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

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- Apr 28th 2008, 05:13 PMEric08Approximate 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, 05:31 PMo_O
Make the following substitution:

$\displaystyle u = 4+ 3x$

$\displaystyle du = 3dx \: \: \Rightarrow \: \: dx = \frac{du}{3}$

Also, we must change our limits of our integral:

$\displaystyle u(8) = 4 + 3(8) = 28$

$\displaystyle u(2) = 4 + 3(2) = 10$

So substituting all these in:

$\displaystyle = \int_{u(2)}^{u(8)}\! \sqrt{u} \frac{du}{3} \: = \: \frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du$ - Apr 28th 2008, 05:42 PMEric08
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, 05:47 PMo_O
This is a must-know integral: $\displaystyle \int x^{n} dx = \frac{x^{n+1}}{n+1} + C$ where $\displaystyle n \neq -1$ and is some constant.

Do you not see how that applies to the last integral I gave you? - Apr 28th 2008, 06:00 PMEric08
is the answer 6.28?

- Apr 29th 2008, 05:00 PMEric08
I think i figured this out.. is the correct answer 25.8976?

- Apr 29th 2008, 05:08 PMo_O
From where we last left off:

$\displaystyle \frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du = \frac{1}{3} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \bigg|_{10}^{28} \: \: = \: \: \frac{2}{9} u^{\frac{3}{2}} \bigg|_{10}^{28} \: \: = \: \: \frac{2}{9} \left( 28^{\frac{3}{2}} - 10^{\frac{3}{2}} \right) \: \approx \: 25.898$

Just as you had ;) - Apr 29th 2008, 05:12 PMKrizalid
That's the correct answer Eric08.

See QuickMath, pretty useful for cheking answers. - Apr 29th 2008, 06:19 PMThePerfectHacker