Transcript Document

Substitution Method
Integration
When one function is not the derivative of the
other e.g.
ò
x(4x -1) 6 dx
u = 4 x -1
x is not the derivative
of
(4x
-1)
and
u+1
Rearranging x =
x is a4 variable
1
Differentiating dx = du
4
Let
Substitute
æ u+ 1ö 6 1
1
ò çè 4 ÷øu 4 du = 16
ò (u
7
+ u )du
6
1
Integrate
16
8
7
1
u
u
7
6
ò (u + u )du = 16 ( 8 + 7 ) + c
Substitute
ò
u= 4x -1
8
7
1
(4x
-1)
(4
x
-1)
x(4x -1) 6 dx = (
+
)+c
16
8
7
ò
8
7
(4
x
-1)
(4x
-1)
x(4x -1) 6 dx =
+
)+c
128
112
Example 2
ò
x -1
dx
x+ 4
xLet
- 1 is not the
u derivative
= x + 4of x +4 and it contains a variable
x=u -4
2
Rearranging
Differentiating
Substitute
ò
dx = 2u du
u2 - 4 -1
2
´ 2u du = 2 (u - 5)du
u
ò
Integrating and substituting back in for u
3
u
2 ò (u - 5)du = 2( - 5u) + c
3
3
1
2
= (x + 4) 2 -10(x + 4) 2 + c
3
2
Delta Exercise 12.8
The definite integral
Example 1
2x
ò 2 1+ x2 dx
4
As 2x is the derivative, use inverse chain rule to integrate
[
2x
2
dx
=
ln1+
x
ò 2 1+ x2
4
Substitute x = 4
]
4
2
= ln17 - ln5 = 1.224
Substitute x = 2
Example4x2divided by 2x = 2
Solving
x = 1/2
ò
2
1
4x + 3
dx
2x -1
Substitute x = 1/2
into 4x + 3 to get 5
Divide the top by the bottom
ò
2
1
4x + 3
dx =
2x -1
5
ò1 (2 + 2x -1)dx
2
é
2
ln 2x -1 ù
5
5ln 3
5ln1
ò1 (2 + 2x -1)dx = êë2x + 5 2 úû = (4 + 2 ) - (2 + 2 )
1
2
= 4.747
Example 3
ò
3
0
x 1+ x dx
Use substitution
Let
u = 1+ x
Rearrange
x = u2 -1
Differentiating
ò
When x = 3,
When x = 0,
u =2
u =1
dx = 2u du
Substituting
3
0
4
2
(u
-1)u´
2udu
=
2
(u
u
du
ò1
ò
1
2
5
3
é2u 2u ù
2
4
2
2 ò (u - u du = ê
= 7.733
ú
1
3 û1
ë 5
x 1+ xdx =
2
2
2
Delta Exercise 12.9
Areas under curves
To find the area under the curve
between a and b…
…we could break the area up into
rectangular sections. This would
overestimate the area.
…or we could break the area up
like this which would
underestimate the area.
The more sections we divide the area up into,
the more accurate our answer would be.
If each of our sections was infinitely narrow,
we would have the area of each section as
Area = y ´ ¶x = y ´ dx
y
The total area would be the sum of all these areas
between a and b.
ò y dx
is the sum all the areas of infinitely narrow
width, dx and height, y.
ò
b
a
y dx = area under the curve between a and b
As the value of dx decreases, the area of the
rectangle approaches y x dx
y
0
dx
The area of this triangle is 3 units squared
2
The equation of the line is y = x
3
2
If we sum all
rectangles
y
0
dx
3
é1 2ù
2
ò 0 y dx = ò 0 3 x dx = êë 3 x úû = 3 - 0 = 3
0
3
3
3
The area of this triangle is 3 units squared
2
The equation of the line is y = - x
3
0
dx
If we sum all
rectangles
3
y
The area is 3
but the integral is -3
2
é 1 2ù
2
ò 0 y dx = ò 0 - 3 x dx = êë- 3 x úû = -3 - 0 = -3
0
3
3
3
http://rowdy.mscd.edu/~talmanl/
MathAnim.html
2011 Level 2
2011 Level 2
ò (x
2
-1
3
- 3x + 4 ) dx
2
éx
ù
3
= ê - x + 4xú = 6.75
ë4
û-1
4
2
2010 Level 2
2010 Level 2
ò
2
0
3
2
x
+
2x
- 8x) dx
(
é x 2x
ù
2
=ê +
- 4x ú = -6.6
3
ë4
û0
4
3
2
• Area cannot be
negative
• Area = 6.67 units2
Combination
Integral is
positive
-6
-1
Integral is
negative
8
To find the area under the curve, we must integrate
between -6 and -1 and between 8 and -1 separately and
add the positive values together.
-6
-1
8
Area =
ò
-1
f
(x)
+
-6
ò-6 (x - x - 50x - 48)dx +
-1
3
2
ò
8
f
(x)
-1
3
2
(x
x
- 50x - 48)dx
ò-1
8
2011 Level 2
2011 Level 2
é x3 ù
m+2 2
ò m x dx = êë 3 úû
m
m+2
1
1 3
3
= ( m+ 2 ) - m
3
3
8
1
2
= 2m + 4m+ = 5
3
6
m = 0.5, m ¹ -2.5
2010 Question 1c
2010 Question 1c
ò
1
-1
ae2 xdx
é ae ù
=ê
ú
ë 2 û-1
2x 1
ae2 ae-2
=
2
2
2012
2012
4x -1
ò 2 x+ 3 dx
6
13
= ò 4dx
2
x+ 3
6
= éë4x -13ln x+ 3 ùû
6
2
= 8.359
2012
2012
• First find the x-value
of the intersection
point
x a
= 2 Þ x=a
2
a x
2012
æa xö
ò1 çè x2 - a2 ÷ø dx
a
é -a x ù
=ê - 2ú
ë x 2a û1
2
a
3
1
= - + a+ 2
2
2a
2010 Question 1e
2010 Question 1e
• Find intersection
points
1
x = px Þ x =
3 p
-2
1
x = 8x Þ x =
2
-2
2010 Question 1e
ò 0 ( 8x - px)dx+ ò
1/2
1
2
é 2 px2 ù
= ê4x ú
2 û0
ë
1
3 p
1
2
-2
x
( - px)dx
1
3 p
é -1 px2 ù
+ê ú
2 û1
ëx
2
3 pö æ
æ pö æ 3
pö
= ç1- ÷ + ç- p ÷ - ç-2 - ÷
è 8ø è
2 ø è
8ø
33 p
= 32
Looking at areas a different way
As the value of dy decreases, the area of the
rectangle approaches x x dy
The equation of the line is
3
y=- x+ 3
4
3
Rearrange
dy
x
0
4
Definite Integral is
ò
3
0
x dy
4( 3 - y)
4
x=
=4- y
3
3
2 ù3
é
3æ
4 ö
2y
= ò ç 4 - y÷dy = ê4 y ú =6
0è
3 ø
3 û0
ë
Areas between two curves
y= x
y =x
2
A typical rectangle in the upper
Solving these
section
Equations gives
1
x=y
y=1
x = y2
x-x
Area =(x - x )dy
Area for this section is
1
ò 0 (y - y )dy = 6
1
2
dy
A typical rectangle in the lower
section
x=y
x = y2
x -x
Area =(x - x )dy
Area for this section is
5
ò-1(y - y)dy = 6
0
2
Total area is equal to 1
dy
y= x
Example 2
2
A typical rectangle
dx
y-y
Area = (y - y)dx
y =1- x
2´
ò
0.707
0
0.707
2
(y2 - y1 )dx = 2 ´
ò
0.707
0
Area
(1- x - x )dx = 2 ´
2
2
ò
0.707
0
1
(1)dx = 2
= 1.414
2
Practice
More practice
Delta Exercise 16.2, 16.3, 16.4
Worksheet 3 and 4
Area in polar: extra for experts