weird integral precalculus info: precalculus covers basic calculus second semester like derivatives and integral
So we got this one on final review $\displaystyle I=\int_{|x|=4}(x+1)^3\,dx.$
So I was thinking, first of all, why is there no upper bound, then I thought maybe it just meant integrate at $x=-4$ and $x=4$, and then I realized that made no sense either. Then I realized complex numbers have the absolute value (distance from $0$) property, so this might have just been a circle of radius $4$.
So using that logic, $\displaystyle I=\int_0^{2π}(4\cos(\theta)+4i\sin(\theta)+1)^3\,d\theta$.
Is this logic correct or nah?
Because from here the integration should be simple (even if it is a bit bashy)
 A: At the end of the day: I'm not sure how to interpret this either. In the context of complex analysis, this expression would be interpreted as a contour integral around the circle $|x| = 4$ in the complex plane. The expression
$$ \displaystyle I=\int_0^{2π}(4\cos(\theta)+4i\sin(\theta)+1)^3\,d\theta $$
you wrote is almost what a contour integral is, but you need to replace $d\theta$ with $dx$ or, more correctly, with
$$ \frac{dx}{d\theta} \, d\theta = \frac{d}{d\theta} \Big(4\cos\theta+4i\sin\theta \Big)\,d\theta $$
because the original integral is in terms of $x$, but your bounds are $0$ and $2\pi$ which are in terms of $\theta$.
However, this is probably not what the question is asking. For one, it would be a question from complex analysis, or the calculus of complex numbers, which is something I doubt you'd be dealing with in ordinary real calculus. Moreover, the integral is actually trivial to evaluate using tools from complex analysis (it's zero), which would make it a somewhat silly question to ask in a complex analysis setting.
My best guess is the expression
$$ \displaystyle I=\int_{|x|=4}(x+1)^3\,dx $$
is meant to mean the real integral from $-4$ to $+4$ (as opposed to, as you say, "at" $-4$ and $+4$ which doesn't make sense to me either) because the equation $|x|=4$ has exactly two real solutions: $\pm 4$, and you need two bounds to evaluate a real definite integral. But it's only a guess. If it's correct, I'd like to have a word with whoever invented that terrible notation...
