Indefinite Integrals & Substitution Rule

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Homework Help Overview

The discussion revolves around the topic of indefinite integrals and the substitution rule in calculus. Participants are examining the integration of a function involving a square root and the appropriate substitutions to apply.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the process of substitution, specifically the transformation of variables and the integration of the resulting expression. There are questions about the correctness of applying the power rule and the relationship between functions and their anti-derivatives.

Discussion Status

The discussion includes attempts to clarify the integration process and check the validity of the steps taken. Some participants express uncertainty about their calculations and seek confirmation of their reasoning, while others provide feedback on the application of integration techniques.

Contextual Notes

There is mention of potential confusion regarding the integral of du/sqrt(u) and the proper use of substitution, indicating that participants are grappling with foundational concepts in integration.

MitsuShai
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Homework Statement






2. The attempt at a solution

I
 
Last edited:
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It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?
 
Dick said:
It's not even close. Since when does the integral of du/sqrt(u) have an ln in it? And shouldn't you be using du=(-5)dx to replace dx with du?

I forgot about the square root.
how about this?

I= ∫ dx/sqrt(2−5x)
= ∫ [1/sqrt(2−5x)]dx
u= 2-5x
du= -5dx
(-1/5)du = dx

∫ (-1/5)du/sqrt(u)
(-1/5) ∫ du/u^(1/2)
(-1/5) 1/ (2/3)u^(3/2)
- 1/[(10/3)u^(3/2)]
- 1/[(10/3)(2-5x)^(3/2)]
 
Last edited:
MitsuShai said:
I forgot about the square root.
how about this?

I= ∫ dx/sqrt(2−5x)
= ∫ [1/sqrt(2−5x)]dx
u= 2-5x
du= -5dx
(-1/5)du = dx

∫ (-1/5)du/sqrt(u)
(-1/5) ∫ du/u^(1/2)
(-1/5) 1/ (2/3)u^(3/2)
- 1/[(10/3)u^(3/2)]
- 1/[(10/3)(2-5x)^(3/2)]

du/sqrt(u) is u^(-1/2)du. You've still got the power rule wrong.
 
Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?
 
Hurkyl said:
Do remember there is a way to check your work -- what relationship is a function supposed to have with its anti-derivatives?

Yeah, I already know that.

I don't trust my math, that's why I want to know if I'm doing this correctly.
 
Last edited:
ok nevermind I got it...
 

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