Question

Determine whether each integral is convergent or divergent. Evaluate

1) ∫ from -infinity to infinity (xe^-x^2) dx

answer comes out to be 0 but I need step by step guidance

2) ∫ from -3 to 6 (x^-4) dx

Answer 1

Answer for #2 is divergent so I assume answer comes out to be infinity? How do I arrive to that answer? Thanks.

Answer 2

your function is
\[\frac{1}{x^4}\] which is not defined at x = 0

Answer 3

so you have to compute
\[\lim_{t\to 0^+}\int_t^6 \frac{1}{x^4}dx\]

Answer 4

anti derivative of
\[x^{-4}\] is
\[-\frac{1}{3}x^{-3}=-\frac{1}{x^3}\] so you need
\[\lim_{t\to 0^+}-\frac{1}{x^3}\] which does not exist

Answer 5

So integral of 1/x^4 then evaluate from t to 6 then plug in t for 0? May I ask why 0 is + and not just left as 0? OR is there a difference?

Answer 6

I see you didn't evaluate from t to 6. Was it not necessary?

Answer 7

you have to break up your integral in to two parts, from -3 to 0 and from 0 to 6 since the function is not defined at 0, you need the limit as t goes to zero actually twice, from the left and from the right

Answer 8

it is true that you can evaluate at \(\frac{1}{x^3}\) at \(x=6\) this is a finite number for sure, but i don't care because the limit as t goes to zero is infinite, so we will not get a number out of it