We can sometimes use integration by parts to find antiderivatives of products. If f and g are two functions, let us abbreviate

d dx f(x) by D(f),     and

g(x) dx by I(g),

assuming these exist. Then integration by parts tells us that

f.g dx

=

f.I(g)

-

D(f).I(g) dx,

assuming that all the derivatives and integrals exist.

It sometimes happens that the product D(f)^.I(g) is easier to integrate than the original product f^.g, and this is where integration by parts comes in handy.

1.	x ex dx	=	xI(ex ) - D(x).I(ex ) dx
		=	xex - 1.ex dx
		=	xex - ex + C
2.	x2 sin x dx	=	x2I(sin x) - D(x2).I(sin x) dx
		=	-x2cos x + 2x cos x dx
	We must now do integration by parts again to evaluate the integral on the right
		=	-x2cos x + 2x. sin x - 2sin x dx
		=	-x2cos x + 2xsin x + 2cos x + C

3.		9x e3x+2	dx	=
4.		4x(x+4)5	dx	=

The tabular metod method makes integration by parts fast and easy to do -- especially when you have to perform it repeatedly (as in the second example above). This particular version of the tabular method was developed by Dan Rosen at Hofstra University.

Q How does it work?
A To compute f^.g dx, just set up the following table, in which we obtain the functions in the "D" column by repeatedy differentiating f, and the those in the "I" column by repeatedly integrating g. (The operations on the left are alternating plusses and minuses).

D I + f g - D(f) I(g) + D2(f) I2 (g)   . . .     . . .   ± Dn (f) In (g)

This process in continued until either:

• the function on the left becomes zero (as it will if f is a polynomial)
• the product of the functions on the bottom row can be integrated, or
• the product of the functions on the bottom is just a constant multiple of the product of the functions on the top row

To read the table, just muiltiply the functions connected by arrows, and add or subtract the products using the plusses and minuses on the left:

f.g dx =

f.I(g) - D(f) .I2(g) + D2(f) .I3(g) - . . . ±

Dn (f) .In(g) dx

Note If the bottom left term is zero (as it will be if f is a polynomial) then the integral on the right goes away (it is also zero), and we can simply write down the answer, as in the the first example on the right.

1. To compute 3x²e^x/2 dx, we put the 3x² in the "D" column (because it is a polynomial and so it will vanish after repeated differentiation, and so we must put the e^x/2 in the "I" column:

D I + 3x2 ex/2 - 6x 2ex/2 + 6 4ex/2 - 0 8ex/2

We can now read off the answer:

	3x2ex/2 dx	=	3x2.2ex/2 - 6x .4ex/2 + 6 .8 ex/2 + C
		=	(6x2 - 24x + 48)ex/2 + C

2. To compute ln x dx, first write ln x as the product (1)(ln x) and put ln x in the "D" column (we can't integrate it yet) and 1 in the "I" column. Since we will never get zero by repeatedly differentiating ln x, we stop as soon as we can integrate the product in the bottom row, which is after one step:

D I + ln x 1 - 1/x x

Thus,

	ln x dx	=	(ln x)(x) - (1/x)(x) dx
		=	x ln x - 1 dx
		=	x ln x - x + C

3.		(x2+1)e-x dx	=
4.		x ln x dx	=

If f(x) ≥ g(x) for all x in [a, b], then the area of the region between the graphs of f and g and between x = a and x = b (shown colored in the figure below) is given by

A =

b     a

[f(x) - g(x)] dx.

If we want to find the area between the graphs of f(x) and g(x), but the graphs cross, we follow this procedure:

(a) Find all points of intersection by solving f(x) = g(x) for x. This either determines the interval over which we will integrate, or breaks up a given interval into regions between the intersection points.

(b) Find the area of each region between intersection points by integrating the difference of the larger and the smaller function. (If we accidentally take the smaller minus the larger, the answer will be the negative of that area, so we just take the absolute value.)

(c) Add together the areas we found in (b) to get the total area.

Turquoise area =		c     a	[f(x) - g(x)] dx
Blue area =		d     c	[g(x) - f(x)] dx
Gold area =		b     d	[f(x) - g(x)] dx

1. To compute the area shown in the following figure,

we compute

A =	1     0 [(2x2-x4) - (x-1)] dx.
=	2x3 3 - x5 5 - x2 2 + x 1     0
=	29 30 .

2. In the figure below, the graphs cross at x = 1/2. To compute the area, we need to compute A and B seperately, and then add then up.

A =	1/2     0 [(1-x2) - 3x2] dx	=	1 3
B =	1     1/2 [3x2 - (1-x2)] dx	=	2 3

Therefore, the total area is 1.

Average Value of a Set of Numbers

The average, or mean, of a set of n numbers, {y₁, y₂, ... y_n }, is defined to be their sum, divided by n:

x

=

y1 + y2 + ... + yn n

.

The average, or mean, of a function f(x) on an interval [a, b] is defined as

f

=

1 b-a

b     a

f(x) dx

Average Value of a Set of Numbers

Here is a set of 5 numbers and their mean. Alter the list any way you want to see the mean change. (Press the tab key or move to another cell in order to get your browser to register any change you make.)

x

=

Average Value of a Function

Let f(x) = 3x². Then the average value of f on the interval [0, 1] is

f	=	1 b-a		b     a	f(x) dx
	=	1 1-0		1     0	3x2 dx
	=	1

The following little utility will compute (accurate to five decimal places) the average of the function you enter over any interval [a, b]. Use it to check your homwork! (You can use any valid formulas, as well as "pi" for and "e" for the base of natural logarithms.)

Moving Average of a Sequence of Numbers

The n-unit moving average of a sequence of numbers is the average of each number together with the the preceding n-1 numbers.

We can compute the n-unit moving averages starting with the nth number. (See the example opposite).

Moving Average of a Function

The a-unit moving average of the function f is given by

Future value = FV =

b     a

R(t)er(b-t) dt.

Moving Average of a Sequence of Numbers

Here is a table showing the 3-unit moving average of a sequence of closing stock prices; each number in the sequence of moving averages is the average of that day's closing price and the preceding two. Complete the table by filling in the blanks.

Day	1	2	3	4	5	6	7
Price	23	21	22	23	27	25	25
Moving Average			22	22	24

   

Moving Average of a Function The 2-unit moving average of f(x) = x² is

f

=

1 2

x     x-2

t2 dt  =  x2 2x + 4/3.

Here are the graphs of f and its 2-unit moving average.

If demand for an item is given by p = D(q), the selling price is p*, and q* is the corresponding demand (so that D(q*) = p*), then the consumers' surplus is the difference between what consumers are willing to pay and their actual expenditure: It therefore represents the total amount saved by consumers who paid p* per unit (but were willing to pay more according to the demand curve).

CS =	q*     0 (D(q) -p*) dq

Graphically, it is the area between the graphs of p = D(q) and p = p*, as shown below.

Producers' Surplus

The producers' surplus is the extra amount earned by producers who charged p* per unit (but were willing to charge less) and is given by

PS =	q*     0 (p* - S(q)) dq

where S(q*) = p*. Graphically, it is the area of the region between the graphs of p = p* and p = S(q) for 0 ≤ q ≤ q* as in the following figure.

Total Value of a Continuous Income Stream

If the rate of receipt of income is R(t) dollars per unit of time, then the total income received from time t = a to t = b is

Total value = TV =

A =

b     a

R(t) dt.

Future Value of a Continuous Income Stream

If the rate of receipt of income from time t = a to t = b is R(t) dollars per unit of time and the income is deposited as it is received in an account paying interest r per unit of time, compounded continuously, then the amount of money in the account at time t = b is

Future value = FV =

b     a

R(t)er(b-t) dt.

Present Value of a Continuous Income Stream

If the rate of receipt of income from time t = a to t = b is R(t) dollars per unit of time and the income is deposited as it is received in an account paying interest r per unit of time, compounded continuously, then the value of the income stream at time t = a is

Present value = PV =

b     a

R(t)er(a-t) dt.

An improper integral is an integral in which (A) one or both of the limits of integration is infinite, or (B) in which the integrand becomes infinite somewhere in the range of integration.

(A) Improper Integral with an Infinite Limit of Integration

These are integrals of the form

+∞     a

f(x) dx,

b     -∞

f(x) dx,   or

+∞     -∞

f(x) dx.

First, make sure that the integrand does not become infinite for any value of x in the range of integration. If it does, see below. Otherwise, use the following formulas.

+∞     a

f(x) dx   =

lim M→+∞

M     a

f(x) dx,

provided the limit exists. If the limit exists, we say that ∫_a^+∞f(x) dx converges. Otherwise, we say that ∫_a^+∞f(x) dx diverges.

Similarly, we define

b     -∞

f(x) dx   =

lim M→-∞

b     M

f(x) dx,

provided the limit exists. Finally, we define

+∞     -∞

f(x) dx   =

a     -∞

f(x) dx   +

+∞     a

f(x) dx

for some convenient a, provided both integrals on the right converge.

(B) Improper Integral in which the Integrand Becomes Infinite

Some improper integrals look quite innocent, since they are of the form ∫_a^bf(x) dx, where a and b are finite. However, it may happen that f(x) is not defined for one or more values of x in the interval [a, b]. Thus, check whether the integrand is infinite for any value of x with a ≤ x ≤ b. If f(x) is never infinite for such values of x, then the integral is not improper and can be evaluated by the Fundamental Theorem of Calculus. Otherwise, we have the following cases.

1. If f(x) is defined for all x with a < x ≤ b but approaches ±∞ as x approaches a, we define

b     a

f(x) dx   =

lim r→a+

b     r

f(x) dx,

2. Similarly, if f(x) is defined for all x with a ≤ x < b but approaches ±∞ as x approaches b, we define

b     a

f(x) dx   =

lim r→b-

r     a

f(x) dx,

3. If f(x) is defined for all x with a ≤ x ≤ b except for a single number c, and approaches ±∞ as x approaches c, we define

b     a

f(x) dx   =

c     a

f(x) dx   +

b     c

f(x) dx,

4. If f(x) is infinite at more than one point, we break the integral into two or more pieces so that each of the pieces falls into Case (1) or Case (2).

In all cases, convergence of the integral in question requires all the associated integrals to exist (and be finite).

(C) Integral Improper in More Than One Way

It may happen that one (or both) of the end-points is infinite and where f(x) is also infinite at some point of the range of integration.

In this case, break up the integral into two or more pieces so that each piece falls under either (A) or (B) above.

In all cases: If an improper integral I is expressed as a sum of two or more improper integrals,

I = J + K + . . . ,

(1) if one or more of the integrals J, K, . . . diverges, so does I, and

(2) if all of the integrals J, K, . . . converge, then so does I. Moreover, I converges to the sum of the individual integrals.

(A) Improper Integral with an Infinite Limit of Integration

1.

+∞     1

dx x2

=

lim M→+∞

M     1

dx x2

=

lim M→+∞

(1 - 1/M) = 1

Thus, the integral converges to 1.

 

2.

+∞     1

dx x

=

lim M→+∞

M     1

dx x

=

lim M→+∞

(ln M - ln 1) = +∞

This integral diverges to +∞.

(B) Improper Integral in which the Integrand Becomes Infinite

1.

1     0

dx x1/2

=

lim r→0+

1     r

dx x1/2

=

lim r→0+

(2 - 2r1/2) = 2

Thus, the integral converges to 2.

 

2.

1     0

dx x2

=

lim r→0+

1     r

dx x2

=

lim r→0+

(1/r - 1) = +∞

This integral diverges to +∞.

(C) Integral Improper in More Than One Way

+∞     0

dx x2

=

1     0

dx x2

+

+∞     1

dx x2

Looking at the right-hand side, we see that the first integral diverges, and the second converges. Since one or more summand diverges, the whole integral diverges.

A differential equation is an equation that involves the derivative of an unknown function f(x). To solve a differential equation means to find that unknown function. The key step in solving a differential equation is integration, and this results in an arbitrary constant. The solution for f(x) that contains the arbitrary constant is called the general solution. We obtain a particular solution by choosing a value for the constant, or by calculating its value using additional information about f(x). Two kinds of differential equations that are easy to solve are these.

(a) Elementary Elementary differential equations have the form

dy dx

=

f(x).

y = f(x) dx.

dy dx

=

f(x)g(y).

dy g(y)

=

f(x)dx