Problem 50 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{0}^{9} \frac{d x}{(x-1)^{1 / 3}}$$

Step-by-Step Solution

Answer

Answer: The value of the definite integral is $\frac{9}{2}$.

Step 1: Identify the Function to Integrate

The given integral function is: $$\int_{0}^{9} \frac{dx}{(x-1)^{1/3}}$$

Step 2: Use Substitution

Let's use substitution: $$u = (x - 1)^{1/3} \implies u^3 = x - 1 \implies x = u^3 + 1$$ Now, find the differential, \(dx\): $$\frac{dx}{du} = 3u^2 \implies dx = 3u^2 du$$ Update the integration limits as well: When \(x = 0\), \(u = (-1)^{1/3} = -1\) When \(x = 9\), \(u = (8)^{1/3} = 2\) Now, substitute \(u\) and \(dx\) back into the integral: $$\int_{-1}^{2} \frac{3u^2}{u} du$$

Step 3: Simplify the Integral

Simplify the integrand: $$\int_{-1}^{2} 3u du$$

Step 4: Find the Antiderivative

Now, find the antiderivative of \(3u\): $$\int 3u du = \frac{3}{2}u^2 + C$$

Step 5: Evaluate the Definite Integral

Plug in the integration limits: $$\left[\frac{3}{2}u^2\right]_{-1}^{2} = \frac{3}{2}(2^2) - \frac{3}{2}(-1)^2 = 3(4) - \frac{3}{2} = \frac{9}{2}$$ The definite integral converges, and the value of the integral is: $$\int_{0}^{9} \frac{dx}{(x-1)^{1 / 3}} = \frac{9}{2}$$