# Arc Length

so I did

## Homework Equations

s=∫ba√(1+[f'(x)]2dx

## The Attempt at a Solution

y=2/3x3/2+7
y'=x1/2 0≤x≤1
=∫6√(1+(x1/2)2)dx
=[2/3(1+x)3/2]7

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Mark44
Mentor

so I did

## Homework Equations

s=∫ba√(1+[f'(x)]2dx

## The Attempt at a Solution

y=2/3x3/2+7
y'=x1/2 0≤x≤1
Why do you have this inequality?
=∫6√(1+(x1/2)2)dx
=[2/3(1+x)3/2]7

What is that 7 doing?

I saw it in an example for the inequality.. The 7 is for the a & b part I don't know how to place the 0 on the bottom.

Mark44
Mentor
I saw it in an example for the inequality.. The 7 is for the a & b part I don't know how to place the 0 on the bottom.
Your inequality is saying that the interval is [0, 1]. It's actually [0, 6].

Here's how your integral looks using LaTeX.
$$\int_0^6 \sqrt{1+x}~dx$$

Here's how to write that integral in LaTeX.

[ tex]\int_0^6 \sqrt{1+x}~dx[ /tex]

(The extra spaces in the tex and /tex tags prevent the browser from rendering the code inside.)

Here's the antiderivative with limits of integration shown.
$$\left. \frac{2}{3}(1 + x)^{3/2}\right|_0^6$$

The LaTeX for that.
[ tex]\left. \frac{2}{3}(1 + x)^{3/2}\right|_0^6[ /tex]

Mark44
Mentor
Good to know. That's what I got, too.