# quick question involving definite integral

• April 10th 2010, 04:17 PM
Archduke01
quick question involving definite integral
http://euler.vaniercollege.qc.ca/web...5160a177d1.png

I'm uncertain about something. I thought this was a simple problem where you simply bring up the denominator and apply the general power rule, but that doesn't seem to be right... Can anyone point me in the right direction?
• April 10th 2010, 04:19 PM
mr fantastic
Quote:

Originally Posted by Archduke01
http://euler.vaniercollege.qc.ca/web...5160a177d1.png

I'm uncertain about something. I thought this was a simple problem where you simply bring up the denominator and apply the general power rule, but that doesn't seem to be right... Can anyone point me in the right direction?

Substitute u = 4x + 1.
• April 10th 2010, 04:24 PM
Archduke01
$\frac {1}{4} \int_6^{20} \frac {4}{\sqrt{4x+1}}$

I'm sorry for asking, but how do I proceed from there? Or atleast tell me why it isn't correct if it isn't.
• April 10th 2010, 04:33 PM
mr fantastic
Quote:

Originally Posted by Archduke01
$\frac {1}{4} \int_6^{20} \frac {4}{\sqrt{4x+1}}$

I'm sorry for asking, but how do I proceed from there? Or atleast tell me why it isn't correct if it isn't.

Have you been taught the substitution technique.

Otherwise you will just have to use the following result which I'm sure is somewhere in your class notes or textbook:

$\int (ax + b)^n \, dx = \frac{1}{a(n+1)} (ax + b)^{n + 1}$ provided $n \neq -1$ (and I have ignored the arbitrary constant since it's not needed for a definite integral).
• April 10th 2010, 04:48 PM
Homeomorphism
For the first problem: Let $u = \sqrt{4x+1}$, then $\dfrac{du}{dx} = \dfrac{2}{\sqrt{(4x+1)}} \Rightarrow dx = \dfrac{\sqrt{4x+1}}{2}\;{du}$ so that $\int_{x=6}^{x=20}\dfrac{dx}{\sqrt{4x+1}} = \int_{u=5}^{u=9}\left(\dfrac{1}{u}\right)\left(\df rac{u}{2}\right)\;{du}$ and finish off.

For the second problem: Let again $u = \sqrt{4x+1}$, then $\dfrac{du}{dx} = \dfrac{2}{\sqrt{(4x+1)}} \Rightarrow dx = \dfrac{\sqrt{4x+1}}{2}\;{du}$ so that $\dfrac{1}{4}\int_{x=6}^{x=20}\dfrac{4}{\sqrt{4x+1} }\;{dx} = \dfrac{1}{4}\int_{u=5}^{u=9}\left(\dfrac{4}{u}\rig ht)\left(\dfrac{u}{2}\right)\;{du}$ and finish off.
• April 10th 2010, 04:51 PM
Archduke01
Where did you get the 2 from?
• April 10th 2010, 04:58 PM
mr fantastic
Quote:

Originally Posted by Archduke01
Where did you get the 2 from?

Did you read my previous reply or did you decide it was easier to focus on a complete solution that is probably beyond your current level of comprehension?

Do you understand that, in the context of my previous reply, n = -1/2?
• April 10th 2010, 05:05 PM
Archduke01
Thank you, but I want to try the substitution method since I've never heard of your second one.
• April 10th 2010, 05:14 PM
mr fantastic
Quote:

Originally Posted by Archduke01
Thank you, but I want to try the substitution method since I've never heard of your second one.

Well, I'll ask again. Have you been taught the substitution method? If so, where are you having trouble making the substitution u = 4x + 1 (which brings us all the way back to post #2)?