In class Quiz
Evaluate the definite integrals. Show all your steps and
use correct notation!
!
2.
4
1
t + t dt
Hint: You do NOT need u-substitution.
"
3.
!
4
0
3
cos x sin x dx
Hint: You do need u-substitution.
Do this by both method I and II
4. Find the area under the curve sin x + 2 from x = 0 to x = !
Problem 2
!
4
t + t dt
1
4
=
!
4
1
1
2
2
3
2
t
t
t + t dt = +
2 3
2
2
t
2
= + t
2 3
3 4
2
1
1
" 4 2 2 23 % " 12 2 23 % "
16 % " 1 2 %
=$
+ 4 ' ( $ + 1 ' = $8 + ' ( $ + '
3 & # 2 3&
# 2 3 & #2 3 & #
73
=
6
Problem 3
Method I:
Indefinite Integral:
!
u = sin x
du = cos xdx
=!
"
Method II:
"
!
4
0
cos x sin 3 x dx = ! sin 3 x cos x dx
!
4
0
1 4
1
u du = u + C = (sin x)4 + C
4
4
3
!
4
1
3
cos x sin x dx = (sin x)4
4
0
1
=
16
!
4
cos x sin 3 x dx = " sin 3 x cos x dx
0
="
!
sin( )
4
sin(0)
u du = "
2
2
3
0
1 4
3
u du == u
4
2
2
0
1
=
16
Problem 4
Since the graph of sin x + 2 Is always above x-axis, the area is
"
!
0
sin x + 2 dx
!
0
= (# cos x + 2x) |
= # cos ! + 2! # (# cos 0 + 2 * 0)
= 2! + 2
Lesson 14
Improper Integrals
Introduction
How much area is under y = e–x for x > 0?
For any finite extent, this is easy:
From 0 to 1:
From 0 to 5:
From 0 to 10:
1
#
#e
0
5
0
#
10
0
!x
e dx = ! e
!x
!x
1
0
5
= !e !1 + e0 " .6321
dx = ! e ! x = !e !5 + e0 " .9933
0
!x
e dx = ! e
In fact, the area from x = 0 to x = b is
!x
"
10
0
b
0
= !e !10 + e0 " .9995
!x
!x
e dx = !e
b
0
= !e !b + e 0 = 1! e! b
So as b increases, the integral approaches 1.
Goes to 0 as b
In fact, the area added by extending the limit a little further goes to infinity.
gets smaller each time. The total area approaches 1.
The Improper Integral
Integrals with one or both limits infinite are referred to as
improper integrals.
"
!
a
f (x)dx
#
a
!"
f ( x) dx
#
!
"!
f ( x) dx
are all considered improper.
With the preceding ideas, we give meaning to integrals with infinite
limits:
will mean
will mean
Example
=1
First find the integral up to b:
!
b
1
b
1
1
1 1
1
dt
=
"
=
"
+
=
1"
b
t2
t1
b 1
Now we must take the limit as b goes to infinity.
If you can’t remember what this limit does, some numerical work
may help remind you.
If b = 1, 1 – 1/b = 0.
If b = 10, 1 – 1/b = 0.9
If b = 100, 1 – 1/b = 0.99
In fact, 1/b goes to zero, so
the integral approaches 1.
Example
#
2
Goes to 0 as a goes to
negative infinity
3x
e dx
!"
3x 2
e
e dx = lim
$!" e dx = alim
#!" $a
a #!" 3
2
2
3x
3x
a
e3(2)
e3( a ) e6
e6
=
! lim
= ! 0=
a
#!"
3
3
3
3
Caution The limit may not exist. (Often the limit goes to
infinity or negative infinity.)
"
!
0
2
x dx
!
b
0
x dx = x
2
1
3
3b
=
b
"
0
=
b
0
1
3
3
1
3
3
so
"
!
0
x 2 dx = !
This does not approach a limit!
(We sometimes say the integral is to indicate precisely why the
integral does not exist. We may also say the integral diverges.)
Example
Evaluate the following integral:
First evaluate
"
b !2x
1
#
" !2x
1
e dx
e dx Your answer will be in terms of b.
Then take the limit as b ! "
1
1 !2
Your answer should be: 2e2 !!or!! 2 e
#
"
1
e
!2 x
b
dx = lim # e
b$" 1
!2 x
!2 x b
e
dx = lim
b$" !2
1
1% 1
1 (
= ! ' 2b ! 2(1) *
2&e
e )
1 %%
1 (
1 (
1%
1(
1
= ! ' ' lim 2b * ! 2(1) * = ! ' 0 ! 2 * = 2
2 & & b$" e ) e )
2&
e ) 2e
Example
Evaluate the following integral:
First evaluate
"
b
1
#
"
1
2xe
! x2
dx
2xe! x dx Your answer will be in terms of b.
2
Then take the limit as b ! "
#
"
1
2xe
! x2
b
dx = lim # 2xe
b$" 1
!u
= lim (!e )
b$"
b2
1
! x2
b
dx = lim # e
b$" 1
1(
% 1
= ! ' b2 ! *
&e
e )
1 1(
1( 1
%
%
= ! ' lim b2 ! * = ! ' 0 ! * =
&
& b$" e
e)
e) e
! x2
b2
2xdx = lim # 2 e!u du
b$" 1
Both Limits Infinite
Integrals with both limits infinite are a little tricky.
The only way to be sure of the answer is to split them into two
improper integrals and solve each separately:
But if we’re sure the limits exist, we can approximate such
an integral by extending both limits at once:
Summary
• We recognize as improper integrals with
one or both limits infinite.
• We evaluate these by taking the limit as the
upper/lower limit increases/decreases.
• Improper integrals may diverge (fail to
exist).
• Integrals with both limits infinite are a little
tricky.