14
Further Integration Techniques
and Applications of the Integral
Copyright © Cengage Learning. All rights reserved.
14.1
Integration by Parts
Copyright © Cengage Learning. All rights reserved.
Integration by Parts
Integration by parts is an integration technique that
comes from the product rule for derivatives.
We start with a little notation to simplify things while we
introduce integration by parts. If u is a function, denote its
derivative by D(u) and an antiderivative by I(u). Thus, for
example, if u = 2x2, then
D(u) = 4x
and
I(u) =
[If we wished, we could instead take I(u) =
+ 46, but we
usually opt to take the simplest antiderivative.]
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Integration by Parts
Integration by parts
If u and v are continuous functions of x, and u has a
continuous derivative, then
Quick Example
4
Derivation of Integration by Parts Formula
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Derivation of Integration by Parts Formula
As we mentioned, the integration-by-parts formula comes
from the product rule for derivatives. We apply the product
rule to the function uI(v)
D[u  I(v)] = D(u)I(v) + uD(I(v))
= D(u)I(v) + uv
because D(I(v)) is the derivative of an antiderivative of v,
which is v. Integrating both sides gives
u  I(v)
A simple rearrangement of the terms now gives us the
integration-by-parts formula.
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Example 1 – Integration by Parts: Tabular Method
Calculate
Solution:
First, the reason we need to use integration by parts to
evaluate this integral is that none of the other techniques of
integration that we’ve talked about up to now will help us.
Furthermore, we cannot simply find antiderivatives of
x and ex and multiply them together.
This integral can be found by integration by parts. We want
to find the integral of the product of x and ex.
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Example 1 – Solution
cont’d
We must make a decision: Which function will play the role
of u and which will play the role of v in the
integration-by-parts formula? Because the derivative of x is
just 1, differentiating makes it simpler, so we try letting x be
u and letting ex be v.
We need to calculate D(u) and I(v), which we record in the
following table.
The table is read as
+x · ex −∫1 · ex dx
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Example 1 – Solution
cont’d
Below x in the D column, we put D(x) = 1; below ex in the
I column, we put I(ex) = ex.
The arrow at an angle connecting x and I(ex) reminds us
that the product xI(ex) will appear in the answer; the plus
sign on the left of the table reminds us that it is +xI(ex) that
appears.
The integral sign and the horizontal arrow connecting
D(x) and I(ex) remind us that the integral of the product
D(x)I(ex) also appears in the answer; the minus sign on the
left reminds us that we need to subtract this integral.
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Example 1 – Solution
cont’d
Combining these two contributions, we get
The integral that appears on the right is much easier than
the one we began with, so we can complete the problem:
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Derivation of Integration by Parts Formula
Integrating a Polynomial Times a Function
If one of the factors in the integrand is a polynomial and the
other factor is a function that can be integrated repeatedly,
put the polynomial in the D column and keep differentiating
until you get zero. Then complete the I column to the same
depth, and read off the answer.
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