Embed Size (px)
Citation preview
Copyright © Cengage Learning. All rights reserved.
Integration by Parts and
Present Value
6.1
2 2
Warm-Up: Find f’(x)
1. F(x) = ln(x+1)
2. F(x) = 𝑒𝑥3
3. F(x) = 𝑥2𝑒𝑥
4. F(x) = ln (x2 – 1)
5. F(x) = 𝑒−𝑥2
6. F(x) = xe-2x
3 3
Objectives, Day #1
Students will be able to:
– use integration by parts to evaluate indefinite and definite
integrals.
– Homework: p. 383:
Skills warm-up: 7-10
Exercises: 5-13 ODD
# u dv
5 x e3xdx
7 lnx x3dx
9 ln2x 1dx
11 x2 e-x
13 lnx x1/2
4 4
Integration by Parts Formula
A way to integrate a product is to write it in the form
If u and v are differentiable function of x, then
.functionanother of aldifferentifunction one
. duvuvdvu
5 5
Following is the guidelines for integration by parts:
Integration by Parts
6 6
Example 1 – Integration by Parts
Find xex dx.
Solution:
To apply integration by parts, you must rewrite the original
integral in the form u dv. That is, you must break xex dx
into two factors—one “part” representing u and the other
“part” representing dv. There are several ways to do this.
7 7
Example 1 – Solution
The guidelines for integration by parts suggest the first
option because dv = ex dx is the most complicated portion
of the integrand that fits a basic integration formula and
because the derivative of u = x is simpler than x.
Next, you can apply the integration by parts formula as
shown.
cont’d
8 8
Example 1 – Solution
You can check this result by differentiating.
cont’d
9 9
You try: Using Integration by Parts
Evaluate .5 3 dxxe xdxedvxu x3,5Let
,5dxdu dxev x3
xev 3
3
1
duvuvdvu
dxeexdxxe xxx 5
3
1
3
155 333
dxexe xx 33
3
15
3
5C
exe
xx
)
33(5
3
5 33
Ce
xex
x 9
5
3
5 33
10 10
Example 2 – Integration by Parts
Find x2 ln x dx.
Solution:
In this case, x2 is more easily integrated than In x.
Furthermore, the derivative of In x is simpler than In x. So,
you should choose dv = x2 dx.
11 11
Example 2 – Solution
Next, apply the integration by parts formula.
cont’d
12 12
Example 3 Solving an Initial Value Problem
Solve the differential equation dy/dx = xlnx subject to the
initial condition y = -1 when x = 1
.ln xdxx xdxdvxu ,lnLet
dxxvdxx
du ,1
2
2xv
It is typically better to let u = lnx
dxx
xxxdvu
1
2)
2(ln
22
dxxxx
2
1ln)
2(
2
Cx
xx
22
1ln)
2(
22
13 13
Cx
xx
4
ln)2
(22
C4
11ln)
2
1(1
22
C4
101
C
4
3
4
3
4ln)
2(
22
xx
xy
14 14
Example 4 – Using Integration by Parts Repeatedly
Find x2ex dx.
Solution:
The factors x2 and ex are both easy to integrate. Notice,
however, that the derivative of x2 becomes simpler,
whereas the derivative of ex does not. So, you should let
u = x2 and let dv = ex dx.
15 15
Example 4 – Solution
Next, apply the integration by parts formula.
This first use of integration by parts has succeeded in
simplifying the original integral, but the integral on the right
still doesn’t fit a basic integration rule.
To evaluate that integral, you can apply integration by parts
again. This time, let u = 2x and dv = ex dx.
cont’d
16 16
Example 4 – Solution
Next, apply the integration by parts formula.
cont’d
17 17
Example 4 – Solution
You can confirm this result by differentiating.
cont’d
18 18
Before starting the exercises in this section, remember that
it is not enough to know how to use the various integration
techniques. You also must know when to use them.
Integration is first and foremost a problem of recognition—
recognizing which formula or technique to apply to obtain
an antiderivative.
Often, a slight alteration of an integrand will necessitate the
use of a different integration technique.
Integration by Parts
19 19
Here are some examples.
As you gain experience in using integration by parts, your
skill in determining u and dv will improve.
Integration by Parts
20 20
The following summary lists several common integrals with
suggestions for the choices of u and dv.
Integration by Parts
21 21
Closure
Use integration by parts.
𝑙𝑛𝑥 𝑑𝑥
Hint: u = lnx and dv = 1dx
du = (1/x)dx v = x
(lnx)(x) - 𝑥1
𝑥𝑑𝑥 = xlnx- 1𝑑𝑥
= xlnx –x + C
22 22
Use integration by parts to find indefinite and
definite integrals.
Find the present value of future income.
HW: p. 383: 6-12 EVEN, p. 384: 74, 75
Objectives, Day #2
23 23
Warm-Up
Use integration by parts repeatedly to solve the indefinite integral.
𝑥3 𝑒𝑥dx
u = x3
du = 3x2 dx
dv = ex dx
v = ex
u = x2 𝑥3 𝑒𝑥 - 3 𝑒𝑥𝑥2𝑑𝑥
du = 2x dx
dv = ex dx
v = ex
. duvuvdvu
𝑥3 𝑒𝑥 - 𝑒𝑥3𝑥2𝑑𝑥
𝑥3 𝑒𝑥 - 3(𝑥2𝑒𝑥 − 𝑒𝑥2𝑥)𝑑𝑥
24 24
u = x
du = dx
dv = ex dx
v = ex
𝑥3 𝑒𝑥 - 3𝑥2𝑒𝑥 + 6 𝑒𝑥𝑥𝑑𝑥
𝑥3 𝑒𝑥 - 3𝑥2𝑒𝑥 + 6(𝑥𝑒𝑥 − 𝑒𝑥𝑑𝑥)
𝑥3 𝑒𝑥 - 3𝑥2𝑒𝑥 + 6𝑥𝑒𝑥 − 6𝑒𝑥 + 𝐶
25 25
Example 5
Evaluate the indefinite integral:
𝑙𝑛𝑥𝑑𝑥𝑒
1
u = lnx and dv = 1dx
du = (1/x)dx v = x
(lnx)(x) - 𝑥1
𝑥𝑑𝑥 = xlnx- 1𝑑𝑥
= 𝑒
1 xlnx –x; (elne –e) - (1ln1 – 1) = 0 – (-1) = 1
26 26
The present value of a future payment is the amount that
would have to be deposited today to produce the future
payment.
What is the present value of a future payment of $1000 one
year from now? Because of inflation, $1000 today buys
more than $1000 will buy a year from now.
Present Value
27 27
The definition below considers only the effect of inflation.
Ignoring inflation, the equation for present value also
applies to an interest-bearing account, where the annual
interest rate r is compounded continuously and c is an
income function in dollars per year.
Present Value
28 28
Present Value
You have just won $1,000,000 in a state lottery. You will be
paid an annuity of $50,000 a year for 20 years. When the
annual rate of inflation is 6%, what is the present value
of this income?
The income function is c(t) = 50,000
Actual income: 50,000𝑑𝑡20
0
20
0
50,000𝑡= 50,000(20) – 50,000(0) = 1,000,000.
However, you did not receive this entire amount right now,
so the present value is: 50,000𝑒−.06𝑡𝑑𝑡20
0 =
20
0 50,000𝑒−.06𝑡
−.06 = $582,338
29 29
Example 7 – Finding Present Value
A company expects its income during the next 5 years to
be given by
c(t) = 100,000t, 0 t 5.
Assuming an annual inflation
rate of 5%, can the company
claim that the present value
of this income is at least $1
million?
Figure 6.2(a)
30 30
Example 7 – Solution
The present value is
Using integration by parts, let dv = e–0.05t dt.
31 31
Example 7 – Solution
This implies that
cont’d
32 32
Example 7 – Solution
So, the present value is
Yes, the company can claim that the
present value of its expected income
during the next 5 years is at least
$1 million.
cont’d
Figure 6.2(b)
33 33
Closure, checkpoint 7
A company expects its income during the next 10 years to
be given by c(t) = 20,000t, for t:[0, 10]. Assuming an
annual inflation rate of 5%, what is the present value of
this income?
20,000𝑡𝑒−.05𝑡𝑑𝑡10
0
u = 20,000t dv = 𝑒−.05𝑡𝑑𝑡
du = 20,000 v = -20e-.05t
20,000t(-20e-.05t) - −400,000𝑒−.05𝑡𝑑𝑡10
0
-400,000te-.05t - 8,000,000e-.05t
10
0 = $721,632.08
34 34
Objectives, Day #3
Students will be able to:
– use integration by parts to evaluate indefinite and definite
integrals.
– use rapid repeated integration or tabular method to evaluate
indefinite integrals.
– Choose the best method to evaluate integrals.
p. 383: 18-32 EVEN, p. 384: 48
35 35
Warm-Up Solve the differential equation: dy/dx = x2e4x
(This means you will need to find the anti-
derivative of dy/dx = x2e4x )
.42 dxex xxedvxu 42 ,Let
dxevdxxdu x4,2
4
4xev
xdxee
xdvuxx
244
442
dxxeex x
x4
42
2
1
4
36 36
dxxeex x
x4
42
2
1
4
xedvxu 4,Let
dxevdxdu x4,
4
4xev
dx
eex
ex xxx
442
1
4
4442
C
exeex xxx
1642
1
4
4442
C
exeex xxx
3284
4442
Cexeex xxx
3284
4442
37 37
Rapid Repeated Integration by Parts
AKA: The Tabular Method
Choose parts for u and dv.
Differentiate the u’s until you have 0.
Integrate the dv’s the same number of times.
Multiply down diagonals.
Alternate signs along the diagonals.
38 38
Example 8 Rapid Repeated
Integration by Parts
Evaluate
u and its derivatives dv and its integrals
.2 dxex x
2x xe
x2
2
0
xe
xe
xe
Ceexex xxx 222
39 39
Example 9: Use the Tabular Method to find the area of a
region
y = x3ex
y = 0
x = 0
x = 2
x3ex 𝑑𝑥2
0
40 40
Example 9, continued
U and its
derivatives
dv and its
integrals
x3 ex
3x2 ex
6x ex
6 ex
0 ex
41 41
Example 9 continued…
𝑥3𝑒𝑥 − 3𝑥2𝑒𝑥 + 6𝑥𝑒𝑥 − 6𝑒𝑥 2
0
20.778
42 42
Choosing the best method
p. 383
#17 𝑒4𝑥dx #19 𝑥𝑒4𝑥dx
43 43
Choosing the best method
#25 𝑥
𝑒𝑥4
Rewrite as a product
44 44
Choosing the best method
#27 𝑥 𝑙𝑛𝑥 2 Must use integration by parts when x in
numerator
45 45
Choosing the best method
#29 𝑙𝑛𝑥 2
𝑥 When “x” or “ax” is in denominator, and there is
a “lnx” in numerator, can use u substitution.
46 46
Choosing the best method
#31 𝑙𝑛𝑥
𝑥2 dx Since there is an x2 in
denominator, we must use
integration by parts.
