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**Unformatted text preview: **Lecture 31 1 Lecture 31: (Sec. 6.2)
Integration by Substitution, Part I
ex. How can we evaluate 6x(3x2 − 2)5 dx? Lecture 31 2 To see why we can use substitution:
Let F be an antiderivative of a given function f .
Then if g is a diﬀerentiable function of x, d
[F (g (x))] =
dx Lecture 31 We have the following result:
du
f (u) dx =
dx 3 Lecture 31 4 Evaluate each integral:
ex. ex. (3x2 + 1)(x3 + x − 2)15 dx 3 − 5 x4 (4 − x 5 )2 dx Lecture 31 These are examples of the
General Power Rule of Integration
Let u be a diﬀerentiable function of x. Then
du
un dx =
dx NOTE: 5 Lecture 31 6 But consider the following example:
Evaluate 5x x2 − 1 dx Lecture 31 7 ex. Evaluate:
1) 5
dx
(1 + 2x)2 2) x2
(6 − 3x)(1 + 2x − ) dx
2 Lecture 31 8 Now consider the following examples:
ex. x3(2x4 − 3)5 dx ex. x(2x4 − 3)5 dx Lecture 31 9 Sometimes we can substitute further to evaluate an
integral:
√
ex. Evaluate:
x x + 1 dx Lecture 31 10 Application
ex. The marginal cost of producing x units of a
400x
new product is C (x) = √
(where x is in
2 + 64
x
hundreds), and ﬁxed costs are $4000. Find the total
cost of producing 600 items. ...

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