View Full Version : Integral Help
Stratosphere
Aug18-09, 05:33 PM
1. The problem statement, all variables and given/known data
As I was reviewing some of my previuosly learned calculus I came across somthing that I had either forgoten how to do or was never taught. How do you take the integral of somthing like this
\int \sqrt{\frac{9}{4}x+1}
I don't know how to start with this one. Do I use U substitution?
CFDFEAGURU
Aug18-09, 06:09 PM
First rewrite the integral as
dy=\int(\frac{9}{4x}+1)^{1/2} dx
Now use the u substitution.
Thanks
Matt
Stratosphere
Aug18-09, 06:16 PM
Using U substitution I got ((9/4x+1)^1/2)/(27/8) I don't think I did this right.
CFDFEAGURU
Aug18-09, 06:29 PM
I don't believe that is correct.
Try this.
Let u = \frac{9}{4x}+1
Now find du/dx.
Stratosphere
Aug18-09, 06:40 PM
Du/dx is =36/16x^{2}
CFDFEAGURU
Aug18-09, 06:41 PM
Can you show your steps as to how you got that result please?
Stratosphere
Aug18-09, 06:48 PM
u=\frac{9}{4x}+1
\frac{0*4x-9*4}{(4x)^{2}} So I made a mistake with the negative? The answer should be -36/16x^2.
CFDFEAGURU
Aug18-09, 07:34 PM
Yes, you got it. Now perform the rest of the u substitution procedure and your done.
Elucidus
Aug18-09, 11:40 PM
1. The problem statement, all variables and given/known data
As I was reviewing some of my previuosly learned calculus I came across somthing that I had either forgoten how to do or was never taught. How do you take the integral of somthing like this
\int \sqrt{\frac{9}{4}x+1}
I don't know how to start with this one. Do I use U substitution?
I think the discussion got off track somewhere. The radicand above is \frac{9}{4}x+1 not \frac{9}{4x}+1. Or was the original statement an error?
If the original is correct, then it can be easilly integrated by substitution - try u=\frac{9}{4}x+1.
(This looks like a Stewart's Calculus exercise.)
--Elucidus
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