# Thread: quick question involving definite integral

1. ## quick question involving definite integral

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?

2. Originally Posted by Archduke01

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.

3. $\frac {1}{4} \int_6^{20} \frac {4}{\sqrt{4x+1}}$

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.

4. 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).

5. 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.

6. Where did you get the 2 from?

7. 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?

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

9. 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)?