Need help walking through evaulation of 30+5s

Hi all, I'm very new to calculus and have been coping well so far but one thing has me stumped.

(30+5s)ds = 70 where s = 0 to 2

I've got the answer off a calculation website but I can't get my workings to reach this conclusion.

Heres what I have so far

30+(5s^2/2)

30+(5/2 x s^2)

subsituting with '2'

30+(5/2 x 2^2) = 40

substituting with '0'

30+(5/2 x 0^2) = 30

but I currently believe it is 2 minus 0 to get the result i.e 40 - 30

I'd like to just add the 2 results and get 70 but why am I adding instead of minusing?

Thanks in advance for any help at all.

John

Re: Need help walking through evaulation of 30+5s

How did you manage to ask this question without using the notation or the word "integral"?

The antiderivative of 30 is 30s.

Re: Need help walking through evaulation of 30+5s

Thanks very much, understanding now.

And apologies for what must have seemed like a lazy question but I'm just getting to grips with the wording and I'm not sure how to enter the integral symbol but I'm going to give it a go...

\int

Hope that was right and thanks again for your help, really appreciate it.

John

Re: Need help walking through evaulation of 30+5s

ok - clearly that didn't work! Would you mind helping me out with the symbol for future use please?

Re: Need help walking through evaulation of 30+5s

This forum uses LaTeX language to enter formulas. See the LaTeX Help subforum for details. Wrap you LaTeX code in [TEX] ... [/TEX] tags (tags are not case sensitive). E.g., [TEX]\int x^2\,dx=\frac{x^3}{3}+C[/TEX] gives . You can see the code in other people's posts if you click "Reply With Quote" button under the post. If you click "Go Advanced" button under your reply, you get a text area with a toolbar with many buttons on top. One of them has the sign; it wraps the TEX tags around the selected text.

Re: Need help walking through evaulation of 30+5s

cheers