Using the method of substitution, I = = [* (62 – 6)³ dz = ["* ƒ(u) du - where u= du = a= b= f(u) = The value of the original integral is I = dz

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Using the method of substitution,
I =
where
U=
·[³ (6x − 6)³ dx
- 6)³ dx
-
du =
a =
b=
f(u) =
= = [² f(u) du
The value of the original integral is I =
dx
Transcribed Image Text:Using the method of substitution, I = where U= ·[³ (6x − 6)³ dx - 6)³ dx - du = a = b= f(u) = = = [² f(u) du The value of the original integral is I = dx
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