Question

Evaluate the integral:
\[\large \int\limits_{-1}^{-8} \frac{ x-x^2 }{ 2 \sqrt[3]{x} }\]

Answer 1

It's just integrating power functions.

Answer 2

something like this right?
\[= \frac{ x }{ 2(x)^{\frac{ 1 }{ 3 }} } - \frac{ x^2 }{ x^\frac{ 1 }{ 3 } }\]
\[= \frac{ \frac{ x^2 }{ 2*x^{\frac{ 4 }{ 3 }} } } {\frac{ 4 }{ 3 } } - \frac{\frac{ x^3 }{ 3x^{\frac{ 4 }{ 3} } }}{\frac{ 4 }{ 3 }}\]

Answer 3

Actually, I just noticed I could simplify it first..

Answer 4

something like \[
\frac{x}{\sqrt[3]{x}} = x^{1-1/3} = x^{2/3}
\]

Answer 5

yeah, overthinking this simplified to this, and then integrate it..
\[\large \frac{ x^{2/3} }{ 2 } - \frac{ x^{5/3} }{ 2 }\]

Answer 6

Yes

Answer 7

\[= \frac{ 5x^{5/3} }{ 6 } - \frac{ 8x^{8/3} }{ 6 }\]

Answer 8

\[=\frac{ 5(-8)^{5/3} }{ 6 } - \frac{ 8(-8)^{8/3} }{ 6 } -\left( \frac{ 5(-1)^{5/3} }{ 6 } - \frac{ 8(-1)^{8/3} }{ 6 } \right)\]
?

Answer 9

I think my calculus teacher purposely gives these problems a lengthy amount of work -.-

Answer 10

is this right so far?

Answer 11

Yeah.

Answer 12

I like how your calc teacher makes you integrate backwards.

Answer 13

He's ancient.

Answer 14

I think i made an error with my integration though..

Answer 15

\[\int\limits_{}^{} \frac{ x^{2/3} }{ 2 } = \frac{ 3x^{5/3} }{ 10 }\]
right?

Answer 16

is your last post the problem?

Answer 17

no, the problem is the definite integral, i'm just doing part of the integration again because i think it was done incorrectly..

Answer 18

integral of [x^(2/3)]2 dx?

Answer 19

oh okay

Answer 20

yeah, i checked with wolfram, that part was wrong.
so uh:

\[\huge \int\limits_{-8}^{-1} \frac{ (x-x^2) }{ 2\sqrt[3]{x} } =>\frac{ x }{ 2\sqrt[3]{x} } - \frac{ x^2 }{ 2\sqrt [3]{x} } =\]
\[\huge \frac{ x^{2/3} }{ 2 } - \frac{ x^{5/3} }{ 2 }\]
\[\huge = \frac{ 3x^{5/3} }{ 10 } - \frac{ 3x^{8/3} }{ 16 }\]

Answer 21

\[\large = \frac{ 3(-1)^{5/3} }{ 10 } - \frac{ 3(-1)^{8/3} }{ 16 } - \left( \frac{ 3(-8)^{5/3} }{ 10 } - \frac{ 3(-8)^{8/3} }{ 16 } \right)\]
evaluating this part is going to be a pain..

Answer 22

how so?

Answer 23

i'm a lazy person, thats how

Answer 24

good enough reason
come up with a technique to make it easier

Answer 25

i'll leave that to the geniuses