I'm having some trouble with limits at infity.
The integral is e^(-x)cos(x) dx evaluated from zero to infinity, so the antiderivative is 1/2 e^(-x) ( -cos(x) + sin (x) ). Then since it's from zero to infinity I made it into a limit: lim b-- infinity from 0 to b. Long story short, I don't know how to evaluate it at infinity.

Question

I'm having some trouble with limits at infity.
The integral is e^(-x)cos(x) dx evaluated from  zero to infinity, so the antiderivative is 1/2 e^(-x) ( -cos(x) + sin (x) ). Then since it's from zero to infinity I made it into a limit: lim b--> infinity from 0 to b. Long story short, I don't know how to evaluate it at infinity.

watchmath · Answer

ok so you have
$$
\lim_{b\to \infty} \frac{1}{2}-\frac{1}{2}e^{-b}(-\cos b+\sin b)
$$
right?

anonymous · Answer

yeah once you plug in the b and 0, right? $$\frac{ 1 }{ 2 } e^-b (-\cos b + \sin b) - (-1/2)$$