# Approximate value of integral

• Apr 28th 2008, 06:13 PM
Eric08
Approximate value of integral
Find the length "L" of a certain curve given by L= http://www.mathhelpforum.com/math-he...63d30917-1.gif

- Im having a hard time with this, any help with be very appreciated
-thanks
• Apr 28th 2008, 06:31 PM
o_O
Make the following substitution:
$u = 4+ 3x$
$du = 3dx \: \: \Rightarrow \: \: dx = \frac{du}{3}$

Also, we must change our limits of our integral:
$u(8) = 4 + 3(8) = 28$
$u(2) = 4 + 3(2) = 10$

So substituting all these in:
$= \int_{u(2)}^{u(8)}\! \sqrt{u} \frac{du}{3} \: = \: \frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du$
• Apr 28th 2008, 06:42 PM
Eric08
Im sorry im very new to integrals, and need to see all the steps involved in solving this problem so I have an idea on what to do on future problems.
-Thank you
• Apr 28th 2008, 06:47 PM
o_O
This is a must-know integral: $\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$ and is some constant.

Do you not see how that applies to the last integral I gave you?
• Apr 28th 2008, 07:00 PM
Eric08
• Apr 29th 2008, 06:00 PM
Eric08
I think i figured this out.. is the correct answer 25.8976?
• Apr 29th 2008, 06:08 PM
o_O
From where we last left off:
$\frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du = \frac{1}{3} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \bigg|_{10}^{28} \: \: = \: \: \frac{2}{9} u^{\frac{3}{2}} \bigg|_{10}^{28} \: \: = \: \: \frac{2}{9} \left( 28^{\frac{3}{2}} - 10^{\frac{3}{2}} \right) \: \approx \: 25.898$

• Apr 29th 2008, 06:12 PM
Krizalid