Questions: So, the particular antiderivative of C'(x) is C(x)=200 x^(2/3). Now, apply Part II of the Fundamental Theorem of Calculus to evaluate the integral ∫ from 200 to 300 of (400/3 sqrt[3]x) dx. Round any intermediate calculations, if needed, to no less than six decimal places, and round your final answer up to two decimal places.

So, the particular antiderivative of C'(x) is C(x)=200 x^(2/3).
Now, apply Part II of the Fundamental Theorem of Calculus to evaluate the integral ∫ from 200 to 300 of (400/3 sqrt[3]x) dx. Round any intermediate calculations, if needed, to no less than six decimal places, and round your final answer up to two decimal places.

Transcript text: So, the particular antiderivative of $\mathrm{C}^{\prime}(\mathrm{x})$ is $\mathrm{C}(\mathrm{x})=200 \mathrm{x}^{\frac{2}{3}}$. Now, apply Part II of the Fundamental Theorem of Calculus to evaluate the integral $\int_{200}^{300} \frac{400}{3 \sqrt[3]{\mathrm{x}}} \mathrm{dx}$. Round any intermediate calculations, if needed, to no less than six decimal places, and round your final answer up to two decimal places. BETA Al Tutor Back to Practice. Previous Page Next Page Display Step Answer Submit © 2025 Hawkes Learning Next • Previous

Solution

Solution Steps

Step 1: Rewrite the integral

The given integral is $\int_{200}^{300} \frac{400}{3 \sqrt[3]{\mathrm{x}}} \mathrm{dx}$. We can rewrite this as $\int_{200}^{300} \frac{400}{3} x^{-1/3} dx$.

Step 2: Find the antiderivative

The antiderivative of $\frac{400}{3} x^{-1/3}$ is $\frac{400}{3} \cdot \frac{x^{2/3}}{2/3} + C = 200x^{2/3} + C$. Let $C(x) = 200x^{2/3}$.

Step 3: Apply the Fundamental Theorem of Calculus

We need to evaluate $C(300) - C(200)$.

$C(300) = 200(300)^{2/3} \approx 200(43.8178046) \approx 8763.56092$ $C(200) = 200(200)^{2/3} \approx 200(34.19951893) \approx 6839.903786$

$C(300) - C(200) \approx 8763.56092 - 6839.903786 \approx 1923.657134$