Problem 35 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{0}^{8} \frac{d x}{\sqrt[3]{x}}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral of the function \(\frac{1}{\sqrt[3]{x}}\) over the interval \([0, 8]\). Answer: The integral of the given function over the interval \([0, 8]\) is 6.

Step 1: Rewrite the function

Rewrite the function \(\frac{1}{\sqrt[3]{x}}\) as \(x^{-\frac{1}{3}}\) so that it is easier to find the antiderivative. $$\int_{0}^{8} \frac{d x}{\sqrt[3]{x}} = \int_{0}^{8} x^{-\frac{1}{3}} dx$$

Step 2: Find the antiderivative

To find the antiderivative, we use the power rule for integration, which states that \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\), where n ≠ -1. Here, \(n = -\frac{1}{3}\), so: $$\int x^{-\frac{1}{3}} dx = \frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1} + C = \frac{3}{2} x^{\frac{2}{3}} + C$$

Step 3: Evaluate the antiderivative at the limits

Now we need to evaluate the antiderivative at the limits 0 and 8: $$\frac{3}{2} \cdot 8^{\frac{2}{3}} = \frac{3}{2} \cdot 4 = 6$$ $$\frac{3}{2} \cdot 0^{\frac{2}{3}} = 0$$

Step 4: Subtract the lower limit evaluation from the upper limit evaluation

Finally, we subtract the value of the antiderivative at the lower limit (0) from the value at the upper limit (8) to obtain the result of the integral: $$\int_{0}^{8} x^{-\frac{1}{3}} dx = 6 - 0 = 6$$ The integral of the given function over the interval \([0, 8]\) is 6.