Evaluate the improper integral?

int_2^3 1/(x-3)^(1/3)dx

2 Answers
Feb 27, 2018

-3/2

Explanation:

The problem is up the upper bound, as at x=3, the integrand has a discontinuity (vertical asymptote) due to division by 0.

Therefore, we want to replace the upper bound with t and take the limit of the integral as t->3^-. The reason we're taking the left-hand limit is because we're working with the upper bound, which we can only ever approach from the left in our interval of integration.

Lim_(t->3^-)int_2^tdx/(x-3)^(1/3)

There's not much to evaluating the actual integral. intdx/(x-3)^(1/3) could be rewritten as int(du)/u^(1/3)=intu^(-1/3)du=3/2u^(2/3)=3/2(x-3)^(2/3), where u=x-3, du=dx. We're leaving out the constant of integration because we're going to be taking a definite integral;

Lim_(t->3^-)3/2(x-3)^(2/3) where we must first evaluate 3/2(x-3)^(2/3) from 2 to t, which gives us:

3/2(t-3)^(2/3)-3/2(2-3)^(2/3)=3/2(t-3)^(2/3)-3/2

Because 3/2(-1)^(2/3)=3/2root(3)(-1^2)=3/2sqrt(1)=3/2

Lim_(t->3^-)3/2(t-3)^(2/3)-3/2=3/2(0^(2/3))-3/2=0-3/2=-3/2

Feb 27, 2018

int_2^3 dx/(x-3)^(1/3) = -3/2

Explanation:

As the integrand function is continuous in the interval x in [2,3) we can evaluate the integral as:

int_2^3 dx/(x-3)^(1/3) = lim_(t->3^-) int_2^t dx/(x-3)^(1/3)

int_2^3 dx/(x-3)^(1/3) = lim_(t->3^-) int_2^t (d(x-3))/(x-3)^(1/3)

int_2^3 dx/(x-3)^(1/3) = lim_(t->3^-) 3/2[(x-3)^(2/3)]_2^t

int_2^3 dx/(x-3)^(1/3) = lim_(t->3^-) 3/2[(t-3)^(2/3)-1]

int_2^3 dx/(x-3)^(1/3) = -3/2