Substitution rule

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Homework Statement


Screenshot2012-02-02at65434PM.png


here is the answer

Screenshot2012-02-02at65428PM.png



The Attempt at a Solution



My book doesn't do a good job of explaining the substitution rule. here is their explanation:

Screenshot2012-02-02at61313PM.png


using the solution manual and looking at how they got the answer to other questions, i've written down my own method in english that i can understand. the method has worked for two other problems, but it broke down with the above problem.

i still don't understand what integrate with respect to u means.

here's my method:

1. substitute one part of the integral with u, find the derivative of that, z,
2. multiply the whole integral by the reciprocal of z, so that z equals 1
3. find the antiderivative of the remaining integral
4. replace u by g(x) in the result

the derivative of g(x) is

step 1. 4y^3 + 8y

multiply 12 by y^3 + 2y

step 2. 12y^3 + 2y

multiply the reciprocal of 1 with the result of step 2

step 3. (12y^3 + 24y)/(4y^3 + 8y)

simplify step 3

step 4. 3 + 3 = 6

we now have

step 5. 6u^2

take the antiderivative of 6u^2

step 6. (6u^3)/3

simplify

step 7 2u^3

plug in g(x) into u

step 8. 2(y^4 + 4y^2 + 1)^3

The book says that the answer is

(y^4 + 4y^2 + 1)^3
 

Answers and Replies

  • #2
I don't understand what you're doing from step 2 to step 4, why are you multiplying by the reciprocal?

Can you explain what you're doing at that step and why? Where did the 3+3 from step 4 come from?
 
  • #3
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That was the method I came up with to get the other answers and it worked.

It looks like the book just eliminates (y^3 + 2y) why?

I don't understand what integrate with respect to u means.

If you just eliminate y^3 + 2y, then add 4 + 8, which is 12 and multiply the reciprocal of that to 12, you will get the answer, but I would like to know why.
 
  • #4
HallsofIvy
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Your integral is [itex]12\int (y^4+ 4y^2+ 1)(y^3+ 2y)dy[/itex] and you want to use the substitution [itex]u= y^4+ 4y^2+ 1[/itex]. By differentiating, [itex]du= (4y^2+ 8y)dy[/itex] or [itex]du= 4(y^2+ 2y)dy[/itex] which is the same thing as [itex](1/4)du= (y^2+ 2y)dy[/itex]. So you replace the "[itex]y^4+ 4y^2+ 1[/itex]" in the integral with u and the "[itex](y^2+ 2y)dy[/itex]" with [itex](1/4)du[/itex]

That gives [itex]12\int u (1/4)du= 3\int dudu[/itex]


That's all it is, just replace the formulas in x with the things in u that are equal to them.
 

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