How do you find the integral of #int (dx / x^(2/3)) # from 1 to -15?

Question

How do you find the integral of #int  (dx / x^(2/3)) # from 1 to -15?

William Abdalla · Accepted Answer

#int_1^-15  (dx / x^(2/3)) = int_1^-15 x^(-2/3) dx # is an improper integral.  The integrand is not defined at #0#.  Early in the treatment of integration, improper integrals are not (yet) defined.. When improper integral have been defined, we do the following:
To (attempt to) evaluate the integral, we have to split it at #0# #int_1^-15  x^(-2/3) dx = int_1^0  x^(-2/3) dx +int_0^-15  x^(-2/3) dx # # = lim_(brarr0^+) int_1^b  x^(-2/3) dx + lim_(ararr0^-) int_a^-15  x^(-2/3) dx # Note that #int x^(-2/3) dx = 3x^(1/3) = 3root(3)x#, So for the first integral we get: #int_1^0  x^(-2/3) dx = lim_(brarr0^+) int_1^b  x^(-2/3) dx# # = lim_(brarr0^+) [3root(3)x]_1^b# # = lim_(brarr0^+) [3root(3)b-3root(3)1] = 0 - 3 = -3# And for the second integral we get: #int_0^-15  x^(-2/3) dx =  lim_(ararr0^-) int_a^-15  x^(-2/3) dx # # =  lim_(ararr0^-) [3root(3)x]_a^-15# # =  lim_(ararr0^-) [3root(3)(-15)- 3root(3)a] = -3root(3)(15) - 0=-3root(3)(15)# Both improper integrals converge, so we get: #int_1^-15  x^(-2/3) dx = int_1^0  x^(-2/3) dx +int_0^-15  x^(-2/3) dx # # = (-3) + (-3root(3)15)# # = -3 -3root(3)15)# Note on limits of integration #int_a^b f(x) dx# is read "the integral from a to b" Recall that #int_a^b f(x) dx = -int_b^a f(x) dx#. So if the intended question was #int_-15^1  x^(-2/3) dx# #" "#(which is the integral from -15 to 1),  then the answer should be  #int_-15^1  x^(-2/3) dx = 3+3root(3)15#. And the details of the solution should be changed appropriately.

Christopher Allers · Answer

undefined To find the integral of $ \int_{1}^{-15} \frac{dx}{x^{2/3}} $, we can integrate the function over the given interval using the definite integral formula:

$$ \int_{1}^{-15} \frac{dx}{x^{2/3}} $$

First, simplify the integral by rewriting $ x^{-2/3} $ as $ \frac{1}{x^{2/3}} $. Then integrate:

$$ \int_{1}^{-15} x^{-2/3} \, dx = \int_{1}^{-15} \frac{1}{x^{2/3}} \, dx $$

The integral of $ x^{-2/3} $ is $ \frac{3x^{1/3}}{1/3} + C $, where $ C $ is the constant of integration. Simplify this to $ 9x^{1/3} + C $.

Now plug in the upper and lower limits:

$$ \left[ 9x^{1/3} \right]_{1}^{-15} $$

$$ = 9(-15)^{1/3} - 9(1)^{1/3} $$

$$ = 9(-15^{1/3}) - 9(1) $$

$$ = -27 $$

Therefore, the integral of $ \int_{1}^{-15} \frac{dx}{x^{2/3}} $ from 1 to -15 is -27.