Solution to Incorrect Homework Equation

[chapsticks]

Homework Statement

so I did

Homework Equations

s=∫_b^a√(1+[f'(x)]²dx

The Attempt at a Solution

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

my answer came out wrong

 

[Mark44]

Homework Statement

so I did

Homework Equations

s=∫_b^a√(1+[f'(x)]²dx

The Attempt at a Solution

y=2/3x^3/2+7
y'=x^1/2 0≤x≤1

Why do you have this inequality?

=∫⁶√(1+(x^1/2)²)dx
=[2/3(1+x)^3/2]⁷

my answer came out wrong

What is that 7 doing?

 

[chapsticks]

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]

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]

 

[chapsticks]

I got it correct

 

[Mark44]

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