Question

How do you evaluate the integral `int 1/(x+1)^(1/3)dx` from -2 to 0?

Answer 1

0

the interval is placed symmetrically about the 2 sided singularity so the areas seem to net off

Explanation:

`int_(-2)^0 1/(x+1)^(1/3)dx`

first note the singularity at x = -1, so through caution i'll do it like this .... probs OTT

`= int_(-2)^(x to -1^-) (x+1)^(-1/3)dx + int_(x to -1^+)^(0) (x+1)^(-1/3)dx`

by power rule

`=3/2 [ (x+1)^(2/3) ]_(-2)^(x to -1^-) + 3/2[ (x+1)^(2/3) ]_(x to -1^+)^(0)`

`=3/2 [ 0 - (-1)^(2/3) ] + 3/2[ (1)^(2/3) - 0]`

`= 0`