Can someone explain to me why this is with finding an integral

  • Thread starter nickb145
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  • #1
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Alright, I know how to find the integral of a regular function, But an inverse, is throwing me off

Can someone explain why 1/Sqrt(7t)dt is 2sqrt(t/7)

I'm probably missing something, but why does the variable end up as a numerator?

I know the inverse is 1/(r+1)x^(r+1)

So I work it out as
1/(-.5+1) which goes out to 2

then the rest
(7t)^-1/2+1
(7t)^1/2
(what am I missing here)
=2sqrt(t/7)+C
 

Answers and Replies

  • #2
Ray Vickson
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Alright, I know how to find the integral of a regular function, But an inverse, is throwing me off

Can someone explain why 1/Sqrt(7t)dt is 2sqrt(t/7)

I'm probably missing something, but why does the variable end up as a numerator?

I know the inverse is 1/(r+1)x^(r+1)

So I work it out as
1/(-.5+1) which goes out to 2

then the rest
(7t)^-1/2+1
(7t)^1/2
(what am I missing here)
=2sqrt(t/7)+C
So, you are saying that
[tex] \int \frac{1}{\sqrt{7t}} \, dt = 2 \sqrt{t/7} + C[/tex]
You can check by taking the derivative of your answer with respect to t and see if it gives you back the integrand. That is something you should always do, and when you do it you will have answered your own question.
 
  • #3
RUber
Homework Helper
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It always helps me to think of
##\sqrt{7t}=(7t)^{1/2}##. Or in your case,
##\int (7t)^{-1/2} dt##
 

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