Here's the question you clicked on:
ggrree
Alright, this is an integration calculus problem, multiple choice. (Written out in the next post)
let K = \[\int\limits_{-2}^{3} \sin (x ^{1/3})\] which of the following is then true? 1<k<or=2 2<k<or=3 3<k 0<k<1 k<or=0
I understand that since sin is an odd function, we can disregard the integral from -2 to 2, as it equals zero. but I don't understand what to do from there. it leaves me with: \[\int\limits_{2}^{3} \sin x^{1/2}\]
sorry! in my last post I meant to write the integral of sin (x ^1/3), not x^1/2
How can you do the substitution if the derivative of x^1/3 isn't there?