Pure C4
Revision Notes
March 2013
Contents Core 4 1 Algebra ................................................................................................................. 3 Partial fractions ........................................................................................................................................................ 3 2 Coordinate Geometry .......................................................................................... 5 Parametric equations ................................................................................................................................................ 5 Conversion from parametric to Cartesian form ...................................................................................................................... 6 Area under curve given parametrically................................................................................................................................... 7 3 Sequences and series ......................................................................................... 8 Binomial series (1 + x)n for any n .......................................................................................................................... 8 4 Differentiation .................................................................................................... 10 Relationship between
and
......................................................................................................................... 10 Implicit differentiation ........................................................................................................................................... 10 Parametric differentiation ...................................................................................................................................... 11 Exponential functions, a x ...................................................................................................................................... 11 Related rates of change .......................................................................................................................................... 12 Forming differential equations ............................................................................................................................... 12 5 Integration .......................................................................................................... 13 Integrals of ex and ............................................................................................................................................ 13 Standard integrals ................................................................................................................................................... 13 Integration using trigonometric identities.............................................................................................................. 13 Integration by ‘reverse chain rule’ ......................................................................................................................... 14 Integrals of tan x and cot x .................................................................................................................................................... 15 Integrals of sec x and cosec x ................................................................................................................................................ 15 Integration using partial fractions .......................................................................................................................... 16 Integration by substitution, indefinite .................................................................................................................... 16 Integration by substitution, definite ....................................................................................................................... 18 Choosing the substitution ...................................................................................................................................................... 18 Integration by parts................................................................................................................................................. 19 Area under curve .................................................................................................................................................... 20 Volume of revolution ............................................................................................................................................. 20 Volume of revolution about the x–axis ................................................................................................................................. 21 Volume of revolution about the y–axis ................................................................................................................................. 21 Parametric integration ............................................................................................................................................ 21 Differential equations ............................................................................................................................................. 22 Separating the variables ........................................................................................................................................................ 22 Exponential growth and decay ............................................................................................................................... 23 C4 14/04/13
1
6 Vectors ................................................................................................................ 24 Notation .................................................................................................................................................................. 24 Definitions, adding and subtracting, etcetera ........................................................................................................ 24 Parallel and non - parallel vectors......................................................................................................................................... 25 Modulus of a vector and unit vectors ................................................................................................................................... 26 Position vectors ..................................................................................................................................................................... 27 Ratios ..................................................................................................................................................................................... 27 Proving geometrical theorems .............................................................................................................................................. 28 Three dimensional vectors ..................................................................................................................................... 29 Length, modulus or magnitude of a vector ........................................................................................................................... 29 Distance between two points ................................................................................................................................................ 29 Scalar product ........................................................................................................................................................ 29 Perpendicular vectors ............................................................................................................................................................ 30 Angle between vectors .......................................................................................................................................................... 31 Vector equation of a straight line .......................................................................................................................... 31 Intersection of two lines......................................................................................................................................... 32 2 Dimensions ......................................................................................................................................................................... 32 3 Dimensions ......................................................................................................................................................................... 33 7 Appendix ............................................................................................................. 34 Binomial series (1 + x)n for any n – proof ........................................................................................................... 34 Derivative of xq for q rational ............................................................................................................................ 34 for negative limits ..................................................................................................................................... 35 Integration by substitution – why it works ............................................................................................................ 36 Parametric integration ............................................................................................................................................ 36 Separating the variables – why it works ................................................................................................................ 37 Index ............................................................................................................................. 38 2
C4 14/04/13 SDB
1
Algebra
Partial fractions
1) You must start with a proper fraction: i.e. the degree of the numerator must be less than
the degree of the denominator.
If this is not the case you must first do long division to find quotient and remainder.
2) (a)
Linear factors (not repeated)
......
A
≡
+ ........
(ax − b)(. . . . . . )
(ax − b)
(b)
Linear repeated factors (squared)
......
A
B
≡
+
+ ........
2
2
(ax − b)
(ax − b) (. . . . . . )
(ax − b)
(c)
Quadratic factors)
......
Ax + B
≡
+ ........
(ax + b)(. . . . . . )
(ax 2 + b)
2
or
......
Ax + B
≡
+ ........
( ax 2 + bx + c )(. . . . . . )
( ax 2 + bx + c )
5 − x + 2x 2
(1 − x)(1 + x 2 )
Example:
Express
Solution:
The degree of the numerator, 2, is less than the degree of the denominator, 3, so
in partial fractions.
we do not need long division and can write
5 − x + 2x 2
(1 − x)(1 + x 2 )
A
Bx + C
+
1− x
1+ x2
≡
multiply both sides by (1–x)(1+x2)
⇒
5 – x + 2x2 ≡ A(1 + x2) + (Bx + C)(1 – x)
⇒
5 – 1 + 2 = 2A
⇒
A=3
clever value!, put x = 1
⇒
5=A+C
⇒
C=2
easy value, put x = 0
⇒
2=A–B
⇒
B=1
equate coefficients of x2
⇒
5 − x + 2x 2
(1 − x)(1 + x 2 )
≡
3
x+2
+
.
1− x
1+ x2
N.B. You can put in any value for x, so you can always find as many equations as you
need to solve for A, B, C, D . . . .
C4 14/04/13
3
Example:
x 2 − 7 x + 22
(2 x − 1)( x − 3) 2
Express
in partial fractions.
Solution:
The degree of the numerator, 2, is less than the degree of the denominator,
3, so we do not need long division and can write
x 2 − 7 x + 22
(2 x − 1)( x − 3) 2
≡
A
B
+
2x − 1
( x − 3) 2
+
C
x−3
multiply by denominator
⇒
x2 – 7x + 22 ≡ A(x – 3)2 + B(2x – 1) + C(2x – 1)(x – 3)
⇒
9 – 21 + 22 = 5B
⇒
B=2
clever value, put x = 3
⇒
¼ – 7/2 + 22 = 2.52A ⇒
A=3
clever value, put x = ½
⇒
22 = 9A – B + 3C
⇒
x 2 − 7 x + 22
(2 x − 1)( x − 3) 2
Example:
⇒
≡
C = –1
3
2
+
2x − 1
( x − 3) 2
x3 + x 2 − 9x − 3
x2 − 9
Express
easy value, put x = 0
−
1
x−3
in partial fractions.
Solution:
Firstly the degree of the numerator is not less than the degree of the
denominator so we must divide top by bottom.
x2 – 9 ) x3
x3
⇒
+ x2
x2
x2
– 9x
– 9x
x 3 + x 2 − 9x − 3
x2 − 9
–
3
–
–
3
9
6
= x+1 +
( x+1
6
x −9
2
Factorise to give x2 – 9 = (x – 3)(x + 3) and write
6
x −9
2
4
≡
6
A
B
≡
+
( x − 3)( x + 3)
x−3
x+3
multiplying by denominator
⇒
6 ≡ A(x + 3) + B(x – 3)
⇒
6 = 6A
⇒
A=1
clever value, put x = 3
⇒
6 = –6B
⇒
B = –1
clever value, put x = –3
⇒
x 3 + x 2 − 9x − 3
= x+1 +
x2 − 9
1
1
−
x−3
x+3
C4 14/04/13 SDB
2
Coordinate Geometry
Parametric equations
If we define x and y in terms of a single variable (the letters t or θ are often used) then this
variable is called a parameter: we then have the parametric equation of a curve.
Example: y = t 2 – 3, x = 2 + t is the parametric equation of a curve. Find
(i)
the points where the curve meets the x-axis,
(ii)
the points of intersection of the curve with the line y = 2x + 1.
Solution:
(i)
The curve meets the x-axis when y = 0 ⇒ t 2 = 3 ⇒ t = ±√3
⇒
curve meets the x-axis at (2 – √3, 0) and (2 + √3, 0).
(ii)
Substitute for x and y in the equation of the line
⇒
t 2 – 3 = 2(2 + t) + 1
⇒
t2 – 2t – 8 = 0
⇒
t = –2 or 4
⇒
the points of intersection are (0, 1) and (6, 13).
⇒
(t – 4)(t + 2) = 0
Example: Find whether the curves y = t2 – 2, x = 2t + 3 and y = t + 3, x = 3t – 4 intersect.
If they do give the point of intersection, otherwise give reasons why they do not
intersect.
Solution:
⇒
If they intersect there must be a value of t which makes their x-coordinates
equal, so find this value of t
2t + 3 = 3t – 4 ⇒ t = 7
But for this value of t the y-coordinates are 47 and 10.
So we cannot find a value of t to give equal x- and equal y-coordinates and therefore
the curves do not intersect.
C4 14/04/13
5
Conversion from parametric to Cartesian form
Eliminate the parameter (t or θ or …) to form an equation between x and y only.
Example: Find the Cartesian equation of the curve given by y = t 2 – 3, x = t + 2.
Solution:
x=t+2
t = x – 2, and y = t 2 – 3
⇒
y = (x – 2)2 – 3,
⇒
which is the Cartesian equation of a parabola with vertex at (2, –3)
With trigonometric parametric equations the formulae
sin2 t + cos2 t = 1
sec2t – tan2 t = 1
and
will often be useful.
Example: Find the Cartesian equation of the curve given by
y = 3sin t + 2, x = 3cos t – 1.
Solution:
Re-arranging we have
sin t =
y−2
, and
3
2
cos t =
2
⇒
⎛ y − 2⎞
⎜
⎟
⎝ 3 ⎠
⇒
(x + 1)2 + (y – 2)2 = 9
⎛ x + 1⎞
+ ⎜
⎟
⎝ 3 ⎠
x +1
, which together with sin2 t + cos2 t = 1
3
= 1
which is the Cartesian equation of a circle with centre (–1, 2) and radius 3.
Example: Find the Cartesian equation of the curve given by y = 3tan t, x = 4sec t.
Hence sketch the curve.
Re-arranging we have tan t =
Solution:
y
, and
3
sec t =
x
,
4
which together with sec2 t – tan2 t = 1
⇒
x2
42
−
y2
32
= 1
4 y
2
which is the standard equation of a hyperbola
with centre (0, 0)
and x-intercepts (4, 0), (–4, 0).
x
−6
−4
−2
2
4
6
−2
−4
6
C4 14/04/13 SDB
Area under curve given parametrically
We know that the area between a curve and the x-axis is given by A = ∫ y dx
⇒
dA
= y.
dx
dA dA dx
=
×
dt dx dt
But, from the chain rule
⇒
dA
dx
=y
dt
dt
Integrating with respect to t
dx
∫ y dt dt .
⇒ A =
Example: Find the area between the curve y = t 2 – 1, x = t 3 + t, the x-axis and the lines
x = 0 and x = 2.
Solution:
The area is A =
.
?
⇒
A =
we should write limits for t, not x
?
Firstly we need to find y and
y = t2 – 1 and
dx
in terms of t.
dt
dx
= 3t2 + 1.
dt
Secondly we are integrating with respect to t and so the limits of integration must be for
values of t.
x = 0 ⇒ t = 0, and
x = 2 ⇒ t3 + t = 2 ⇒ t3 + t – 2 = 0 ⇒
(t – 1)(t2 + t + 2) = 0
⇒
t = 1.
so the limits for t are from 0 to 1
⇒ A =
1
∫
1
0
y
dx
dt =
dt
∫
1
0
(t2 – 1) (3t2 + 1) dt
=
∫
=
⎡ 3t 5
⎤
2t 3
−
− t ⎥ = − 1 151
⎢
3
⎣ 5
⎦0
0
3t4 – 2t2 – 1 dt
1
Note that in simple problems you may be able to eliminate t and find ∫ y dx in the usual
manner. However there will be some problems where this is difficult and the above technique
will be better.
C4 14/04/13
7
3
Sequences and series
Binomial series (1 + x)n for any n
n(n − 1)
n(n − 1)(n − 2)
× x2 +
× x 3 + ....
2!
3!
(1 + x)n = 1 + nx +
This converges provided that |x| < 1.
Example: Expand (1 + 3x) –2, giving the first four terms, and state the values of x for
which the series is convergent.
Solution:
(1 + 3x) –2
= 1 + (–2) × 3x +
(−2) × (−3)
(−2) × (−3) × (−4)
× (3x)2 +
× (3x)3 + ...
2!
3!
= 1 – 6x + 27x2 – 108x3 + ...
This series is convergent when |3x| < 1 ⇔ |x| < 1/3.
Example: Use the previous example to find an approximation for
Solution:
Notice that
1
.
0 ⋅ 9997 2
1
= 0⋅9997 –2 = (1 + 3x)–2 when x = – 0⋅0001.
0 ⋅ 9997 2
So writing x = – 0⋅0001 in the expansion 1 – 6x + 27x2 – 108x3
0⋅9997 –2 ≈ 1 + 0⋅0006 + 0⋅00000027 + 0⋅000000000108 = 1⋅000600270108
The correct answer to 13 decimal places is 1⋅0006002701080
not bad eh?
1
Example: Expand ( 4 − x ) 2 , giving all terms up to and including the term in x3, and state
for what values of x the series is convergent.
Solution:
As the formula holds for (1 + x)n we first re-write
(4 − x)
1
2
⎛
1⎛ x⎞
= 2 × ⎜1 + ⎜ − ⎟ +
⎜
2 ⎝ 4⎠
⎝
1
1
2
1
x⎞2
x⎞2
⎛
⎛
= 4 ⎜1 − ⎟ = 2 × ⎜1 − ⎟
4⎠
⎝
⎝ 4⎠
1
2
× − 12
2!
⎛ x⎞
× ⎜− ⎟
⎝ 4⎠
2
+
1
2
× − 12 × − 32
3!
and now we can use the formula
⎛ x⎞
× ⎜− ⎟
⎝ 4⎠
3
⎞
+ ...⎟..
⎟
⎠
x
x2
x3
−
−
.
= 2 –
4
64 512
This expansion converges for
8
x
4
<1
⇔
|x| < 4.
C4 14/04/13 SDB
Example: Find the expansion of
Solution:
⇒
First write in partial fractions
3 x −1
2
1
, which must now be written as
=
+
x
+
3
x
−
2
x − x−6
2
2
1
−
x
3(1 + 3 )
2(1 − 2x )
=
=
3 x −1
in ascending powers of x up to x2.
x − x−6
2
2
3
=
2
3
(1 + 3x ) −1 −
1
2
(1 − 2x ) −1
2
2
⎛
⎞
⎛
⎞
⎜1 + (−1)⎛⎜ x ⎞⎟ + (−1)(−2) ⎛⎜ x ⎞⎟ ⎟ − 1 ⎜1+ (−1)⎛⎜ − x ⎞⎟ + (−1)(−2) ⎛⎜ − x ⎞⎟ ⎟
⎜
2! ⎝ 3 ⎠ ⎟⎠
2 ⎜⎝
2! ⎝ 2 ⎠ ⎟⎠
⎝3⎠
⎝ 2 ⎠
⎝
1
17 x
11x 2
−
−
.
6
36
216
C4 14/04/13
9
4
Differentiation
Relationship between
1 y = 3x2
So
and
1 dy
= 6x
dx
⇒
⇒
dx
1
=
.
dy
6x
Implicit differentiation
This is just the chain rule when we do not know explicitly what y is as a function of x.
Examples: The following examples use the chain rule (or implicit differentiation)
sin
5
sin
10
5
3
cos using the product rule
3
2
2
dy ⎞
d 2
d 2
⎛
(
(
x + 3 y ) = 3 (x 2 + 3 y ) ×
x + 3 y ) = 3 (x 2 + 3 y ) × ⎜ 2 x + 3 ⎟
dx
dx
dx ⎠
⎝
Example: Find the gradient of, and the equation of, the tangent to the curve
x2 + y2 – 3xy = –1 at the point (1, 2).
Solution:
Differentiating x2 + y2 – 3xy = –1 with respect to x gives
2x + 2y
⇒
dy ⎛
dy ⎞
– ⎜ 3 y + 3x ⎟ = 0
dx ⎝
dx ⎠
dy
3y − 2x
=
dx
2 y − 3x
⇒
dy
= 4
dx
when x = 1 and y = 2.
Equation of the tangent is y – 2 = 4(x – 1)
⇒
10
y = 4x – 2.
C4 14/04/13 SDB
Parametric differentiation
Example: A curve has parametric equations x = t 2 + t, y = t 3 – 3t.
(i)
Find the equation of the normal at the point where t = 2.
(ii)
Find the points with zero gradient.
Solution:
(i)
When t = 2, x = 6 and y = 2.
dx
dy
= 3t 2 − 3 and
= 2t + 1
dt
dt
⇒
3
2
3
1
9
5
when t = 2
Thus the gradient of the normal at the point (6, 2) is
and its equation is y – 2 =
–5
/9 (x – 6)
⇒
–5
/9
5x + 9y = 48.
dy
3t 2 − 3
=
= 0
dx
2t + 1
(ii)
gradient = 0 when
⇒
3t2 – 3 = 0
⇒
t = ±1
⇒
points with zero gradient are (0, 2) and (2, –2).
Exponential functions, ax
y = ax
⇒
ln y = ln ax = x ln a
⇒
1 dy
= ln a
y dx
⇒
d x
a
dx
C4 14/04/13
( )
⇒
dy
= y ln a
dx
= a x ln a
11
Related rates of change
We can use the chain rule to relate one rate of change to another.
Example:
A spherical snowball is melting at a rate of 96 cm3 s –1 when its radius is 12 cm.
Find the rate at which its surface area is decreasing at that moment.
Solution:
We know that V = 4/3 πr3 and that A = 4π r2.
Using the chain rule we have
dr
dV
dV dr
=
×
= 4π r 2 ×
,
dt
dr
dt
dt
⇒
dV
dt
⇒
96 = 4 × π × 122 ×
⇒
dr
1
=
dt
6π
since
dV
= 4π r 2
dr
since
dA
= 8π r
dr
dr
dt
= 4π r2 ×
dr
dt
Using the chain rule again
dA
dA dr
dr
=
×
= 8π r ×
,
dt
dr dt
dt
dA
1
= 8 × π × 12 ×
= 16 cm 2 s −1
dt
6π
⇒
Forming differential equations
Example: The mass of a radio-active substance at time t is decaying at a rate which is
proportional to the mass present at time t. Find a differential equation connecting the
mass m and the time t.
dm
means the rate at which the mass is increasing so in this
dt
case we must consider the rate of decay as a negative increase
Solution:
Remember that
⇒
dm
∝ –m
dt
⇒
dm
= – km, where k is the (positive) constant of proportionality.
dt
1
⇒
ln m = −
kt 2 + ln A
12
C4 14/04/13 SDB
5
Integration
Integrals of ex and
∫e
dx = e x + c
x
⌠ 1 dx = ln | x | + c
⎮
⌡x
for a further treatment of this result, see the appendix
3
⌠ x + 3x
dx
Example: Find ⎮
⌡ x2
3
3
⌠ x + 3x
dx = ⌠
dx = ½ x2 + 3 ln x + c.
⎮x+
⎮
2
⌡
x
⌡ x
Solution:
Standard integrals
x must be in RADIANS when integrating trigonometric functions.
∫ f ( x)
f (x)
f (x)
dx
∫ f ( x)
dx
x n +1
n +1
sin x
– cos x
1
x
ln ⏐x⏐
cos x
sin x
ex
ex
sec x tan x
sec x
n
x
2
sec x
tan x
cosec x cot x
– cosec x
cosec2 x
– cot x
Integration using trigonometric identities
Example: Find
∫ cot
2
x dx .
cot2 x = cosec2 x – 1
Solution:
⇒
∫
=
– cot x – x + c.
C4 14/04/13
cot 2 x dx =
∫
cosec 2 x − 1 dx
13
Example: Find
∫ sin
2
x dx.
sin2 x = ½ (1 – cos 2x)
Solution:
⇒
∫ sin
=
½ x – ¼ sin 2x + c.
2
∫
x dx =
1
2
(1 − cos 2 x) dx
You cannot change x to 3x in the above result to find
Example: Find
2
2
3x dx .
see next example
3x dx .
sin2 3x = ½ (1 – cos 2 × 3x) = ½ (1 – cos 6x)
Solution:
⇒
∫ sin
∫ sin
∫ sin
=
2
=
=
∫
1
2
(1 − cos 6 x) dx
½ x – 1/12 sin 6x + c.
Example: Find
Solution:
3x dx =
∫ sin 3x cos 5x
dx.
Using the formula 2 sin A cos B = sin(A + B) + sin(A – B)
∫ sin 3x cos 5x dx
1
/2 ∫ sin 8x + sin (–2x)
dx =
1
– /16 cos 8x + ¼ cos 2x + c.
1
/2 ∫ sin 8x – sin 2x dx
Integration by ‘reverse chain rule’
Some integrals which are not standard functions can be integrated by thinking of the chain
rule for differentiation.
Example: Find
Solution:
∫ sin
∫ sin
4
4
3x cos 3x dx .
3x cos 3x dx
If we think of u = sin 3x, then the integrand looks like u 4
constants, which would integrate to give 1/5 u5
du
if we ignore the
dx
so we differentiate u5 = sin5 3x
to give
(
d
sin 5 3x
dx
)
(
)
= 5 sin 4 3x × 3 cos 3x = 15 sin 4 3x cos 3x
which is 15 times what we want and so
14
C4 14/04/13 SDB
∫ sin
4
3x cos 3x dx =
Example: Find
∫
Solution:
x
∫
( 2 x − 3)
2
x
( 2 x − 3)
2
sin 5 3x + c
1
15
dx
dx
If we think of u = (2x2 – 3), then the integrand looks like
constants, which would integrate to ln ⏐u⏐
1 du
if we ignore the
u dx
so we differentiate ln ⏐u⏐ = ln ⏐2x2 – 3⏐
(
d
ln 2 x 2 − 3
dx
to give
)
=
1
4x
× 4x =
2x − 3
(2 x 2 − 3)
2
which is 4 times what we want and so
∫
x
( 2 x − 3)
In general
2
∫
dx = ¼ ln ⏐2x2 – 3⏐ + c.
f ' ( x)
dx = ln f ( x) + c
f ( x)
Integrals of tan x and cot x
∫ tan x dx =
sin x
∫ cos x dx
and we now have
∫
= −∫
− sin x
dx ,
cos x
f ' ( x)
dx = ln f ( x) + c
f ( x)
⇒
∫ tan x dx = – ln ⏐cos x⏐ + c
⇒
∫ tan x dx = ln ⏐sec x⏐ + c
cot x can be integrated by a similar method to give
∫ cot x dx = ln ⏐sin x⏐ + c
Integrals of sec x and cosec x
∫ sec x dx =
sec x (sec x + tan x)
∫ sec x + tan x dx =
sec2 x + sec x tan x
∫ sec x + tan x dx
The top is now the derivative of the bottom
and we have
⇒
C4 14/04/13
∫
f ' ( x)
dx = ln f ( x) + c
f ( x)
∫ sec x dx = ln ⏐sec x + tan x⏐ + c
15
and similarly
∫ cosec x dx = –ln ⏐cosec x + cot x⏐ + c
Integration using partial fractions
For use with algebraic fractions where the denominator factorises.
6x
dx
x + x−2
6x
in partial fractions.
Solution: First express
2
x +x−2
6x
A
B
6x
≡
≡
+
2
( x − 1)( x + 2)
x −1
x+2
x +x−2
Example: Find
⇒
∫
2
6x ≡ A(x + 2) + B(x – 1).
put x = 1
⇒
A = 2,
put x = –2
⇒
B=4
6x
dx =
x +x−2
∫
=
2 ln |x – 1| + 4 ln |x + 2| + c.
2
∫
2
4
+
dx
x −1
x+2
⇒
Integration by substitution, indefinite
(i)
(ii)
Use the given substitution involving x and u, (or find a suitable substitution).
du
dx
or
, whichever is easier and re-arrange to find dx in terms
dx
du
of du, i.e dx = …... du
Find either
(iii)
Use the substitution in (i) to make the integrand a function of u, and use your
answer to (ii) to replace dx by ...... du.
(iv)
Simplify and integrate the function of u.
(v)
Use the substitution in (i) to write your answer in terms of x.
∫
x 3 x 2 − 5 dx using the substitution u = 3x2 – 5.
Example:
Find
Solution:
(i)
u = 3x2 – 5
du
= 6x
⇒
dx
(ii)
(iii)
du
6x
dx =
We can see that there an x will cancel, and √3
∫
x 3 x 2 − 5 dx =
∫
x u
du
6x
=
∫
u
6
5 √
du
3
1
(iv)
16
=
1
u 2 du
6∫
=
1 u2
× 3
6
2
+ c
C4 14/04/13 SDB
3
(3x 2 − 5) 2
(v) =
9
Example:
Find
Solution:
(i)
+ c
1
dx using the substitution x = tan u.
1+ x2
∫
x = tan u.
(ii)
dx
= sec 2 u
du
(iii)
∫
(iv)
=
∫
sec 2 u
du
sec 2 u
=
∫
du
=
tan –1 x + c.
(v)
1
dx =
1+ x2
Example: Find
Solution:
(ii)
(iii)
=
x −4
2
2
sec 2 u du
u
since 1 + tan2 u = sec2 u
u + c
dx using the substitution u2 = x2 – 4.
u2 = x2 – 4.
(i)
We know that
2u
1
∫ 1 + tan
3x
∫
dx = sec2 u du.
⇒
( )
d 2
du
u = 2u
so differentiating gives
dx
dx
du
= 2x
dx
⇒
dx =
u
du .
x
We can see that an x will cancel and
∫
3x
x −4
2
dx =
∫
du
=
x 2 − 4 = u so
3x
u
×
du
u
x
(iv)
=
∫3
(v)
=
3 x2 − 4 + c
3u + c
A justification of this technique is given in the appendix.
C4 14/04/13
17
Integration by substitution, definite
If the integral has limits then proceed as before but remember to change the limits from
values of x to the corresponding values of u.
Add
(ii) (a) Change limits from x to u, and
new
(v)
Put in limits for u.
Example: Find
Solution:
∫
(i)
6
2
x 3x − 2 dx using the substitution u = 3x – 2.
u = 3x – 2.
du
= 3
dx
(ii)
du
3
dx =
⇒
(ii) (a) Change limits from x to u
x = 2 ⇒ u = 3 × 2 – 2 = 4, and x = 6 ⇒ u = 3 × 6 – 2 = 16
6
∫
(iv)
=
1
9
=
3
⎡ 52
u2
1 ⎢u
+ 2× 3
9 ⎢ 5
2
⎢⎣ 2
(v)
2
x 3x − 2 dx =
=
1
9
∫
[
16
4
2
5
u
3
2
∫
4
+ 2u
× 1024 +
1
u+2
×u2
3
16
(iii)
4
3
1
2
⎤
⎥
⎥
⎥⎦
du
3
du
16
4
× 64 ] –
1
9
[
2
5
× 32 + 43 × 8] = 52⋅4 to 3 S.F.
Choosing the substitution
In general put u equal to the ‘awkward bit’ – but there are some special cases where this will
not help.
1
√2
√4
put u = x2 + 1
put u = x – 2
5 put u = 2x + 5
u2 = 2x + 5
put u or u2 = 4 – x2
only if n is ODD
put x = 2 sin u
only if n is EVEN (or zero)
this makes √4
18
or
= √4 cos
= 2 cos u
C4 14/04/13 SDB
There are many more possibilities – use your imagination!!
Integration by parts
The product rule for differentiation is
To integrate by parts
(i)
(ii)
(iii)
dv
dx
du
find v and
dx
substitute in formula and integrate.
choose u and
Example: Find
Solution:
∫
x sin x dx
(i)
Choose u = x, because it disappears when differentiated
dv
= sin x
and choose
dx
du
= 1 and
(ii)
u=x ⇒
dx
dv
= sin x ⇒ v = ∫ sin x dx = − cos x
dx
dv
du
(iii)
∫ u dx dx = uv − ∫ v dx dx
⇒
∫
=
–x cos x + sin x + c.
x sin x dx = –x cos x –
∫ 1× (− cos x)
Example: Find
∫ ln x
Solution:
It does not look like a product, u
(i)
dx
dx
dv
, but if we take u = ln x and
dx
dv
dv
= 1 then u
= ln x × 1 = ln x
dx
dx
(ii)
C4 14/04/13
u = ln x ⇒
du
1
=
and
dx
x
dv
= 1 ⇒ v =x
dx
19
∫
(iii)
ln x ×1 dx = x ln x −
=
∫
x×
1
dx
x
x ln x – x + c.
Area under curve
We found in Core 2 that the area under the curve is written
.
as the integral
We can consider the area as approximately the sum of the
rectangles shown.
If each rectangle has width δx and if the heights of the
rectangles are y1, y2, ..., yn
a
b
then the area of the rectangles is approximately the area
under the curve
b
∑ yδ x
a
≅
∫
b
a
y dx
and as δx → 0 we have
b
∑ yδ x
→
a
∫
b
a
y dx
This last result is true for any function y.
Volume of revolution
y = f (x)
y
y
b
a
x
x
x+
δx
If the curve of y = f (x) is rotated about the x–axis then the volume of the shape formed can
be found by considering many slices each of width δx: one slice is shown.
The volume of this slice (a disc) is approximately π y2δx
⇒ Sum of volumes of all slices from a to b ≈
b
∑π y
2
δx
a
20
C4 14/04/13 SDB
and as δx → 0 we have (using the result above
b
∑ yδ x
→
a
b
∑π y
2
b
∫ πy
δx →
2
a
a
∫
b
a
y dx )
dx .
Volume of revolution about the x–axis
Volume when y = f (x), between x = a and x = b, is rotated about the x–axis
∫
is V =
b
π y 2 dx .
a
Volume of revolution about the y–axis
Volume when y = f (x) , between y = c and y = d, is rotated about the y–axis
∫
is V =
d
π x 2 dy .
c
Volume of rotation about the y-axis is not in the syllabus but is included for completeness.
Parametric integration
When x and y are given in parametric form we can find integrals using the techniques in
integration by substitution.
think of ‘cancelling’ the ‘dt’s
See the appendix for a justification of this result.
Example: If x = tan t and y = sin t, find the area under the curve from t = 0 to t = π/4.
Solution:
∫y
The area =
dx for some limits on x =
∫
y
dx
dt .
dt
We know that y = sin t, and also that
dx
= sec 2 t
dt
x = tan t ⇒
⇒
area =
∫
=
π
4
0
∫y
dx =
∫
sin t sec t dt =
2
y
∫
π
4
0
dx
dt
dt
tan t sec t dt
π
[sec t ] 04
=
=
2 −1
To find a volume of revolution we need
C4 14/04/13
∫πy
2
dx and we proceed as above writing
21
Example:
The curve shown has parametric equations
y
x = t2 − 1, y = t3.
The section between x = 0 and x = 8 above the x-axis
is rotated about the x-axis through 2π radians. Find
the volume generated.
Solution:
V =
∫
8
0
20
10
π y 2 dx .
5
10
x
x = 0 ⇒ t = ±1 and x = 8 ⇒ t = 3,
but the curve is above the x-axis ⇒ y = t3 > 0 ⇒ t > 0, ⇒ t = +1, or 3
dx
= 2t
dt
also y = t3, x = t2 − 1 ⇒
∫ πy
⇒
V =
=
⎡ t8 ⎤
2π ⎢ ⎥
⎣ 8 ⎦1
2
dx
dt =
dt
3
π
=
4
(3
8
∫
3
1
π t 6 2t dt
=
2π
∫
3
1
t 7 dt
)
−1
Differential equations
Separating the variables
Example: Solve the differential equation
Solution:
dy
= 3 y + xy .
dx
dy
= 3 y + xy = y(3 + x)
dx
We first ‘cheat’ by separating the x s and y s onto different sides of the equation.
⇒
1
dy = (3 + x ) dx and then put in the integral signs
y
⇒
∫
1
dy =
y
∫
3 + x dx
⇒ ln y = 3x + 1/2 x2 + c.
See the appendix for a justification of this technique.
22
C4 14/04/13 SDB
Exponential growth and decay
Example:
A radio-active substance decays at a rate which is proportional to the mass of the
substance present. Initially 25 grams are present and after 8 hours the mass has
decreased to 20 grams. Find the mass after 1 day.
Solution:
Let m grams be the mass of the substance at time t.
dm
is the rate of increase of m so, since the mass is decreasing,
dt
dm
= − km
dt
1 dm
=
m dt
⇒
1
dm =
m
⇒
∫
∫
⇒
ln ⏐m⏐ = −kt + ln ⏐A⏐
⇒
ln
⇒
m = Ae − kt .
−k
− k dt
= –kt
When t = 0, m = 25 ⇒ A = 25
⇒
m = 25e–kt.
When t = 8, m = 20
⇒
20 = 25e–8k
⇒
e–8k = 0⋅8
⇒
–8k = ln 0⋅8
⇒
k = 0⋅027892943
So when t = 24,
m = 25e–24 × 0⋅027892943 = 12⋅8.
Answer 12⋅8 grams after 1 day.
C4 14/04/13
23
6
Vectors
Notation
The book and exam papers like writing vectors in the form
a = 3i – 4j + 7k.
It is allowed, and sensible, to re-write vectors in column form
⎡ 3 ⎤
i.e. a = 3i – 4j + 7k = ⎢⎢ − 4⎥⎥ .
⎢⎣ 7 ⎥⎦
Definitions, adding and subtracting, etcetera
A vector has both magnitude (length) and direction. If you always think of a vector as a
translation you will not go far wrong.
Directed line segments
The vector
is the vector from A to B,
B
(or the translation which takes A to B).
This is sometimes called the
displacement vector from A to B.
A
Vectors in co-ordinate form
Vectors can also be thought of as column vectors,
B
⎡7 ⎤
thus in the diagram AB = ⎢ ⎥ .
⎣ 3⎦
3
A
Negative vectors
is the 'opposite' of
7
⎡− 7⎤
= ⎢ ⎥.
⎣ − 3⎦
and so
Adding and subtracting vectors
(i) Using a diagram
Geometrically this can be done using a triangle (or a parallelogram):
Adding:
a
b
a
a+b
24
a+b
b
C4 14/04/13 SDB
The sum of two vectors is called the resultant of those vectors.
Subtracting:
b
a–b
a
(ii) Using coordinates
⎡a ⎤ ⎡ c ⎤
⎡a + c⎤
⎢b ⎥ + ⎢d ⎥ = ⎢b + d ⎥
⎣ ⎦ ⎣ ⎦
⎣
⎦
and
⎡a ⎤ ⎡ c ⎤
⎡a − c⎤
⎢b ⎥ − ⎢d ⎥ = ⎢b − d ⎥ .
⎣ ⎦ ⎣ ⎦
⎣
⎦
Parallel and non - parallel vectors
Parallel vectors
Two vectors are parallel if they have the same direction
⇔ one is a multiple of the other.
Example: Which two of the following vectors are parallel?
⎡ 6 ⎤ ⎡− 4⎤ ⎡2⎤
⎢− 3⎥, ⎢ 2 ⎥, ⎢1⎥.
⎣ ⎦ ⎣ ⎦ ⎣ ⎦
Solution:
⎡6⎤
⎡6⎤
⎡ − 4⎤
− 3 ⎡ − 4⎤
× ⎢ ⎥ and so ⎢ ⎥ is parallel to ⎢ ⎥
Notice that ⎢ ⎥ =
2 ⎣2⎦
⎣− 3⎦
⎣− 3⎦
⎣ 2 ⎦
⎡2⎤
⎡ − 4⎤
but ⎢ ⎥ is not a multiple of ⎢ ⎥ and so cannot be parallel to the other two vectors.
⎣1 ⎦
⎣ 2 ⎦
⎡4⎤
Example: Find a vector of length 15 in the direction of ⎢ ⎥ .
⎣− 3⎦
⎡4⎤
Solution: a = ⎢ ⎥ has length
⎣− 3⎦
a = ⏐a⏐ =
42 + 32 = 5
⎡4⎤
⎡ 12 ⎤
and so the required vector of length 15 = 3 × 5 is 3a = 3 × ⎢ ⎥ = ⎢ ⎥ .
⎣− 3⎦
⎣− 9⎦
C4 14/04/13
25
Non-parallel vectors
If a and b are not parallel and if α a + β b = γ a + δ b, then
αa – γ a = δ b – β b ⇒ (α – γ) a = (δ – β) b
but a and b are not parallel and one cannot be a multiple of the other
⇒ (α – γ) = 0 = (δ – β)
⇒ α = γ and δ = β.
Example: If a and b are not parallel and if
b + 2a + β b = α a + 3b – 5a, find the values of α and β.
Solution:
Since a and b are not parallel, the coefficients of a and b must ‘balance out’
⇒
2= α–5 ⇒
α = 7 and 1 + β = 3
⇒ β = 2.
Modulus of a vector and unit vectors
Modulus
The modulus of a vector is its magnitude or length.
7
then the modulus of
√7
3
3
then the modulus of c = c = ⏐c⏐ =
Or, if c =
3
5
If
3 √58
5
√34
Unit vectors
A unit vector is one with length 1.
⎡− 12⎤
Example: Find a unit vector in the direction of ⎢
⎥.
⎣ 5 ⎦
⎡− 12⎤
2
2
a = ⎢
has length ⏐a⏐= a = 12 + 5 = 13 ,
⎥
⎣ 5 ⎦
⎡ −1312 ⎤
⎡− 12⎤
=
and so the required unit vector is 131 × a = 131 × ⎢
⎢5 ⎥ .
⎥
⎣ 5 ⎦
⎣ 13 ⎦
Solution:
26
C4 14/04/13 SDB
Position vectors
If A is the point (–1, 4) then the position vector of A is the vector from the origin to A,
⎡− 1⎤
= a = ⎢ ⎥.
usually written as
⎣4⎦
y
For two points A and B the position vectors are
= a and
b–a
a
= b
B
go from A → O → B
To find the vector
b
= –a + b = b – a
giving
A
O
x
Ratios
Example: A, B are the points (2, 1) and (4, 7). M lies on AB in the ratio 1 : 3. Find the
coordinates of M.
Solution :
= ⇒
⇒
2
6
B
2
0·5
6
1·5
2
0·5
1
1·5
2·5
2·5
3
M
1
A
M is (2⋅5, 2⋅5)
O
C4 14/04/13
27
Proving geometrical theorems
Example:
In a triangle OBC let M and N be the midpoints of OB and OC.
Prove that BC = 2 MN and that BC is parallel to MN.
Solution:
as b, and
Write the vectors
= ½
= ½
Then
and
as c.
O
= ½b
= ½ c.
, go from M to O using –½ b and
To find
then from O to N using ½ c
⇒
= –½ b + ½ c
⇒
= ½c –½b
M
N
b
c
B
, go from B to O using –b and
Also, to find
then from O to C using c
⇒
C
= –b + c = c – b .
But
= –½ b + ½ c = ½ (c – b) = ½ BC
⇒ BC is parallel to MN
and BC is twice as long as MN.
Example:
P lies on OA in the ratio 2 : 1, and Q lies on OB in the ratio 2 : 1. Prove that PQ
is parallel to AB and that PQ = 2/3 AB.
A
1
Let a =
Solution:
⇒
, and b =
a,
P
b−a
2
b
O
and
2
2
= /3 b − /3 a
2
Q
1
= 2/3 (b − a)
= 2/3
28
⇒
PQ is parallel to AB and
⇒
PQ = 2/3 AB.
C4 14/04/13 SDB
B
Three dimensional vectors
Length, modulus or magnitude of a vector
The length, modulus or magnitude of the vector
= ⏐a⏐ = a =
⎡ a1 ⎤
= ⎢⎢ a2 ⎥⎥ is
⎢⎣ a3 ⎥⎦
a12 + a2 2 + a32` ,
a sort of three dimensional Pythagoras.
Distance between two points
To find the distance between A, (a1, a2, a3) and B, (b1, b2, b3) we need to find the length of
the vector
.
= b – a
⇒ ⏐
⎡ b1 ⎤
⎡ a1 ⎤
⎡ b1 − a1 ⎤
⎢
⎥
⎢
⎥
= ⎢b2 ⎥ − ⎢ a2 ⎥ = ⎢⎢b2 − a2 ⎥⎥
⎢⎣ b3 ⎥⎦
⎢⎣ a3 ⎥⎦
⎢⎣ b3 − a3 ⎥⎦
⏐ = AB =
(b
1
− a1 ) + ( b2 − a2 ) + ( b3 − a3 )
2
2
2
Scalar product
a . b = ab cos θ
where a and b are the lengths of a and b
b
and θ is the angle measured from a to b.
Note that
(i)
a . a = aa cos 0o = a2
(ii)
a (b + c) = a b + a c .
(iii)
a.b = b.a
.
.
θ
a
.
since cosθ = cos (–θ)
In co-ordinate form
⎡ a1 ⎤ ⎡ b1 ⎤
a . b = ⎢ ⎥ . ⎢ ⎥ = a1b1 + a 2 b2 = ab cos θ
⎣a 2 ⎦ ⎣b2 ⎦
or
C4 14/04/13
⎡ a1 ⎤ ⎡ b1 ⎤
a . b = ⎢⎢a 2 ⎥⎥ . ⎢⎢b2 ⎥⎥ = a1b1 + a 2 b2 + a 3 b3 = ab cos θ .
⎢⎣ a 3 ⎥⎦ ⎢⎣b3 ⎥⎦
29
Perpendicular vectors
If a and b are perpendicular then θ = 90o and cos θ = 0
thus a perpendicular to b ⇒ a . b = 0
and a . b = 0 ⇒ either a is perpendicular to b
or
a or b = 0.
Example: Find the values of λ so that a = 3i – 2λj + 2k and
b = 2i + λj + 6k are perpendicular.
Since a and b are perpendicular a . b = 0
Solution:
⎡ 3 ⎤ ⎡2⎤
⇒ ⎢⎢ −2λ ⎥⎥ . ⎢⎢ λ ⎥⎥ = 0
⎢⎣ 2 ⎥⎦ ⎢⎣ 6 ⎥⎦
⇒
6 − 2 λ 2 + 12 = 0
⇒ λ2 = 9 ⇒ λ = ± 3.
⎡1⎤
⎡3⎤
⎢
⎥
Example: Find a vector which is perpendicular to a, ⎢− 1⎥ , and b, ⎢⎢1⎥⎥ .
⎢⎣ 2 ⎥⎦
⎢⎣1⎥⎦
Solution:
⎡ p⎤
Let the vector c, ⎢⎢ q ⎥⎥ , be perpendicular to both a and b.
⎢⎣ r ⎥⎦
⎡ p⎤
⇒ ⎢⎢ q ⎥⎥
⎢⎣ r ⎥⎦
⎡1⎤
⎢− 1⎥ = 0 and
⎢ ⎥
⎢⎣ 2 ⎥⎦
.
⎡ p⎤
⎢q ⎥
⎢ ⎥
⎢⎣ r ⎥⎦
.
⎡3⎤
⎢1⎥ = 0
⎢ ⎥
⎢⎣1⎥⎦
⇒ p − q + 2r = 0 and 3p + q + r = 0.
Adding these equations gives 4p + 3r = 0.
Notice that there will never be a unique solution to these problems, so having
eliminated one variable, q, we find p in terms of r, and then find q in terms of r.
30
− 3r
4
⇒
q=
5r
4
⇒
p=
⇒
⎡ −43r ⎤
⎢ ⎥
c is any vector of the form ⎢ 54r ⎥ ,
⎢⎣ r ⎥⎦
C4 14/04/13 SDB
and we choose a sensible value of r = 4 to give
⎡ − 3⎤
⎢ 5 ⎥.
⎢ ⎥
⎢⎣ 4 ⎥⎦
Angle between vectors
Example: Find the angle between the vectors
= 4i – 5j + 2k
Solution:
and
= –i + 2j – 3k,
to the nearest degree.
First re-write as column vectors (if you want)
⎡ − 1⎤
⎢2⎥
⎢ ⎥
⎢⎣ − 3⎥⎦
⎡ 4 ⎤
a = ⎢⎢ − 5⎥⎥ and b =
⎢⎣ 2 ⎥⎦
a = ⏐a⏐ =
4 2 + 5 2 + 2 2 = 45 = 3 5 ,
.
a . b = ab cos θ
⇒
14
–20 = 3 5 × 14 cosθ
⇒
cos θ =
⇒
θ = 143o to the nearest degree.
3 70
12 + 2 2 + 32 =
⎡ − 1⎤
⎢ 2 ⎥ = –4 – 10 – 6 = –20
⎢ ⎥
⎢⎣ − 3⎥⎦
⎡ 4 ⎤
and a . b = ⎢⎢ − 5⎥⎥
⎢⎣ 2 ⎥⎦
− 20
b = ⏐b⏐ =
= – 0.796819
Vector equation of a straight line
⎡ x⎤
r = ⎢⎢ y ⎥⎥ is usually used as the position vector of a
⎣⎢ z ⎦⎥
R
general point, R.
l
A
In the diagram the line l passes through the point A
and is parallel to the vector b.
r
To go from O to R first go to A, using a, and then
from A to R using some multiple of b.
C4 14/04/13
a
O
b
31
⇒ The equation of a straight line through the point A and parallel to the vector b is
r = a + λ b.
Example: Find the vector equation of the line through the points M, (2, –1, 4),
and N, (–5, 3, 7).
Solution:
We are looking for the line through M (or N) which is parallel to the vector
⎡− 5⎤
⎡2⎤
⎡− 7 ⎤
⎢
⎥
⎢
⎥
= n – m = ⎢ 3 ⎥ − ⎢− 1⎥ = ⎢⎢ 4 ⎥⎥
⎢⎣ 7 ⎥⎦
⎢⎣ 4 ⎥⎦
⎢⎣ 3 ⎥⎦
⎡2⎤
⎡− 7⎤
⎢
⎥
equation is r = ⎢− 1⎥ + λ ⎢⎢ 4 ⎥⎥ .
⎢⎣ 4 ⎥⎦
⎢⎣ 3 ⎥⎦
⇒
Example:
Show that the point P, (–1, 7, 10), lies on the line
⎡1 ⎤
⎡− 1⎤
⎢
⎥
r = ⎢3⎥ + λ ⎢⎢ 2 ⎥⎥ .
⎢⎣4⎥⎦
⎢⎣ 3 ⎥⎦
Solution:
⇒
The x co-ord of P is –1 and of the line is 1 – λ
–1 = 1 – λ ⇒ λ = 2.
In the equation of the line this gives y = 7 and z = 10
⇒
P, (–1, 7, 10) does lie on the line.
Intersection of two lines
2 Dimensions
Example: Find the intersection of the lines
⎡2⎤
⎣3⎦
⎡ −1⎤
⎥,
⎣2⎦
l1 , r = ⎢ ⎥ + λ ⎢
and
⎡1⎤
⎡1⎤
l2, r = ⎢ ⎥ + μ ⎢ ⎥ .
⎣ 3⎦
⎣ −1⎦
Solution: We are looking for values of λ and μ which give the same x and y
co-ordinates on each line.
32
Equating x co-ords
⇒
2–λ =1+μ
equating y co-ords
⇒
3 + 2λ = 3 – μ
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.
⇒
Adding
⇒
5+λ=4
⇒ λ = –1
⇒
μ = 2
lines intersect at (3, 1).
3 Dimensions
This is similar to the method for 2 dimensions with one important difference – you can not be
certain whether the lines intersect without checking.
You will always (or nearly always) be able to find values of λ and μ by equating
x coordinates and y coordinates but the z coordinates might or might not be equal and
must be checked.
Example: Investigate whether the lines
l 1,
⎡2⎤
⎡− 1⎤
⎢
⎥
r = ⎢1⎥ + λ ⎢⎢ 2 ⎥⎥
⎢⎣3⎥⎦
⎢⎣ 1 ⎥⎦
and
⎡− 3⎤
l 2, r = ⎢⎢ 1 ⎥⎥ + μ
⎢⎣ 5 ⎥⎦
⎡1⎤
⎢3⎥
⎢ ⎥
⎢⎣1⎥⎦
intersect
and if they do find their point of intersection.
Solution: If the lines intersect we can find values of λ and μ to give the same x, y and z
coordinates in each equation.
Equating x coords
⇒
2 – λ = –3 + μ,
I
equating y coords
⇒
1 + 2λ = 1 + 3μ,
II
equating z coords
⇒
3 + λ = 5 + μ.
III
2 × I + II ⇒ 5 = –5 + 5μ ⇒ μ = 2, in I ⇒ λ = 3.
We must now check to see if we get the same point for the values of λ and μ
In l 1, λ = 3 gives the point (–1, 7, 6);
in l 2, μ = 2 gives the point (–1, 7, 7).
The x and y co-ords are equal (as expected!), but the z co-ordinates are different and so
the lines do not intersect.
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33
7
Appendix
Binomial series (1 + x)n for any n – proof
Suppose that
f (x) = (1 + x)n = a + bx + cx2 + dx3 + ex4 + …
put x = 0, ⇒ 1 = a
⇒ f ′(x) = n(1 + x)n − 1 = b + 2cx + 3dx2 + 4ex3 + …
put x = 0, ⇒ n = b
⇒ f ′′(x) = n(n − 1)(1 + x)n − 2 = 2c + 3 × 2dx + 4 × 3ex2 + …
put x = 0, ⇒ n(n − 1) = 2c ⇒
⇒ f ′′′(x) = n(n − 1)(n − 2)(1 + x)
!
n−3
= 3 × 2d + 4 × 3 × 2ex + …
put x = 0, ⇒ n(n − 1)(n − 2) = 3 × 2d ⇒
!
Continuing this process, we have
and
!
, etc.
!
giving f (x) = (1 + x)n
= 1 + nx +
+
+
!
… +
!
+
!
+…
!
+…
!
Showing that this is convergent for ⏐x⏐ < 1, is more difficult!
Derivative of xq for q rational
Suppose that q is any rational number,
Then
⁄
y = xq =
, where r and s are integers, s
⇒ ys = xr
= r × xr − 1
Differentiating with respect to x ⇒ s ×
⇒
⇒
34
=
≠ 0.
= qx q − 1
since q =
since ys = xr and y = xq
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which follows the rule found for xn, where n is an integer.
for negative limits
We know that the ‘area’ under any curve, from
x = a to x = b is approximately
y
, 0
δx
a
b
If the curve is above the x-axis, all the y values are positive, and if a < b then all values of δ x
are positive, and so the integral is positive.
1
ln|x|
Example: Find
∫
−3
−1
ln
ln
1
dx .
x
Solution: The integral wanted is shown as A′
in the diagram.
y=1/x
1
By symmetry ⏐A′⏐ = A (A positive)
A
and we need to decide whether the integral
is +A or − A.
−3
−2 A'
−1
1
2
3
−1
From x = −1 to x = −3, we are going in
the direction of x decreasing
⇒ all δ x are negative.
And the graph is below the x-axis,
⇒ the y values are negative,
⇒
y δ x is positive
⇒
∑
⇒
the integral is positive and equal to A.
0 The integral, A′ = A =
⇒ A′ =
∫
−3
−1
∫
3
1
0
1
dx = [ ln x ] 13 = ln 3 − ln 1 = ln 3
x
1
dx = ln 3
x
Notice that this is what we get if we write ln ⏐x⏐ in place of ln x
∫
−3
−1
1
dx = [ ln | x |] −−31 = ln 3 – ln 1 = ln 3
x
As it will always be possible to use symmetry in this way, since we can never have one
positive and one negative limit (because there is a discontinuity at x = 0), it is correct to write
ln ⏐x⏐ for the integral of 1/x .
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35
Integration by substitution – why it works
We show the general method with an example.
1
1
integrand =
√1
Choose u = 1 + x3
But 3
1
3
rearrange to give 3
leave the x2 because it appears in dx
√ √ 1
3
this is the same as writing the integrand in
terms of u, and then replacing dx by
The essential part of this method, writing the integrand in terms of u, and then replacing dx
by
, will be the same for all integrations by substitution.
Parametric integration
This is similar to integration by substitution.
36
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Separating the variables – why it works
We show this with an example.
If y = 6y3 then
and so
18
= 18y2
18
6
Notice that we ‘cancel’ the dx.
= x2 sec y
Example: Solve
Solution:
⇒
= x2 sec y
cos y
= x2
⇒
cos ⇒
cos ⇒
sin y = C4 14/04/13
‘cancelling’ the dx
+ c
37
Index
Binomial expansion
convergence, 8
for any n, 8
proof - for any n, 34
Differential equations
exponential growth and decay, 23
forming, 12
separating the variables, 22
Differentiation
derivative of xq for q rational, 34
exponential functions, ax, 11
related rates of change, 12
Implicit differentiation, 10
Integration
1
/x for negative limits, 35
by parts, 19
by substitution – why it works, 36
choosing the substitution, 18
ex, 13
ln x, 13
parametric, 21
parametric – why it works, 36
reverse chain rule, 14
sec x and cosec x, 15
separating variables – why it works, 37
standard integrals, 13
substitution, definite, 18
substitution, indefinite, 16
tan x and cot x, 15
using partial fractions, 16
using trigonometric identities, 13
volume of revolution, 20
Parameters
area under curve, 7
equation of circle, 6
equation of hyperbola, 6
intersection of two curves, 5
parametric equations of curves, 5
parametric to Cartesian form, 6
trig functions, 6
38
Parametric differentiation, 11
Parametric integration, 36
areas, 21
volumes, 21
Partial fractions, 3
integration, 16
linear repeated factors, 4
linear unrepeated factors, 3
quadratic factors, 3
top heavy fractions, 4
Scalar product, 29
perpendicular vectors, 30
properties, 29
Vectors, 24
adding and subtracting, 24
angle between vectors, 31
displacement vector, 24
distance between 2 points, 29
equation of a line, 31
intersection of two lines, 32
length of 3-D vector, 29
modulus, 26
non-parallel vectors, 26
parallel vectors, 25
position vector, 27
proving geometrical theorems, 28
ratios, 27
unit vector, 26
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