A company finds that the marginal profit, in dollars per foot, from drilling a well that is \( x \) feet deep is given by \( P^{\prime}(x)=2 \sqrt[3]{x} \). Find the profit when a well 300 ft deep is drilled. Set up the integral for the total profit for a well that is 300 feet deep. \( P(300)=\int_{0} \) dx

To find the total profit \( P(300) \) when drilling a well that is 300 feet deep, we need to integrate the marginal profit function \( P'(x) = 2 \sqrt[3]{x} \) from 0 to 300 feet. The integral for the total profit is set up as follows: \[ P(300) = \int_{0}^{300} P'(x) \, dx = \int_{0}^{300} 2 \sqrt[3]{x} \, dx \] Now, we can compute the integral: 1. First, rewrite \( \sqrt[3]{x} \) as \( x^{1/3} \): \[ P(300) = \int_{0}^{300} 2 x^{1/3} \, dx \] 2. Next, we can factor out the constant: \[ P(300) = 2 \int_{0}^{300} x^{1/3} \, dx \] 3. Now, we find the antiderivative of \( x^{1/3} \): \[ \int x^{1/3} \, dx = \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3} \] 4. Now we evaluate the definite integral: \[ P(300) = 2 \left[ \frac{3}{4} x^{4/3} \right]_{0}^{300} \] 5. Plugging in the limits: \[ P(300) = 2 \left( \frac{3}{4} (300)^{4/3} - \frac{3}{4} (0)^{4/3} \right) \] \[ P(300) = 2 \cdot \frac{3}{4} (300)^{4/3} \] \[ P(300) = \frac{3}{2} (300)^{4/3} \] 6. Now, we need to calculate \( (300)^{4/3} \): \[ (300)^{4/3} = (300^{1/3})^4 \] First, calculate \( 300^{1/3} \): \[ 300^{1/3} \approx 6.669 \quad (\text{using a calculator}) \] Then raise it to the power of 4: \[ (300^{1/3})^4 \approx (6.669)^4 \approx 1980.66 \] 7. Finally, multiply by \( \frac{3}{2} \): \[ P(300) \approx \frac{3}{2} \cdot 1980.66 \approx 2970.99 \] Thus, the total profit when drilling a well that is 300 feet deep is approximately \( P(300) \approx 2970.99 \) dollars. In summary, the integral for the total profit is: \[ P(300) = 2 \int_{0}^{300} x^{1/3} \, dx \]