How do you evaluate the integral #int 1/root3x# from -1 to 1?

1 Answer
Jack
Sep 29, 2016

#int_-1^1x^(-1/3) = 0#

Explanation:

Expressing the reciprocal third root of x as #x^(-1/3)# is useful here, as we can simply apply the reverse power rule. It is also important to note that there is a discontinuity at #x=0#, so we must split the integral into two parts:

#int_0^1x^(-1/3) + int_-1^0x^(-1/3)#

By the reverse power rule, we increment the power by one and divide by the exponent:

#3/2x^(2/3)|_0^1 = 3/2 - 0#

#+# #3/2x^(2/3)|_-1^0 = 0 -3/2#

Thus,
#int_-1^1x^(-1/3) = 3/2 -3/2 = 0#

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