Thread: 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

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

The antiderivative of 30 is 30s.

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

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

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.

cheers