# Arc length

## Homework Statement

y = 5 + 8x^3/2 from 0 to 1

## The Attempt at a Solution

I have tried it a few times Keep getting variations of 1+ 12$$\sqrt{12}$$. I would like some to give me a step by step how to work it. Something is killing me on this I am lost.

## Answers and Replies

Mark44
Mentor
Show us what you've tried and we can go from there.

Arc length is the Integral of sqrt(1+(f '(x))^2) so Derivative of the equation is 12x^1/2. Then I believe you are supposed to square that. Thats the same as (sqrt(12x))^2 right? so I tired it like that I also did 12x^1/2 * 12x^1/2= 144x so then I know you finish of the integral. I used U substitution to do that.

Roughly: $$\int{\sqrt{1+144x}}\:dx = \frac{1}{216}(1 + 144x)^{3/2}$$

Mark44
Mentor
Roughly: $$\int{\sqrt{1+144x}}\:dx = \frac{1}{216}(1 + 144x)^{3/2}$$
You're going to have to back up, since your integrand is wrong. Show us what you did to get the part under the radical.j

Edit: Never mind. I misread the exponent as 3, rather than 3/2.

Last edited:
I'm not the original poster dude.

$$s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx.$$

$$f(x) = 5 + 8x^{\frac{3}{2}}$$

$$f'(x) = 12x^{\frac{1}{2}}$$

$$[f'(x)]^{2} = 144x$$

$$s = \int_{0}^{1} \sqrt { 1 + 144x }\, dx = {\frac{1}{216}(1 + 144x)^{3/2}}\left|^{1}_{0}$$

ok i have got to the that point is that answer right? then you just substitute the 1 in and thats the answer? I just want to know that I did it right my online homework is picky about the answer format.
Thanks to all

Mugen, you should be able to check your work (or the posted work) for errors.

Yes, substituting 1 and 0 in is the correct procedure for evaluating this definite integral. Be more confident!

Mark44
Mentor
I'm not the original poster dude.

$$s = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx.$$

$$f(x) = 5 + 8x^{\frac{3}{2}}$$

$$f'(x) = 12x^{\frac{1}{2}}$$

$$[f'(x)]^{2} = 144x$$

$$s = \int_{0}^{1} \sqrt { 1 + 144x }\, dx = {\frac{1}{216}(1 + 144x)^{3/2}}\left|^{1}_{0}$$

Looks fine. I misread the exponent in the original problem as x^3 rather than x^(3/2).

OK I just put it in and it said it was wrong so I dont care anymore. I think its right as do all of you so Im going with the homework having a typo in the answer.
Thanks again

Mark44
Mentor
Your answer should be $$\frac{1}{216}(145^{3/2} - 1)$$
which can also be written as
$$\frac{1}{216}(145\sqrt{145} - 1)$$