INTEGRATION BY PARTS

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INTEGRATION BY PARTS
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INTEGRATION BY PARTS ( Chapter 16 ) If u and v
are differentiable functions, then ? u dv uv
? v du.
There are two ways to integrate by parts the
standard method (above) and column integration.
Both methods will be shown but on the test I will
only ask for the column integration procedure.
First write the expression as a product of two
functions u and dv in such a way that ? dv can be
found.
Example ? xe 6x dx
Choose dv e 6x dx and u x. Then du
1 dx
Same concept as used in Test 4.
You must remember that
Find v by integrating dv v ? dv ? e 6x dx
Substitute into the formula ? u dv uv ? v
du.
Evaluating
Substituting yields
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CONDITIONS FOR INTEGRATION BY PARTS Integration
by parts can be used only if the integrand
satisfies the following conditions. 1. The
integrand can be written as the product of two
factors u and dv. 2. It is possible to
integrate dv to get v and to differentiate u to
get du. 3. The integral ? v du can be found.
COLUMN INTEGRATION
Is a technique that is equivalent to integration
by parts. Create two columns, one entitled D (
for derivative) and the other one entitled I (
for integral ). Under the D write the part of
the problem that will be differentiated. Under
the I write the rest of the problem that will be
integrated.
Working with the column entitled D first, write
the derivative of the function and on each
subsequent line write the derivative of the
preceding function until the answer of 0 is found.
Next form the second column under I in a similar
fashion except you will write an antiderivative
of the preceding function until the second column
has the same number of rows as the column
entitled D.
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Next, connect the terms diagonally with a line.
Then alternately label each line with then
until all lines are labeled. You may write
the signs on each line or in front of each term
of D that has been connected by a diagonal line.
Starting with the first sign multiply the
opposite ends of each diagonal line writing the
answer as a sum or difference until all products
have been formed remembering to write a C at
the end.
D
I
EXAMPLE ? 5x 3 e 4x dx
5x 3

Pay close attention to the order of the animation
in the problem. It gives the order in which the
solution is obtained.

15x 2
30x

30

0
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Even though you might believe you should put the
16x 3 20x beneath the D, you place the Ln 7x
beneath D. The reason is we only know how to
find its derivative. We do not know how to
integrate it.
EXAMPLE ? (16x 3 20x) Ln 7x dx
D
I
Ln 7x
16x 3 20x

Since this can never yield the value of zero by
repeated derivatives we stop.
4x 4 10x 2

Draw diagonal lines with alternating signs as
before. Because there is no zero beneath D draw a
horizontal line. The presence of a horizontal
line indicates the product is to be integrated.
(4x 4 10x 2 ) Ln 7x
(4x 4 10x 2 ) Ln 7x ? ( 4x 3 10x ) dx
(4x 4 10x 2 ) Ln 7x x 4 5x 2 C
Notice the sign has changed. Make certain you
understand why.
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Find the general solution and or particular
solution for the differential equation.
Algebraically separate the problem so that ys
are on one side of the equal sign and xs are on
the other side. Then integrate both sides
remembering to write C. See page 958 problems 1
30.
Example 2
Example 1
In step 3, only a y 2 was factored out of the
right expression even though all the numbers were
divisible by three. We do not do complete
factoring. We are only trying to separate the
variables.
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Example 3
Example 4
Cross multiplying gives
Ln y 5 x C
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Find the particular solution for
Substitute the given values of the variables and
solve for C.
Remember to write the final answer to the
question.
Answer
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Some problems are solved using a table of
integrals. Page A 4 in the back of your text
book contains 17 examples. These are not the
only examples, there are books dedicated to such
examples. These solutions are beyond the scope
of this course and may have taken years to solve.
However people noticed that problems having the
similar characteristics had similar solutions.
What you do is find the example on page A 4
that best matches your problem. You may have to
do some algebraic work first. The table of
integrals then shows how to write the answer to
that example.
Suppose we were trying to solve the following
problem
Upon close inspection this problem is similar to
example 15 on page A 4
The a 2 does not mean we have to actually have
the exponent two. In fact looking at the
examples solution indicates this is not
important because the value of a is not needed.
But if it was necessary, we could consider 7 as
What is important is the sign between x 2 and the
number is a plus sign.
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Next, the example shows there is no number
between the integral sign and the square root
symbol. This can be remedied by rewriting the
problem as
In doing this, we must remember to multiply the
entire
answer by eight not just part of it.
The Table of Integrals states the solution to our
problem is in the form of
Making the indicated substitutions and
remembering to multiply the entire answer by
eight yields the following
Work problems 23 28 on page 925. On the test I
will tell you when to use the Table of Integrals.