How do you test the improper integral #int 2x(x^2-1)^(-1/3)dx# from #[0,1]# and evaluate if possible?

Question

Evelyn Agard · Accepted Answer

#int_0^1 2x(x^2-1)^(-1/3) dx = -3/2#

By substitution #int2x(x^2-1)^(-1/3) dx = 3/2(x^2-1)^(2/3) +C#

So,

#int_0^1 2x(x^2-1)^(-1/3) dx = lim_(brarr1^-)int_0^b 2x(x^2-1)^(-1/3) dx #

# = lim_(brarr1^-)[3/2(x^2-1)^(2/3)]_0^b#

# = lim_(brarr1^-)[3/2(b^2-1)^(2/3)]-[3/2(-1)^(2/3)]#

# = 0-3/2 = -3/2#

Samantha Adkison · Answer

undefined To test the improper integral $\int_0^1 2x(x^2-1)^{-1/3}dx$ for convergence, first check if the integrand is continuous on the interval [0,1]. Since the function $f(x) = 2x(x^2-1)^{-1/3}$ is continuous on (0,1), there is no issue of discontinuity within the interval.

Next, examine if there are any infinite discontinuities at the endpoints of the interval. At $x = 1$, $(x^2-1)^{-1/3}$ is undefined, but $\lim_{x \to 1^-} 2x(x^2-1)^{-1/3} = +\infty$. Therefore, there is an infinite discontinuity at $x = 1$.

To evaluate the integral, split it into two parts:

1. Evaluate $\int_0^1 2x(x^2-1)^{-1/3}dx$ from $0$ to a value just less than $1$.
2. Evaluate $\lim_{b \to 1^-} \int_0^b 2x(x^2-1)^{-1/3}dx$.

The first part is a standard definite integral, which can be computed using substitution.

Let $u = x^2 - 1$, then $du = 2x dx$.
When $x = 0$, $u = -1$, and when $x = 1$, $u = 0$.

So, the integral becomes $\int_{-1}^{0} u^{-1/3} du$, which can be evaluated using the power rule for integration.

The second part involves finding the limit of a definite integral as $b$ approaches $1$ from the left. Since the function approaches infinity at $x = 1$, the improper integral does not converge.

Therefore, the improper integral $\int_0^1 2x(x^2-1)^{-1/3}dx$ is divergent.