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- Thread starter chapsticks
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Mark44

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Why do you have this inequality?## Homework Statement

so I did

## Homework Equations

s=∫_{ba}√(1+[f'(x)]^{2}dx

## The Attempt at a Solution

y=2/3x^{3/2}+7

y'=x^{1/2}0≤x≤1

What is that 7 doing?=∫^{6}√(1+(x^{1/2})^{2})dx

=[2/3(1+x)^{3/2}]^{7}

my answer came out wrong

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Mark44

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Your inequality is saying that the interval is [0, 1]. It's actually [0, 6].

Here's how your integral looks using LaTeX.

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

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.

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

The LaTeX for that.

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

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Mark44

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

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