Download Lecture 5 §7.1 Integration by parts §7.8 Improper integrals

Lecture 5
§7.1 Integration by parts
§7.8 Improper integrals
Lecture 5
Integration
Integration by parts
Recall the product rule for differentiation
d
[f (x)g(x)] = f 0 (x)g(x) + f (x)g 0 (x).
dx
Thus,
Z
f (x)g(x) =
[f 0 (x)g(x) + f (x)g 0 (x)] dx.
Formular for integration by parts:
Z
Z
0
f (x)g (x) dx = f (x)g(x) − f 0 (x)g(x) dx
Let u = f (x) and v = g(x). The formula becomes
Z
Z
udv = uv − vdu
Lecture 5
Integration
Sometimes you need to apply the integration by parts formula
more than once.
R
Example(7.1-1) Compute x sin x dx
R
Exercise(7.1-5) Compute xex/2 dx
R
Example(7.1-3) Compute t2 et dt
Lecture 5
Integration
Definite integrals
Thanks to the Fundamental Theorem of Calculus we can compute
definite integral using the integration by parts formula
Z b
Z b
0
b
f (x)g (x) dx = f (x)g(x)|a −
f 0 (x)g(x) dx
a
a
Lecture 5
Integration
Example(7.1-5). Calculate
1
(Hint: (arctan x)0 = 1+x
2)
R1
Exercise(7.1-24) Compute
R1
0
0
arctan x dx
(x2 + 1)e−x dx
Lecture 5
Integration
LIATE principle of integration by parts
A rule of thumb suggests to choose the function that comes first in
the following list as u:
Logarithmic: ln x
Inverse trigonometric: arcsin x, arccos x, arctan x
Algebraic: x2
Trigonometric: sin x, etc
Exponential: ex
Lecture 5
Integration
Some strategy
Simplify
the integrand,
R
R 2 for example
x(x
+
1)
dx
=
(x
R
R
R + x) dx;
2
tan θ cos θ dθ = sin θ cos θ dθ = 12 sin 2θ dθ.
R
Look for substitution sin 2θ dθ, 2θ = u.
Try integration by parts.
Relate to previous exercises!
Lecture 5
Integration
Example (7.5-5) Solve
R q 1−x
Exercise (7.5-9) Evaluate
1+x
R3
1
dx
x4 ln x dx
Lecture 5
Integration
Infinite intervals
To determine the area which lies under y = 1/x2 , above the x-axis
and to the right of x = 1, we have to compute an integral of this
form
Z ∞
1
dx
x2
1
For all t > 1
Z t
1
1
dx = 1 − < 1
2
t
1 x
Lecture 5
Integration
We have
Z
1
∞
1
dx = lim
t→∞
x2
Lecture 5
Z
1
t
1
dx = 1
x2
Integration
Improper integral of type 1: infinite intervals
If
Rt
a
f (x) dx exists for all t ≥ a, then
Z
∞
Z
t
f (x) dx = lim
t→∞ a
a
f (x) dx
if this limit exists and is a finite number.
Rb
The integral −∞ f (x) dx is defined in a similar way.
R∞
Rb
f (x) dx and −∞ f (x) dx are called convergent if the
limit exists and divergent if it does not.
R∞
Rb
If a f (x) dx and −∞ f (x) dx are convergent, we define
a
Z
∞
Z
f (x) dx =
−∞
t
Z
a
f (x) dx +
Note: a is arbitrary.
Lecture 5
f (x) dx.
−∞
a
Integration
Example (7.8-1) Determine
R∞
1
1
x
dx
Example (7.8-4) For which p is the following integral convergent?
Z ∞
1
dx
p
x
1
Lecture 5
Integration
Improper integral of type 2: discountinuous integrands
We can define the improper integral also when a function f has a
discontinuity (for example an asymptote)
Lecture 5
Integration
If f is continuous in [a, b) and discontinuous at b
Z b
Z t
f (x) dx = lim
f (x) dx
t→b−
a
a
if this limit exists and is a finite number.
If f is continuous in (a, b] and discontinuous at a
Z b
Z b
f (x) dx = lim
f (x) dx
t→a+
a
t
if this limit exists and is a finite number.
Rb
a f (x) dx is called convergent if the limit exists and
divergent if it does not.
Rc
If f has a discontinuity at c ∈ [a, b] and both a f (x) dx and
Rb
c f (x) dx are convergent, we define
Z b
Z c
Z a
f (x) dx =
f (x) dx +
f (x) dx.
a
a
Lecture 5
c
Integration
Example (7.8-5) Find
R5
2
√1
x−2
dx
Example (7.8-5) Evaluate (if possible)
Lecture 5
R3
1
0 x−1
Integration
dx
Comparison Theorem
When you are not interested in the evaluating an integral but only
in knowing whether it converges or not you can use the following
Suppose f and g are continuous functions with f (x) ≥ g(x) ≥ 0
for x ≥ a.
If
R∞
If
R∞
0
0
f (x) dx is convergent, then
g(x) dx is divergent, then
Lecture 5
R∞
0
R∞
0
g(x) dx is convergent
f (x) dx is divergent
Integration
R∞
Exercise (7.8-25) Determine if 1 x(ln1x)3 dx is convergent or
divergent. If it converges evaluate it.
R∞
Exercise (7.8-58) Find the values of p for which e x(ln1x)p dx
converges and evaluate the integral for those values of p.
Lecture 5
Integration