How Do You Calculate y = 5 + 8x^(3/2) from 0 to 1?

[Mugen Prospec]

Homework Statement

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

Homework Equations

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.

 

[Mark44]

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

 

[Mugen Prospec]

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.

 

[miqbal]

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

 

[Mark44]

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.

 

[miqbal]

I'm not the original poster dude.

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

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

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

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

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

 

[Mugen Prospec]

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

 

[miqbal]

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]

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).

 

[Mugen Prospec]

OK I just put it in and it said it was wrong so I don't care anymore. I think its right as do all of you so I am going with the homework having a typo in the answer.
Thanks again

 

[Mark44]

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