# Length of the curve defined by $y=6 x^{3/2} - 7$ from $x=1$ to $x = 9\;$?

Find the length of the curve defined by $y=6 x^{3/2} - 7$ from $x=1$ to $x = 9$.

I need help with this section. I would really appreciate it. Thank you!

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Where did you get stuck? Did you have trouble computing the integral in the arc length formula? – David Mitra Dec 17 '12 at 14:18
I just realized it's arclength :-D I know how to do this. – Ceelos Dec 17 '12 at 14:18
Compute $L=\int_1^9 \sqrt{1+ (y')^2 }\,dx$ – David Mitra Dec 17 '12 at 14:20
@DavidMitra, Yes, thank you (: – Ceelos Dec 17 '12 at 14:20

Use the formula

$$s = \int_{a}^{b}\sqrt{1+y'(x)^2} dx$$

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Thank you, I don't know why I was blanking out. I know how to do this already. Thank you for answering though (-: – Ceelos Dec 17 '12 at 14:20
@Ceelos: You are welcome. – Mhenni Benghorbal Dec 17 '12 at 14:22

We have $$y=6x^{(3/2)} -7$$ Let $\mathcal{L}$ denote the length of the curve. We can calculate $\mathcal{L}$ by \begin{align} \mathcal{L} &= \int_1^9 \sqrt{1+(y')^2} \, dx \\ &= \int_1^9 \sqrt{1+((6x^{(3/2)} -7)')^2} \, dx \\ &= \int_1^9 \sqrt{1+(9x^{(1/2)})^2} \, dx \\ &= \int_1^9 \sqrt{1+81x} \, dx \\ &= \Bigl[\frac{2}{243} (1+81x)^{3/2} \Bigr]_1^9 \approx 156.22 \end{align} See Wolfram Alpha for the integration.

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Hint: Compute $y'$ and then compute the length $\mathcal{l}$:

$$\mathcal{l} =\int_1^9 \sqrt{1 + (y'(x))^2}dx$$

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