x 2 − x) + c

(d)

4 3

(a) − 34

(b)

(c) − 83 (e) I don’t know

(d)

3 8 3 4

+c 3 x2 (e) I don’t know

(c)

4)

5

x4 + c

3

x 2 − 2x 2 + c 1

2 3

∫ ( 3x 2

1

−2

− 8 x −5 ) dx =

© MEI, 18/08/08

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MEI C2 Further calculus Section 2 MC tests 5)

1

∫ (2 x − 1)

3

0

x dx = (b) − 32

(a) 103 (c) 0

(d)

3 28

(e) I don’t know

6) A curve has gradient function

dy 3 and passes through the point (3, 2). = dx x 2

The equation of the curve is 3 +1 x 3 3 (c) y = + 2x 2 (e) I don’t know

3 x 5 3 (d) y = − 2 2x

(a) y =

(b) y = 3 −

7) A curve has gradient function

dy 1 and passes through the point (4, 3). = dx x

The equation of the curve is (a) y =

1 2

(b) y = 2 x − 1

x +2

(c) y = 2 x − 5 (e) I don’t know

(d) y =

8) The area under the graph y = 1 − (a) 118 (c) 1

1 2

x +1

1 between x = -2 and x = -1 is x3

(b)

13 8

(d)

5 4

(e) I don’t know 9) The area enclosed by the graph y = x , the x-axis and the line x = 4 is (a) 43 (c) 16

(b) 12 (d)

16 3

(e) I don’t know

© MEI, 18/08/08

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MEI C2 Further calculus Section 2 MC tests 10) The area enclosed by the graph y =

1 , the coordinate axes, the line x = 2 and the x2

line y = 4 is (a)

1 4

(c) 72 (e) I don’t know

(b)

3 2

(d)

5 2

© MEI, 18/08/08

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MEI Core 2 Further differentiation and integration Chapter assessment 1. Differentiate the following 3 + 2x (i) y = x ( x − 2)(2 x + 1) (ii) y = x4

[3] [3]

2. Find the following indefinite integrals 3 ⎞ ⎛ (i) ∫ ⎜ 2 x − ⎟ dx x⎠ ⎝ 1 ⎞ ⎛ 2 (ii) ∫ ⎜ 2 − 3 ⎟ dx ⎝ x 2x ⎠

[3]

[3]

3. Find the following definite integrals (i)

∫

9

(ii)

∫

2

1

1

(2 − x) x dx

[4]

⎛ 6 8⎞ ⎜ 2 − 3 ⎟ dx x ⎠ ⎝x

[4]

2 at the point where x = -1. x2 (ii) Find the equation of the tangent to the graph at this point. (iii) Find the equation of the normal to the graph at this point.

4. (i)

Find the gradient of the graph y = x −

5. Find the coordinates of the stationary points of the graph y =

passes through the point (1, 2).

[3] [3]

1 1 1 and + − x x 2 x3

determine the nature of each. 6. Find the equation of the graph with gradient function

[2]

[5] dy 1 = − x x which dx x 2

[3]

7. A closed rectangular tank is to have dimensions x m, 2x m and y m, and is to have a capacity of 9 cubic metres. To keep costs down the surface area of the tank is to be as small as possible. (i) Write down expressions for the volume and the surface area of the tank in terms of x and y, and hence find an expression for A in terms of x only. [3] (ii) Find the value of x for which A has a stationary point, and show that this is a minimum point. [4] (iii) Find the dimensions of the tank for which A is as small as possible, and the corresponding value of A. [2]

© MEI, 18/08/08

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MEI C2 Further calculus assessment 8. The diagram shows part of the graph of y = 5 − x 2 −

4 . x2

y

x

Find the area enclosed between the graph and the x-axis.

[5]

Total 50 marks

© MEI, 18/08/08

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