Is the Integration Calculation Correct?

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Homework Help Overview

The discussion revolves around the correctness of an integration calculation involving limits and substitutions in the context of improper integrals. The subject area is calculus, specifically focusing on integration techniques and limit evaluation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to evaluate an integral using substitution and limits, expressing uncertainty about their approach. Some participants question the use of the limit symbol and the implications of the variable changes during substitution. Others provide feedback on the correctness of the algebraic manipulations and raise concerns about the limits of integration.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's calculations. While some affirm the algebraic steps taken, others highlight potential oversights regarding the limits of integration. There is no explicit consensus on the correctness of the final result, but guidance is being offered to clarify the integration process.

Contextual Notes

Participants are navigating the complexities of improper integrals and the implications of variable substitution, with some expressing confusion about the notation used for infinity and limits. The original poster indicates this is an early attempt at such problems, suggesting a learning context.

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[tex]\int_0^* \frac {dz}{z^23Z +2}[/tex]

[tex]=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{z^2+3Z+2}[/tex]

[tex]=\lim_{t\rightarrow \*} \int_0^* \frac{dz}{(z+3/2)^2 -1/4}[/tex]

let u=z+3/2
du=dz

[tex]=\lim_{t\rightarrow \*} \int_0^* \frac{du}{(u)^2 -1/4}[/tex]

[tex]= \frac{1}{2(1/2)}ln|\frac{u-1/2}{u+1/2}[/tex]

[tex]= ln |\frac{z+1}{z+2}| \right]_0^*[/tex]

[tex]= -ln \frac{1}{2}[/tex]

this is one of the first of these I am doing so bear with me if its horrible wrong :rolleyes:

EDIT:* denotes infinite..PS is there a latex for infinite?
 
Last edited:
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I don't know why you used the lim symbol in the first place, and where the t commes in play, but

[tex]= \int_0^* \frac{dz}{z^2+3Z+2}[/tex]
[tex]= \int_0^* \frac{du}{(u)^2 -1/4}[/tex]

using your algebraic manipulations. After that, you used the partial sum decomposition skillfully and unless I'm mistaking, you are right!

Just one minor thing : [tex]-ln \frac{1}{2}[/tex] = ln(2)
 
Last edited:
yyyea..but what?I already had that is it right?
 
It's correct except you forgot to change the lower limit of integration. When z goes from 0 to infinity, then u = z + 3/2 goes from what to what?

You do the infinity sign like [tex]\infty[/tex]
 

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