Can someone please explain to me why integral 1/(x^3) dx evaluated from 1 to infinity is 1/2. I got -1/2x^2 for the antiderivative. I then set made it into a limit where b approaches infinity getting -1/2b^2 +1/2, so -1/2b^2 must equal zero, but why?
and why integral 3/(x^(1/3)) dx evaluated from 1 to infinity diverges?

Question

Can someone please explain to me why integral 1/(x^3) dx evaluated from 1 to infinity is 1/2. I got -1/2x^2 for the antiderivative. I then set made it into a limit where b approaches infinity getting -1/2b^2 +1/2, so -1/2b^2 must equal zero, but why?
and why integral 3/(x^(1/3)) dx evaluated from 1 to infinity diverges?

kmeis002 · Answer

Since b is approaching infinity, the function $$ \frac{1}{2b^2}$$ is approaching zero since the denominator is approaching infinity.

Find the anti derivative of $$ x^{1/3}$$
When you apply your limits at infinity, the term with the b will not disappear and hence it will converge.

kmeis002 · Answer

*diverge instead of converge

anonymous · Answer

Wait, but for $$\int\limits_{1}^{b} \frac{ 3 }{ \sqrt[3]{x} } dx$$ the antiderivative is $$9x^{1/3}$$, so at b it would just go to infinity ?

kmeis002 · Answer

Indeed, that is the definition of divergence.

anonymous · Answer

so could the answer be either infinity or diverges or just diverges because it's like infinity- 9?

kmeis002 · Answer

Well, I suppose you could put infinity if your teacher excepts it, they essentially mean the same thing. Its like when you evaluate a limit and you say it = infinity. Technically the limit doesn't exist, but infinity is nice shorthand.

anonymous · Answer

Ok cool I think I got it, thanks!