91Ó°ÊÓ

For an integral to converge, the function within the integral should approach a finite value as the limits of integration approach the given values. We will first examine the convergence of the function at the lower limit, $$-\infty$$: $$\lim_{x\to -\infty} \frac{1}{\sqrt[3]{2-x}}$$ As x approaches $$-\infty$$, the denominator, $$\sqrt[3]{2-x}$$, will approach a very large value. Thus, the function will approach zero: $$\lim_{x\to -\infty} \frac{1}{\sqrt[3]{2-x}} = 0$$ Now, we will examine the convergence of the function at the upper limit, 0: $$\lim_{x\to 0} \frac{1}{\sqrt[3]{2-x}}$$ As x approaches 0, the denominator, $$\sqrt[3]{2-x}$$, will approach $$\sqrt[3]{2}$$. Thus, the function will approach: $$\lim_{x\to 0} \frac{1}{\sqrt[3]{2-x}} = \frac{1}{\sqrt[3]{2}}$$ Since the function approaches a finite value at both limits of integration, the integral converges. Now, we will proceed to evaluate the integral.

To evaluate the integral, let's use the substitution method. We will make the following substitution: $$u = 2 - x \Rightarrow -du = dx$$ Now, we will change the limits of integration according to the substitution. When $$x=-\infty$$, $$u= 2-(-\infty) = \infty$$. When $$x=0$$, $$u=2-0=2$$. With the substitution, the integral becomes: $$\int_{\infty}^{2} \frac{-d u}{\sqrt[3]{u}} = -\int_{\infty}^{2} u^{-\frac{1}{3}}du$$

Now, we will integrate the function with respect to u: $$-\int u^{-\frac{1}{3}}du = -\frac{3}{2}u^{\frac{2}{3}} + C$$

Now, we substitute x back into the function and apply the limits of integration: $$-\frac{3}{2}(2-x)^{\frac{2}{3}}\Bigg|_{-\infty}^{0} = -\frac{3}{2}(2-0)^{\frac{2}{3}} - \lim_{x\to -\infty} \left(-\frac{3}{2}(2-x)^{\frac{2}{3}}\right)$$

Now, we will evaluate the definite integral: $$-\frac{3}{2}(2)^{\frac{2}{3}} - (-\frac{3}{2}(2-(-\infty))^{\frac{2}{3}}) = -\frac{3}{2}(2)^{\frac{2}{3}}$$ Hence, the value of the integral is: $$\int_{-\infty}^{0} \frac{d x}{\sqrt[3]{2-x}} = -\frac{3}{2}(2)^{\frac{2}{3}}$$