Aim: What is Integration by Substitution?
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Transcript Aim: What is Integration by Substitution?
Aim: What is Integration by Substitution?
Do Now:
Find the derivative of f ( x ) 3 x 2 x
Aim: Integration by Substitution
Course: Calculus
2
3
Chain Rule
If y = f(u) is a differentiable function of u
and u = g(x) is a differentiable function of x,
then y = f(g(x)) is a differentiable function of
x and
dy dy du
dx du dx
or, equivalently,
d
f ( g( x )) f '( g( x )) g '( x )
dx
Think of the composite function as having
2 parts – an inner part and an outer part.
outer
y f ( g ( x )) f ( u)
inner
Aim: Integration by Substitution
Course: Calculus
Do Now
Find the derivative of f ( x ) 3 x 2 x
2
Then f ( x ) u
Let u = 3x –
n 1 du
dy
General Power Rule
n u x
dx
dx
3
2x2
u’
un - 1
n
2
2
3 4x
f ' 3 3x 2x
f ' 3 3x 2x
Aim: Integration by Substitution
2
d
3 x 2 x 2
dx
2
Course: Calculus
3
u - substitution
dy
Find
for y x 2 1
dx
3
d
f ( g( x )) f '( g( x )) g '( x )
dx
y f ( g( x ))
u g( x )
y f ( u)
y x 1
2
3
y u3
u x2 1
dy dy du
dx du dx
3u2 2 x
chain rule
2
3 x 1 2x
2
6x x 1
2
Aim: Integration by Substitution
2
substitute
&
simplify
6x x
2
2
1 dx ?
Course: Calculus
u – substitution in Integration
From the definition of an antiderivative it
follows that:
F '( g( x )) g '( x ) dx F g( x ) C
F u C
Let g be a function whose range is an interval I,
and let f be a function that is continuous on I. If g
is differentiable on its domain and F is an
antiderivative of f on I, then
f ( g( x )) g '( x ) dx F g( x ) C
If u g( x ), then du g '( x )dx and
f (u) du F (u) C
Aim: Integration by Substitution
Course: Calculus
Recognizing the Pattern
g(x) 2 g’(x)
3
1 g(x)
dy
y
x2 1 4
x2 1 2 x
3
dx
inside function g(x)
g(x)
g’(x)
3
1 2
2
2
x 1 C
Evaluate x 1 2 x dx
3
(g(x))2
Recognizing the f '( g( x )) g '( x ) Pattern
g(x) = x2 + 1
g’(x) = 2x
f ( g( x )) ( g( x ))2 f ( x 2 1)
Aim: Integration by Substitution
Course: Calculus
Model Problem
g’ g(x)
cos xdx sin x C
Evaluate 5cos 5 x dx
Recognize the f '( g( x )) g '( x ) Pattern?
f ' uu '
What is the inside function, u? 5x
f ( x ) cos( g( x ))
cos(g(x)) g’
Evaluate cos5 x 5 dx sin5x C
Check
Aim: Integration by Substitution
sin ax
cos axdx a C
kf ( x )dx k f ( x )dx
Course: Calculus
Multiplying/Dividing by a Constant
kf ( x ) dx k f ( x ) dx
g(x)
Evaluate x x 1
2
2
dx
Recognize the f '( g( x )) g '( x ) Pattern?
What is the inside function? x2 + 1
f ( x ) ( g( x ))2
g '( x ) 2 x
problem!
factor of 2 is missing
x x
2
1
2
dx
Aim: Integration by Substitution
k!
1
x 1 2 x dx
2
2
2
Course: Calculus
Multiplying/Dividing by a Constant
x x
2
1
2
Constant
Multiple Rule
1
dx x 1 2 x dx
2
2
1
2
x 1 2 x dx
2
2
2
2
1
g ( x ) g '( x ) dx
2
1 g ( x )
C
2
3
3
1 2
x 1 C
6
3
Integrate
Check
Aim: Integration by Substitution
Course: Calculus
Change of Variable
-u
Recognizing
the Pattern
u 3
1 g(x)
y
x2 1 4
3
g(x)
u 2
dy
x2 1 2 x
dx
u
inside function g(x)
g(x)
u
g’(x)
u’
3
2
1 2
2
x 1 C
Evaluate x 1 2 x dx
3
u2 2
(g(x))
Recognizing the f '( g( x )) g '( x ) Pattern
u = g(x) = x2 + 1
u’ = g’(x) = 2x
f ( g( x )) ( g( x )) f (u) f ( x 1) u
2
Aim: Integration by Substitution
2
Course: Calculus
2
Change of Variables - u
f ( g( x )) g '( x ) dx f (u)du F (u) c
Evaluate
Calculate the
differential
u 2x 1
2 x 1 dx
u 2x 1
du 2dx
du
dx
2
Constant
Multiple Rule
Substitute in
terms of u
Antiderivative
in terms of u
1 12
du
u u du
2
2
1 u3 2
C
2 3 2
1 32
u C
3
1
32
2 x 1 C
3
Aim: Integration by Substitution
Course: Calculus
Antiderivative
in terms of x
Guidelines for Making Change of Variables
1. Choose a substitution u = g(x). Usually it
is best to choose the inner part of a
composite function, such as a quantity
raised to a power.
2. Compute du = g’(x) dx
3. Rewrite the integral in terms of the
variable u.
4. Evaluate the resulting integral in terms
of u.
5. Replace u by g(x) to obtain an
antiderivative in terms of x.
6. Check your answer by differentiating.
Aim: Integration by Substitution
Course: Calculus
Model Problem
Evaluate x 2 x 1 dx
u1
x
2
Substitute in
terms of u
u 2x 1
u 2x 1
du
dx
2
u 1 1 2 du
u
2
2
1
1
12
u 1 u
du u3 2 u1 2 du
4
4
Antiderivative
in terms of u
1 u5 2 u 3 2
C
4 5 2 3 2
Antiderivative
in terms of x
1
1
52
32
2 x 1 2 x 1 C
10
6
Aim: Integration by Substitution
Course: Calculus
Model Problem
Evaluate sin 2 3 x cos 3 x dx
u = sin 3x
du
cos 3 x 3 du cos 3 x 3 dx
dx
du
cos 3 x dx
sin
xdx
cos
x
C
3
cos
ax
2
2 du
sin
axdx
C
sin 3 xacos 3 x dx u
3
1 2
1 u3
u du C
3 3
3
u3
1 3
C sin 3 x C
9
9
Check
Aim: Integration by Substitution
Course: Calculus
General Power Rule for Integration
If g is a differentialbe function of x , then
g ( x )
dx
g( x ) g ' x
n
n 1
C
n1
Equivalently, if u g( x ), then
n 1
u
u du n 1 C
n
n 1
u4
u = 3x – 1
3 3 x 1
4
n 1
du
dx 3 x 1 3 dx
4
u5/5
3 x 1
5
Aim: Integration by Substitution
5
C
Course: Calculus
Model Problems
u=
x2
2 x 1 x
+x
2
u1
x dx x x
u2/2
x
2
2
3x
x 2 dx
3
x
2 x 1 dx
1
2
C
2
u1/2
u = x3 – 2
2
du
du
x 2
3
12
3 x 2dx
u3 2 / 3 2
x
Aim: Integration by Substitution
3
2
32
3 2
C
Course: Calculus
Model Problems
u=1–
4 x
1 2 x
2
u-2
2x2
2
du
4 x dx
2
dx 1 2 x
u-1/(-1)
2
1 2x
2
1
C
1
u2
u = cos x
cos
2
du
x sin x dx cos x sin x dx
2
3
u /3
cos x
3
Aim: Integration by Substitution
3
C
Course: Calculus
Change of Variables for Definite Integrals
If the function u f ( x ) has a continuous
derivative on the closed interval [a , b]
and f is continuous on the range of g , then
f g x g ' x dx
b
g(b)
a
g(a )
u = x2 + 1
f u du
du = 2x dx
3
1
3
1 1 2
2
0 x x 1 dx 2 0 x 1 2 x dx
determine new upper and lower limits of integration
lower limit
When x = 0, u = 02 + 1 = 1
integration
limits for x
and u
upper limit
x = 1, u = 12 + 1 = 2
2
1
1 15
1 u
1 2 3
u du 4
1
2
4 8
2 4 1
2
Aim: Integration by Substitution
Course: Calculus
4
Model Problem
5
x
Evaluate A =
dx
1
2x 1
2
u
1
2
dx u du
x
u 2x 1
u 2x 1
2
determine new upper and lower limits of integration
lower limit
upper limit
When x = 1, u = 1
x = 5, u = 3
3
1
1 u2 1
1 3 2
u du u 1 du
u 2
2 1
3
1
1
16
1 u
u 9 3 1
2
3
2 3
3
1
3
Aim: Integration by Substitution
Course: Calculus
Model Problem
Before substitution
After substitution
7
5
6
4
5
3
4
3
2
2
1
1
2
4
6
2
x
A =
dx
1
2x 1
Area of region
is 16/3
5
=
Aim: Integration by Substitution
4
u2 1
A =
du
1
2
Area of region
is 16/3
3
Course: Calculus
Even and Odd Functions
Let f be integrable on the closed
interval [ a , a ].
1. If f is an even function, the
a
a
f x dx 2 f x dx .
a
0
2. If f is an odd function, the
a
a
f x dx
0
a
f x dx f x dx .
a
0
To prove even, use f(x) = f(-x) and then
substitute u = -x
Aim: Integration by Substitution
Course: Calculus
Model Problem
Evaluate
A =
2
2
sin 3 cos x sin x cos x dx = 0
1
0.5
-1
1
-0.5
-1
f ( x ) sin3 x cos x sin x cos x
f ( x ) sin 3 x cos x sin x cos x
f ( x ) sin3 x cos x sin x cos x f ( x )
Aim: Integration by Substitution
Course: Calculus
Multiplying/Dividing by a Constant
Aim: Integration by Substitution
Course: Calculus
Multiplying/Dividing by a Constant
Aim: Integration by Substitution
Course: Calculus