MAT1A01: Substitution in Integration
Dr Craig
22 May 2013
Video memos for Semester Tests
These are available at:
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and
bit.ly/MAT1A01TEST1
Or YouTube “MAT1A01”
Revision class
There will be a revision class in D1 Lab K01 from
12pm-3pm this Saturday – 25 May.
WebAssign
Revision assignments will be posted on WebAssign.
Do them!
Getting help with your exam revision
Maths Learning Centre: C-Ring 512
My office: C-Ring 533A (Stats Dept corridor)
If you want to be certain I will be there – make an
appointment:
[email protected]
Otherwise, just arrive and if I am free, I will help
you.
Fundamental Theorem of Calculus: Part 1
If f is continuous on [a, b], then the function g
defined by
Z x
g(x) =
f (t) dt
a6x6b
a
is continuous on [a, b] and differentiable on (a, b)
and g 0 (x) = f (x).
In other words:
d
dx
Z
a
x
f (t) dt = f (x)
First, we revise u substitution when using FTC1.
Suppose
√
Z
g(x) =
x
cos x dx
a
First, we revise u substitution when using FTC1.
Suppose
√
Z
x
g(x) =
cos x dx
a
Then
√
1 g (x) = cos( x). √
2 x
0
First, we revise u substitution when using FTC1.
Suppose
√
Z
x
g(x) =
cos x dx
a
Then
√
1 g (x) = cos( x). √
2 x
0
In general, if
Z
u(x)
f (t) dt
g(x) =
a
then
g 0 (x) = f (u(x)).
du
dx
Consider the following example:
Z
p
2x 1 + x2 dx
Consider the following example:
Z
p
2x 1 + x2 dx
To find the integral, we use substitution. We let
u = 1 + x2
Consider the following example:
Z
p
2x 1 + x2 dx
To find the integral, we use substitution. We let
u = 1 + x2
Note that
du
du
= 2x and so dx =
dx
2x
The original integral then becomes
Z
Z
p
√
2x 1 + x2 dx =
u du
The substitution rule
If u = g(x) is a differentiable function whose range
is an interval I and f is continuous on I, then
Z
Z
0
f (g(x)).g (x) dx = f (u) du
The substitution rule is like the chain rule in reverse.
Z
Example: Find
u = cos θ.
cos3 θ sin θ dθ
by letting
Z
Example: Find
cos3 θ sin θ dθ
u = cos θ.
Z
Example: Find
x sin(x2 ) dx
by letting
Z
Example: Find
cos3 θ sin θ dθ
by letting
u = cos θ.
Z
Example: Find
Z
Example: Find
x sin(x2 ) dx
p
(x + 1) 2x + x2 dx.
Integral of tan x
Z
tan x dx
Do we know anything that can be differentiated to
give tan x?
Integral of tan x
Z
tan x dx
Do we know anything that can be differentiated to
give tan x? No, but we can rewrite this integral as
Z
sin x
dx
cos x
Z
Z
tan x dx =
Let u = cos x. Then
du
dx
sin x
dx
cos x
= − sin x.
Z
Z
tan x dx =
Let u = cos x. Then
du
dx
sin x
dx
cos x
= − sin x.
Finally
Z
tan x dx = ln | sec x| + C
Substitution rule for definite integrals
When we use u-substitution to calculate a definite
integral, we will need to change the limits of the
integral.
Substitution rule for definite integrals
When we use u-substitution to calculate a definite
integral, we will need to change the limits of the
integral.
Z 4
√
Example:
2x + 1 dx.
0
Substitution rule for definite integrals
When we use u-substitution to calculate a definite
integral, we will need to change the limits of the
integral.
Z 4
√
Example:
2x + 1 dx. We let u = 2x + 1.
0
du
Now
= 2 and so dx = 12 du.
dx
Substitution rule for definite integrals
When we use u-substitution to calculate a definite
integral, we will need to change the limits of the
integral.
Z 4
√
Example:
2x + 1 dx. We let u = 2x + 1.
0
du
Now
= 2 and so dx = 12 du.
dx
If x = 0, then u = 1 and if x = 4, u = 9. Thus
Z 4
Z 9
√
1√
2x + 1 dx =
u du
0
1 2
Example:
4
Z
√
0
x
dx
1 + 2x
Integrals of Symmetric Functions
Suppose that f is continuous on [−a, a]. Then
1. If f is even (i.e. f (−x) = f (x)), then
Z a
Z a
f (x) dx = 2
f (x) dx
−a
0
Integrals of Symmetric Functions
Suppose that f is continuous on [−a, a]. Then
1. If f is even (i.e. f (−x) = f (x)), then
Z a
Z a
f (x) dx = 2
f (x) dx
−a
0
2. If f is odd (i.e. f (−x) = −f (x)), then
Z a
f (x) dx = 0
−a
Examples of Integrals of Symmetric Functions
Z
2
−2
x2 dx
Examples of Integrals of Symmetric Functions
Z
2
−2
x2 dx = 2
Z
0
2
3 2
8 16
x
2
x dx = 2
=2
=
3 0
3
3
Examples of Integrals of Symmetric Functions
Z
2
x2 dx = 2
−2
Z
0
Z
1
x dx
−1
2
3 2
8 16
x
2
x dx = 2
=2
=
3 0
3
3
Examples of Integrals of Symmetric Functions
Z
2
x2 dx = 2
−2
Z
0
Z
2
3 2
8 16
x
2
x dx = 2
=2
=
3 0
3
3
1
x dx =
−1
x2
2
1
=
−1
1 1
− =0
2 2
Questions?
Questions?
Good luck!