Problem 46 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{1}^{11} \frac{d x}{(x-3)^{2 / 3}}$$

Step-by-Step Solution

Answer

Question: Evaluate the definite integral $$\int_{1}^{11} \frac{d x}{(x-3)^{2 / 3}}$$ Answer: $$6 - 3 \cdot (-2)^{1/3}$$

Step 1: Define the substitution

Let's make the substitution: $$u = x - 3$$ Then, the derivative of u with respect to x is: $$\frac{d u}{d x} = 1$$ Now, we can rewrite the integral in terms of u, and we also need to change the limits of integration.

Step 2: Rewrite the integral in terms of u

First, let's find the new limits. When x = 1, $$u = 1 - 3 = -2$$ When x = 11, $$u = 11 - 3 = 8$$ Now, we change the integral to have u as the variable. Since \(\frac{d u}{d x} = 1\), we have: $$\int_{-2}^{8} \frac{d u}{u^{2 / 3}}$$

Step 3: Find the anti-derivative

Now, we find the anti-derivative of the function \(\frac{1}{u^{2 / 3}}\) with respect to u. We have: $$\int \frac{d u}{u^{2 / 3}} = \int u^{-2/3} d u$$ Applying the power rule, we get: $$\frac{u^{1/3}}{\frac{1}{3}} + C = 3u^{1/3} + C$$

Step 4: Apply the limits and evaluate the definite integral

Now that we have the anti-derivative, we can evaluate the definite integral. We have: $$\left(3u^{1/3} + C\right) \Big|_{-2}^{8} = 3 \cdot 8^{1/3} - 3 \cdot (-2)^{1/3}$$ Now evaluate, $$3 \cdot 2 - 3 \cdot (-2)^{1/3} = 6 - 3 \cdot (-2)^{1/3}$$ Hence, the value of the integral is $$\int_{1}^{11} \frac{d x}{(x-3)^{2 / 3}} = 6 - 3 \cdot (-2)^{1/3}$$