Evaluate the Improper integral [4,13] 1/(x-5)^(1/3)

[ThatOneGuy]

Homework Statement

Evaluate the Improper integral [4,13] 1/(x-5)^1/3

Homework Equations

N/A

The Attempt at a Solution

In step 1, I split the integral into two separate integrals because at x=5, it would be undefined. I made the first limit approach 5 from the left and the second limit approach five from the right.

For step 2, I did u substitution to prepare to integrate. Since I did the substitution, I plugged the old intervals into u = x - 5 to get the new ones.

For step 3, I rewrote both integrals in terms of u du and brought the u^(1/3) from the denominator to the numerator for easier integration. I also added the new intervals in.

In step 4, I integrated both integrals.

Step 5 is where I'm not sure what to do. I plugged in the values and that is what I ended up with but in the first parentheses, the second value has a (-1)^2/3 which is undefined. The (R-5)'s will equal 0 and cancel those fractions out but I don't know how to handle the (-1)^2/3.

 

[Mark44]

Homework Statement

Evaluate the Improper integral [4,13] 1/(x-5)^1/3

Homework Equations

N/A

The Attempt at a Solution

View attachment 81908
In step 1, I split the integral into two separate integrals because at x=5, it would be undefined. I made the first limit approach 5 from the left and the second limit approach five from the right.

For step 2, I did u substitution to prepare to integrate. Since I did the substitution, I plugged the old intervals into u = x - 5 to get the new ones.

For step 3, I rewrote both integrals in terms of u du and brought the u^(1/3) from the denominator to the numerator for easier integration. I also added the new intervals in.

In step 4, I integrated both integrals.

Step 5 is where I'm not sure what to do. I plugged in the values and that is what I ended up with but in the first parentheses, the second value has a (-1)^2/3 which is undefined. The (R-5)'s will equal 0 and cancel those fractions out but I don't know how to handle the (-1)^2/3.

##(-1)^{2/3}## is defined -- its value is 1.

You can work with it either as ##[(-1)^2]^{1/3} = 1^{1/3} = 1## or as ##[(-1)^{1/3}]^2 = (-1)^2 = 1##.