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

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SUMMARY

The discussion focuses on integrating the function e^7x using U-substitution. The correct substitution is u = 7x, leading to du = 7dx. The integral ∫e^u du simplifies to e^u + C, where the factor of 1/7 is accounted for when substituting back to the original variable, resulting in (1/7)e^7x + C. This method highlights the importance of correctly applying U-substitution in integration.

PREREQUISITES
  • Understanding of U-substitution in calculus
  • Familiarity with exponential functions and their integrals
  • Basic knowledge of differential calculus
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Practice integrating more complex exponential functions using U-substitution
  • Explore integration techniques such as integration by parts
  • Learn about definite integrals and their applications
  • Study the properties of exponential growth and decay functions
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Students studying calculus, particularly those learning integration techniques, as well as educators seeking to reinforce concepts of U-substitution and exponential functions.

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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