Integrating e^7x using U-Substitution | Step-by-Step Guide

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Homework Help Overview

The discussion revolves around the integration of the function e^7x using u-substitution. Participants are exploring the steps involved in this integration process, particularly focusing on the substitution and the handling of differential elements.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of u-substitution, with one suggesting u = 7x and du = 7dx. Questions arise about the integration of e^u and the treatment of the differential du in the context of the integral.

Discussion Status

There is ongoing exploration of the integration process, with some participants providing insights into the integration of e^u. Questions about the disappearance of du in the solution and the implications of the substitution are being raised, indicating a productive dialogue without a clear consensus.

Contextual Notes

Some participants express uncertainty about the integration steps and the implications of the u-substitution, suggesting that there may be assumptions or misunderstandings regarding the integration process itself.

Cacophony
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Homework Statement



S e^7x


Homework Equations



no

The Attempt at a Solution



Ok so I am using U-substitution for this problem but I don't know what to do next.

u = 7x, du = 7dx

How do I integrate e^u*du?
 
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Try treating ∫eu du just as if it were ∫ex dx
 
Cacophony said:

Homework Statement



S e^7x


Homework Equations



no

The Attempt at a Solution



Ok so I am using U-substitution for this problem but I don't know what to do next.

u = 7x, du = 7dx

How do I integrate e^u*du?
This is one of the easiest integrations!

##\int e^u~du = e^u + C##
 
What happen's to du? Why does it disappear in the solution?
 
It disappears for the same reason when integrating something like ∫2x dx to get x2 + C
 
Cacophony said:

Homework Statement



S e^7x

Homework Equations



no

The Attempt at a Solution



Ok so I am using U-substitution for this problem but I don't know what to do next.

u = 7x, du = 7dx

How do I integrate e^u*du?
I just want to point out, you are asking "how do I integrate"
\int e^u \,du

Which implies a bit you MAY think that will give you the answer (you could have just omitted the 1/7 to be brief).

But if you did miss it, since
du =7dx \rightarrow dx = \frac{du}{7}
When you replace dx in the original integral with du/7 and 7x in the original integral with u, you get
\int e^u \frac{du}{7}=\frac{1}{7} \int e^u \, du
 
Last edited:

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