Question

How to Integrate (2sin(2pi+2pit)dt ? .(Integration Limits from 0 to 1 ) .Please,I need the final answer asap now :)

Answer 1

Is that
\[ \int_0^1 \sin(2\pi+2\pi t) dt\]
?

Answer 2

\[\sin(2\pi + 2\pi t) = \sin(2\pi t)\]
because sin(x) is periodic with period 2pi, so that's just
\[\int_0^1 \sin(2\pi t)dt = \frac{-\cos(2\pi t)}{2\pi}|^1_0 = 0\]

Answer 3

Please ,Answer Problem #6 from the attached file here

Answer 4

A vector field is conservative if the curl of the field is zero. In 2-D, that means
\[ F = <F_x,F_y>\]
\[\frac{\partial F_y }{\partial x} - \frac{\partial F_x}{\partial y} = 0\]
Here,
\[\frac{\partial F_y}{\partial x} = 2x\]
\[\frac{\partial F_x}{\partial y} = 2x\]
so the field is conservative. If that's the case, it can be written as a potential:
\[\vec{\nabla}f =<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}> = <F_x,F_y>\]
Apparently,
\[\frac{\partial f}{\partial x} = 3 + 2xy \rightarrow f(x,y) = 3x + x^2y+C(y)\]
differentiating wrt y,
\[\frac{\partial f}{\partial y} = x^2 + C'(y) = x^2-3y^2 \rightarrow C'(y) = -3y^2 \rightarrow C= 3y^3 +K\]
which means
\[f(x,y) = 3x + x^2y-3y^2+K\]
The line integral goes from (0,1) to (0,-exp(pi))
so evaluating at the endpoints,
\[\int \vec{\nabla}f \centerdot d\vec{r} = f(0,-e^\pi) - f(0,1) = 3e^{2\pi}-3\]
I hope I didn't make a typo but that's the principle of the thing.

Answer 5

The procedure is correct.Thank you for your help .But The final answer should be (e^3pi+1)

Answer 6

ah, yeah,
\[C(y) = y^3 \neq 3y^3\]
sorry about that.