• Jan 20th 2009, 11:35 AM
cloverz7
I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

2 (-10 + x) Sqrt[5 + x]
-----------------------
3

• Jan 20th 2009, 11:39 AM
Jester
Quote:

Originally Posted by cloverz7
I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

2 (-10 + x) Sqrt[5 + x]
-----------------------
3

FOr $\int_{-1}^{4} \frac{x}{\sqrt{x+5}}\,dx$ try the substitution $u = x+5$
• Jan 20th 2009, 11:50 AM
Moo
Or make the substitution $t=\sqrt{x+5}$ (Nod)

( $dt=\frac{dx}{2t}$)
• Jan 20th 2009, 11:51 AM
cloverz7
If you do that how do you convert the top x to u? would you just add a -5 outside of the integral?
• Jan 20th 2009, 12:33 PM
cloverz7
sorry for the bump, but I really need some help
• Jan 20th 2009, 12:57 PM
cloverz7
I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

2 (-10 + x) Sqrt[5 + x]
-----------------------
3

I've tryed substituting u=5+x and even u=sqrt(5+x) I just can't seem to get anywhere with it, I think I'm doing soemthing wrong with converting the top x to u, Please Help meeeeeeee
• Jan 20th 2009, 01:31 PM
Soroban
Hello, cloverz7!

Quote:

$\int^4_{\text{-}1} \frac{x}{\sqrt{x+5}}\,dx \;=\;\frac{2}{3}(x-10)\sqrt{x+5}\,\bigg]^4_{\text{-}1}$

I don't know how it got to that point.

Let: $u \,=\,\sqrt{x+5}\quad\Rightarrow\quad x \,=\,u^2-5\quad\Rightarrow\quad dx \,=\,2u\,du$

Substitute: . $\int\frac{u^2-5}{u}(2u\,du) \;=\; 2\int(u^2-5)\,du \;=\;2\left(\tfrac{1}{3}u^3 - 5u\right) \;=\;\tfrac{2}{3}u\left(u^2-15\right)$

Back-substitute: . $\tfrac{2}{3}\sqrt{x+5}\,\bigg[\left(\sqrt{x+5}\right)^2 - 15\bigg] \;=\;\tfrac{2}{3}\sqrt{x+5}\,\bigg[x+5 - 15\bigg]$

And we have: . $\frac{2}{3}(x-10)\sqrt{x+5}\,\bigg]^4_{\text{-}1} \qquad\hdots$ Got it?

• Jan 20th 2009, 01:55 PM
DeMath
Quote:

Originally Posted by cloverz7
I have maple and have used other tools online to find the answer, but I don't know the steps taken to get to it, I have a feeling its some algebra tricks that I'm just not remembering, but onto the problem.

Its the integral of X/(sqrt(5+X)) from a=-1 to b=4, The indefinate integral is

2 (-10 + x) Sqrt[5 + x]
-----------------------
3

I've tryed substituting u=5+x and even u=sqrt(5+x) I just can't seem to get anywhere with it, I think I'm doing soemthing wrong with converting the top x to u, Please Help meeeeeeee

Or

$\int\limits_{ - 1}^4 {\frac{x}{{\sqrt {x + 5} }}} dx = \int\limits_{ - 1}^4 {\frac{{x + 5 - 5}}{{\sqrt {x + 5} }}} dx = \int\limits_{ - 1}^4 {\left[ {\frac{{x + 5}}{{\sqrt {x + 5} }} - \frac{5}{{\sqrt {x + 5} }}} \right]} dx =$

$= \int\limits_{ - 1}^4 {\left[ {\sqrt {x + 5} - \frac{5}{{\sqrt {x + 5} }}} \right]} dx = \int\limits_{ - 1}^4 {\left[ {{{\left( {x + 5} \right)}^{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}} - 5{{\left( {x + 5} \right)}^{{{ - 1} \mathord{\left/{\vphantom {{ - 1} 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right]} dx =$

$= \left. {\left[ {\frac{2}{3}{{\left( {x + 5} \right)}^{{3 \mathord{\left/{\vphantom {3 2}} \right.\kern-\nulldelimiterspace} 2}}} - 10{{\left( {x + 5} \right)}^{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2}}}} \right]} \right|_{ - 1}^4 = \ldots = - 12 - \left( { - \frac{{44}}{3}} \right) = \frac{8}{3}.$
• Jan 20th 2009, 02:18 PM
Jhevon
Quote:

Originally Posted by cloverz7
sorry for the bump, but I really need some help

bumping is against the rules, and you have gotten help

if you follow post #2's suggestion (yes, you must change the x to u - 5) then you end up with

$\int_4^9 \frac {u - 5}{\sqrt{u}}~du$

if you follow post #3's suggestion, then you end up with

$\int_2^3 2(t^2 - 5)~dt$

either is fine, it is up to you to do which you prefer. is there anything you do not understand on how we got these new integrals?