Question

Integral of (x-2)^(-1/3) from 1 to 3.
I get 0. But Wolfram gets imaginary numbers!!!!! Why is that? Who is right?

Answer 1

Nah, you cray. It's a simple power rule integral.

Answer 2

\[\int\limits_{1}^{3} (x-2)^{-1/3} dx\]

Answer 3

u = (x-2); du = 1 dx
\[\int\limits_{1}^{3} u^{-1/3} du\]

Answer 4

Can you take it from here?

Answer 5

No. This is improper integral.

Answer 6

FTC does not apply.

Answer 7

Need to change the limits of integration when you change variables

but the indefinite integral will be \[\frac{3}{2}(x-2)^{2/3} + C\]
The problem is evaluating \(-1^{2/3}\)

Answer 8

\[\int\limits_{1}^{3}\left( x-2 \right)^{\frac{ -1 }{ 3 }}dx=\frac{ \left( x-2 \right)^{\frac{ -1 }{ 3 }+1 } }{ \frac{ -1 }{ 3 }+1 }~x~ from~ 1 \to 3\]
\[=\frac{ 3 }{ 2 }\left[ \left( 3-2 \right)^{\frac{ 2 }{ 3 }} -\left( 1-2 \right)^{\frac{ 2 }{ 3 }}\right]\]
\[=\frac{ 3 }{ 2 }\left[ 1^{\frac{ 2 }{ 3 }} -\left( \iota^2 \right)^{\frac{ 2 }{ 3}}\right]\]
\[=\frac{ 3 }{ 2 }-\frac{ 3 }{ 2 }\iota ^{\frac{ 4 }{ 3 }},\]
which is an imaginary number.

Answer 9

I think I should consider i^(4/3) to be 1 simply.

Answer 10

\[\int_1^3 \frac{1}{\sqrt[3]{x-2}} \, dx = -\frac{3}{2}(1+(-1)^{2/3})\]

Since when does \(i^{4/3} =1\)?

Answer 11

\[cube~roots~of~1~are 1,\omega,\omega^2 \]
\[1^4=1,\omega^4=\omega^3\omega=1\omega=\omega,\left( \omega^2 \right)^4=\omega^8=\left( \omega^3 \right)^2\omega^2=1^2\omega^2=\omega^2\]
so you can not consider it as 1.

Answer 12

When using WolframAlpha you have to specify that you're taking the real root. It's designed to find the principal root by default.