Approximate value of integral

**Eric08** · April 28th 2008, 05:13 PM

Find the length "L" of a certain curve given by L=

- Im having a hard time with this, any help with be very appreciated
-thanks

**o_O** · April 28th 2008, 05:31 PM

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:
$= \int_{u(2)}^{u(8)}\! \sqrt{u} \frac{du}{3} \: = \: \frac{1}{3} \int_{10}^{28} u^{\frac{1}{2}}du$

**Eric08** · April 28th 2008, 05:42 PM

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

**o_O** · April 28th 2008, 05:47 PM

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?

**Eric08** · April 28th 2008, 06:00 PM

is the answer 6.28?

**Eric08** · April 29th 2008, 05:00 PM

I think i figured this out.. is the correct answer 25.8976?

**o_O** · April 29th 2008, 05:08 PM

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

Just as you had

**Krizalid** · April 29th 2008, 05:12 PM

That's the correct answer Eric08.

See QuickMath, pretty useful for cheking answers.

**ThePerfectHacker** · April 29th 2008, 06:19 PM

Originally Posted by Krizalid

That's the correct answer Eric08.

See QuickMath, pretty useful for cheking answers.

You can check answers on Graph 4.3 as well. The only problem is it does not always have "nice" answers like QuickMath does.

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