Question

calc 2

Answer 1

integrate cox(x)

Answer 2

-sinx

Answer 3

sorry was asking a question.

Answer 4

@zepdrix this is all you boyyy

Answer 5

lookup ur notes, u must be having arc length formula ?

Answer 6

yes i have it, but i can't integrate it right

Answer 7

Hmm I must be doing something wrong, this one isn't working out very nice :(

Answer 8

\[\Large\bf\sf L\quad=\quad \int\limits ds\]

Answer 9

For arc length yes?
Where \(\Large\bf\sf ds\quad=\quad \sqrt{1+\left(\frac{df}{dx}\right)^2}dx\)

Answer 10

\(\large s = \int \limits _?^?\sqrt{1 + [f'(x)]^2} dx\)

Answer 11

yea

Answer 12

\[\Large\bf\sf L\quad=\quad \int\limits_0^4 \sqrt{1+\sin^2x}~dx\]Hmmmm :(

Answer 13

thats where i got stuck too

Answer 14

:( too

Answer 15

looks evil

Answer 16

So how do we find arc length? Well if we have a function that's like this:



we can see that the function is really a bunch of tiny hyptenuses of triangles added together. So:

\[dx^2+dy^2=ds^2\]

So one individual section will be:

\[ds = \sqrt{dx^2+dy^2}\]
and so if we want all the tiny pieces between two points, we integrate which is just a sum.
\[\int\limits_{}^{} ds =\int\limits_{}^{} \sqrt{dx^2+dy^2}\]

That looks like no integral I've ever seen before! So let's multiply by a fancy version of 1, dx/dx.\[ardxc length =\int\limits_{}^{} \frac{ dx }{ dx } \sqrt{dx^2+dy^2}=\int\limits_{}^{} dx \sqrt{(\frac{ dx }{ dx })^2+(\frac{ dy }{ dx })^2}=\int\limits_{}^{} \sqrt{(1+(\frac{ dy }{ dx })^2}dx\]

Answer 17

Are you supposed to use power series or some sort of numerical integration?

Answer 18

don't think so

Answer 19

Can't use a calculator?