Question

Okay I'm stuck...How do you take the integral from 0 to 2pi of r(t)dt = -cos(t)i+sin(t)j + x^2/2

Answer 1

I already took the derivative, that's how I got -cos(t)i+sin(t)j + x^2/2

Answer 2

I just don't know what is my next step.

Answer 3

You just integrate the vector component by component. So, you'll end up with:
\[-\sin t\hat{i}-\cos t\hat{j}+\frac{x^3}{6}\hat{k}\]

Then you plug in 2pi and 0, and subtract. Looks like \[\frac{8\pi^3}{6}\hat{k}\]

Answer 4

@srossd I already integrated the original function; and that is -cos(t)i+sin(t)j + x^2/2...Now I just need to put 2pi and 0 into it...But my answer came out to pi^k...Is that correct?

Answer 5

(pi^2)k

Answer 6

Oh, I see. It should be (2pi)^2/2 k, so 2pi^2 k

Answer 7

Wait don't you mean pi^2 k?

Answer 8

You divided by 2 correct?

Answer 9

Yeah, but it was a 4 on top.

Answer 10

One other question. So I dont need i or j? I can just write k as my answer?

Answer 11

Ohhh because of the power. I gotcha! :)

Answer 12

Yeah, right on both counts.

Answer 13

Okay so the integral of r(t)dt from 0 to 2pi of cos(t)i+sin(t)j + x^2/2
= 2pi^2 k

Answer 14

Yes, assuming r(t) is the derivative of that function.