1. Techniques of Integration
Chapter 9
Techniques of Integration
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Chapter Outline
Integration by Substitution
Integration by Parts
Evaluation of Definite Integrals
Approximation of Definite Integrals
Some Applications of the Integral
Improper Integrals
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3. Evaluation of Definite Integrals
Section 9.3
Evaluation of Definite Integrals
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Section Outline
The Definite Integral
Evaluating Definite Integrals
Change of Limits Rule
Finding the Area Under a Curve
Integration by Parts and Definite Integrals
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The Definite Integral
where F΄(x) = f (x).
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Evaluating Definite Integrals
EXAMPLE
Evaluate.
SOLUTION
First let u = 1 + 2x and therefore du = 2dx. So, we have
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Evaluating Definite Integrals
CONTINUED
Consequently,
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Change of Limits Rule
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Using the Change of Limits Rule
EXAMPLE
Evaluate using the Change of Limits Rule.
SOLUTION
First let u = 1 + 2x and therefore du = 2dx. When x = 0 we have u = 1 + 2(0) = 1. And when x = 1, u = 1 + 2(1) = 3. Thus
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Finding the Area Under a Curve
EXAMPLE
Find the area of the shaded region.
SOLUTION
To find the area of the shaded region, we will integrate the given function. But we must know what our limits of integration will be. Therefore, we must determine the three x-intercepts of the function.
This is the given function.
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Finding the Area Under a Curve
CONTINUED
Replace y with 0 to find the x-intercepts.
Set each factor equal to 0.
Solve for x.
Therefore, the left-most region (above the x-axis) starts at x = -3 and ends at x = 0. The right-most region (below the x-axis) starts at x = 0 and ends at x = 3. So, to find the area in the shaded regions, we will use the following.
Now let’s find an antiderivative for both integrals. We will use u = 9 – x2 and du = -2xdx.
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Finding the Area Under a Curve
CONTINUED
Now we solve for the area.
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Finding the Area Under a Curve
CONTINUED
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Integration by Parts & Definite Integrals
EXAMPLE
Evaluate.
SOLUTION
To solve this integral, we will need integration by parts. Our calculations can be set up as follows:
Then
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Integration by Parts & Definite Integrals
CONTINUED
Therefore, we have
= 1
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