Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.50. ∫ (from 0 to 9) 1/(x - 1)¹ᐟ³ dx

Accepted Answer

Identify the type of integral and check for any points of discontinuity or singularities within the interval of integration. Here, the integrand is $$\frac{1}{(x - 1$$)^{1/3}$$}$$, and the interval is from 0 to 9. Notice that the function is undefined at $$x = 1$$ because the denominator becomes zero there. Since $$x = 1$$ lies within the interval of integration, split the integral at this point to handle the improper integral properly. Write the integral as the sum of two integrals: $$\$$int_0$$^1 \frac{1}{(x - 1$$)^{1/3}$$} \, dx + \$$int_1$$^9 \frac{1}{(x - 1$$)^{1/3}$$} \, dx. $$Rewrite each integral as a limit approaching the point of discontinuity. For the first integral, express it as $$\lim_{t 	o 1^-} \$$int_0$$^t \frac{1}{(x - 1$$)^{1/3}$$} \, dx. $$For the second integral, express it as $$\lim_{s 	o 1^+} \int_s^9$ \frac{1}{(x - 1$$)^{1/3}$$} \, dx. $$Find the antiderivative of the integrand $$\frac{1}{(x - 1$$)^{1/3}$$}. $$Recall that $$\int (x - a)^n \, dx = \frac{(x - a$$)^{n+1}$$}{n+1} + C$$ for $$n 
eq -1. $$Here, $$n = -\frac{1}{3}$$, so the antiderivative is $$\frac{(x - 1$$)^{2/3}$$}{\frac{2}{3}} + C = \frac{3}{2} (x - 1$$)^{2/3}$$ + C. $$Evaluate each limit by substituting the antiderivative back into the definite integrals and taking the limits as $$t 	o 1^-$$ and $$s 	o 1^+. $$Determine whether these limits converge to finite values or diverge to infinity to conclude if the original integral converges or diverges.

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
50. ∫ (from 0 to 9) 1/(x - 1)¹ᐟ³ dx

Risposta video verificata per un problema simile:

Concetti chiave

Improper Integrals

Integrals with Discontinuous Integrands

Behavior of Functions Near Singularities