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y
ay
re er?
y = x3 + 7
O
x
y = x3 + 4
y = x
y = x3 – 2
Integration
Many small make a
great.
P
Chaucer
? In what way can you s
that these four curves a
all parallel to each oth
Reversing differentiation
In some situations you know the gradient function,
d y
, and want to find the
d x
function itself, y. For example, you might know that
d y
2 x and want to find y.
d x
You know from the previous chapter that if y x
2
then
d y
2 x , but
d x
y x
2
1, y x
2
2 and many other functions also give
d y
2 x.
d x
Suppose that f( x ) is a function with f( x ) 2 x. Let g( x ) f( x ) x
2
.
Then g( x ) f( x ) 2 x 2 x 2 x 0 for all x. So the graph of y g( x ) has zero
gradient everywhere, i.e. the graph is a horizontal straight line.
Thus g( x ) c (a constant). Therefore f( x ) x
2
c.
All that you can say at this point is that if
d y
2 x then y x
2
c where c is
d x
differentiation Reversing
described as an arbitrary constant. An arbitrary constant may take any value.
n 1
!
The rule for integrating x
n
P
Recall the rule for differentiation:
y x
n
d y
nx
n
1
.
d x
Similarly y x
n
1
d y
( n 1) x
n
d x
or y
1
x
n
1
d y
x
n
.
( n 1)
d x
Reversing this, integrating x
n
gives
x
n
1
.
This rule holds for all real values of the power n except –1.
Note
In words: to integrate a power of x , add 1 to the power and divide by the
new power. This works even when n is negative or a fraction.
Differentiating x gives 1, so integrating 1 gives x. This follows the pattern if you
remember that 1 x
0
.
Given that
d y
3 x
2
4 x 3
d x
(i) find the general solution of this differential equation
(ii) find the equation of the curve with this gradient function which
passes through (1, 10).
SOLUTION
(i) By integration, y
3 x
3
4 x
2
3 x c
3 2
x
3
2 x
2
3 x c , where c is a constant.
(ii) Since the curve passes through (1, 10),
10 1
3
2(1)
2
3(1) c
c 4
y x
3
2 x
2
3 x 4.
EXAMPLE 6.
differentiation Reversing
2 2 2 2 x
6
P
A curve is such that d y 3 d x find the equation of the curve. SOLUTION 8
. Given that the point (4, 20) lies on the curve,
x 2 Rewrite the gradient function as d y 3 x 1 8 x
. d x By integration, y 3 2 x 3 8 x
c Dividing by 3 is the same 2 as multiplying by 2 . 3 3 1 y 2 x 3 8 c x Since the curve passes through the point (4, 20), 20 3 8 c 2 ( 4 ) 2 4 20 16 2 c c 6 So the equation of the curve is y 2 x 3 8 6. x The gradient function of a curve is d y 4 x 12. d x (i) The minimum y value is 16. By considering the gradient function, find the corresponding x value. (ii) Use the gradient function and your answer from part (i) to find the equation of the curve. SOLUTION (i) At the minimum, the gradient of the curve must be
zero, 4 x 12 0 x 3.
(ii) d y 4 x 12 d x
y 2 x
2 12 x c. At the minimum point, x 3 and y 16
2 12 3 c
c 34
So the equation of the curve is y 2 x 2 12 x 34. EXAMPLE 6. EXAMPLE 6. Integratio n
6
P
x x 7 A curve passes through the point (4, 1) and its gradient at any point is given by d y 2 x 6. d x (i) Find the equation of the curve. (ii) Draw a sketch of the curve and state whether it passes under, over or through the point (1, 4). 8 A curve passes through the point (2, 3). The gradient of the curve is given by d y 3 x 2 2 x 1. d x (i) Find y in terms of x. (ii) Find the co-ordinates of any stationary points of the graph of y. (iii) Sketch the graph of y against x , marking the co-ordinates of any stationary points and the point where the curve cuts the y axis. 9 The gradient of a curve is given by d y 3 x 2 8 x 5. The curve passes through the point (0, 3). d x [MEI] (i) Find the equation of the curve. (ii) Find the co-ordinates of the two stationary points on the curve. State, with a reason, the nature of each stationary point. (iii) State the range of values of k for which the curve has three distinct intersections with the line y k. (iv) State the range of values of x for which the curve has a negative gradient. Find the x co-ordinate of the point within this range where the curve is steepest. 10 A curve is such that d y d x [MEI] x
. Given that the point (9, 20) lies on the curve,
find the equation of the curve. 11 A curve is such that d y 2 3. Given that the point (2, 10) lies on the d x x 2 curve, find the equation of the curve. 12 A curve is such that d y 1
. Given that the point (1, 5) lies on the
d x x 2 curve, find the equation of the curve. 13 A curve is such that d y 3 x 2 5. Given that the point (1, 8) lies on the d x curve, find the equation of the curve. 14 A curve is such that d y 3 d x 9 and the point (4, 0) lies on the curve. 178 (i) Find the equation of the curve. (ii) Find the x co-ordinate of the stationary point on the curve and determine the nature of the stationary point. Integration
x
P(x, y)
N
M Q
U T
P S
A
y
y + y
Q R
15 The equation of a curve is such that
d y
3
x
. Given that the curve passes
d x
through the point (4, 6), find the equation of the curve.
[Cambridge AS & A Level Mathematics 9709, Paper 12 Q1 November 2009]
16 A curve is such that
d y
4 x and the point P(2, 9) lies on the curve. The
d x
normal to the curve at P meets the curve again at Q. Find
(i) the equation of the curve,
(ii) the equation of the normal to the curve at P,
(iii) the co-ordinates of Q.
[Cambridge AS & A Level Mathematics 9709, Paper 1 Q9 November 2007]
Finding the area under a curve
Figure 6.2 shows a curve y f( x ) and the area required is shaded.
y = f( x )
y
O
a x b
x
Figure 6.
P is a point on the curve with an x co-ordinate between a and b. Let A denote
the area bounded by MNPQ. As P moves, the values of A and x change, so you
can see that the area A depends on the value of x. Figure 6.3 enlarges part of figure
6.2 and introduces T to the right of P.
Figure
6.
x x + x
Finding
the
area
under
a
curve
P
a
y
y = f( x)
Integrating, A x
6
6 x c.
When x 1, the line PQ coincides with the left-hand boundary so A 0
0 1 6 c 6
c 5.
So A x
6
6 x 5.
The required area is found by substituting x 2
A 64 12 5
81 square units.
Note
The term ‘square units’ is used since area is a square measure and the
units are unknown.
Standardising the procedure
Suppose that you want to find the area between the curve y f( x ), the x axis, and
the lines x a and x b. This is shown shaded in figure 6.5.
O a b
x
Figure 6.
●
d A
y f( x ).
dx
● Integrate f( x ) to give A F( x ) c.
● A 0 when x a 0 F( a ) c
c F( a )
A F( x ) F( a ).
● The value of A when x b is F( b ) F( a ).
Notation
F( b ) F( a ) is written as
[F( x )] b.
Finding
the
area
under
a
curve
P
1
y
y = x2 + 1
A
y = x2 + 1
Find the area between the curve y 20 3 x
2
, the x axis and the lines x 1 and x 2.
SOLUTION
f( x ) 20 3 x
2
F( x ) 20 x x
3
a 1 and b 2
Area
[20 x – x 3]
(40 8) (20 1)
13 square units.
Area as the limit of a sum
Suppose you want to find the area between the curve y x
2
1, the x axis and the
lines x 1 and x 5. This area is shaded in figure 6.6.
O 1 5
x
Figure 6.
You can find an estimate of the shaded area, A , by considering the area of
four rectangles of equal width, as shown in figure 6.7.
y
17
10
5
2
182
Figure
6.
0
1 2 3 4 5
x
EXAMPLE 6.
Integratio n
P
1
means ‘the sum of’ so all the Ai are added from
A 1 (given by i = 1) to An
(when i = n).
δA 1
δA 2
δA 3
δA 4 δAn
Notation
This process can be expressed more formally. Suppose you have n rectangles,
6 each of width x. Notice that n and x are related by
n x width of required area.
So in the example above,
n x 5 1 4.
In the limit, as n , x 0, the lower
estimate A and the higher estimate A.
The area A of a typical rectangle may
be written y x where y is the
appropriate
A
i
y
i
x
y
i
x
i i
Figure 6.
y value (see figure 6.9).
So for a finite number of strips, n , as shown in figure 6.10, the area A is given
approximately by
A A
1
A
2
… A
n
or A y
1
x y
2
x … y
n
x.
i n
This can be written as A
A
i
i 1
i n
or A
y
i
x.
i 1
y
y
n
y
4
y
3
y
2
y
1
O
x
Figure 6.
184
In the limit, as n and x 0, the result is no longer an approximation; it is
exact. At this point, A
y
i
x is written A
y d x , which you read as ‘the integral
of y with respect to x ’. In this case y x
2
1, and you require the area for
values of x from 1 to 5, so you can
write A
5
( x
2
1)d x.
Integratio n
P
a
3
1
The limits have now moved to the right of the s
Notice that in the limit:
● is replaced by
● x is replaced by d x
●
is replaced by
, the integral sign (the symbol is the Old English letter S)
● instead of summing for i 1 to n the process is now carried out over a range
of values of x (in this case 1 to 5), and these are called the limits of the
integral. (Note that this is a different meaning of the word limit.)
This method must give the same results as the previous one which used
dA
y ,
and at this stage the notation
F x
b
is used again.
d x
5 3
5
In this case
( x 2 1) d x
x
x
.
1
3
1
Recall that this notation means: find the value of
x
3
x when x 5 (the upper
3
limit) and subtract the value of
x
3
x when x 1 (the lower limit).
3
x 3 5 53
13
1
So the area A is 45
1
square units.
Find the area under the curve y 4 x
3
4 between x 1 and x 2.
SOLUTION
The graph is shown in figure 6.11.
y
The shaded part, A
2
(4 x
3
4) d x
[ x
4
4 x ]
2
(
4
4(2)) – ((–1)
4
4(–1))
27 square units.
4 A
–1 O 2
x
Figure 6.
EXAMPLE 6.
Area
as
the
limit
of
a
sum
1
P
y = x + 3
y = x + 3
ACTIVITY 6.2 Figure 6.13 shows the region bounded by the graph of y x 3, the x axis and
the lines x a and x b.
y
3
O
a b
x
Figure 6.
(i) Find the shaded area, A , by considering it as the difference between the two
trapezia shown in figure 6.14.
(ii) Show that the expression for A you obtained in part (i) may be written as
x
2
(iii) Show that you obtain the same answer for A by integration.
y
b + 3
3
O
a b
x
Figure
6.
y
a + 3
3
O
a
x
Area
as
the
limit
of
a
sum
a
P
b
y = x + 3
x
2
1 0
0 1
5 1
3 0
2 1
x
x
1 Find the following indefinite integrals.
(i)
3 x
2
d x (ii)
(5 x
4
7 x
6
) d x
(iii)
(6 x
2
5) d x (iv)
( x
3
x
2
x 1) d x
(v)
(11 x
10
10 x
9
) d x (vi)
(3 x
2
2 x 1) d x
(vii)
( x
2
5) d x (viii)
5 d x
(ix)
(6 x
2
4 x ) d x (x)
( x
4
3 x
2
2 x 1) d x
2 Find the following indefinite integrals.
(i)
10 x
d x (ii)
(2 x 3 x
) d x
(iii)
(2 x
3
5 x
) d x (iv)
(6 x
2
7 x
) d x
1
(v) 5 x
4
d x
(vi)
1
d x
x
4
(vii)
x d x
3 Evaluate the following definite integrals.
(viii)
2 x
4
4
d x
(i)
2
2 x d x (ii)
3
2 x d x
(iii)
3
3 x
2
d x (iv)
5
x d x
(v)
6
(2 x 1) d x (vi)
2
(2 x 4) d x
(vii)
5
(3 x
2
2 x ) d x (viii)
1
x
5
d x
(ix)
1
( x
4
x
3
) d x (x)
1
x
3
d x
(xi)
4
( x
3
3 x ) d x (xii)
2
5 d x
5 3
4 Evaluate the following definite integrals.
(i)
4
3 x
d x (ii)
4
8 x
d x
1
(iii)
4
1
1
12 x
2
d x
(iv)
2
6
d x
x
3
(v)
2
x
2
3 x 4
d x
(vi)
9
1
d x
x
4
4
EXERCISE 6B
6B Exercise
P
2 1 B A
6
P
y = x x x 5 The graph of y 2 x y is shown here. The shaded region is bounded by y 2 x , the x axis and the lines x 2 and x 3. (i) Find the co-ordinates of the points A and B in the diagram. (ii) Use the formula for the area of a trapezium to find the area of the shaded region. (iii) Find the area of the shaded 2 3 x region as 3 2 x d x , and confirm that your answer is the same as that for part (ii). (iv) The method of part (ii) cannot be used to find the area under the curve y x 2 bounded by the lines x 2 and x 3. Why? 6 (i) Sketch the curve y x 2 for 1 x 3 and shade the area bounded by the curve, the lines x 1 and x 2 and the x axis. (ii) Find, by integration, the area of the region you have shaded. 7 (i) Sketch the curve y 4 x 2 for 3 x 3. (ii) For what values of x is the curve above the x axis? (iii) Find the area between the curve and the x axis when the curve is above the x axis. 8 (i) Sketch the graph of y ( x 2) 2 for values of x between x 1 and x 5. Shade the area under the curve, between x 0 and x 2. (ii) Calculate the area you have shaded. [MEI] 9 The diagram shows the y graph of y 1
x for x 0. The shaded region is bounded by the curve, the x axis and the lines x = 1 and x = 9. Find its area. O 1 9 x Integratio n
