Let's solve this problem step by step.

Part (a) - Perform the Integration

The total cost function is given by:

$C(x) = 2000 \left( 4 + 8 \int_0^x t^{1/4} \, dt \right)$

We need to evaluate the integral $\int_0^x t^{1/4} \, dt$.

1. Compute the integral: $\int t^{1/4} \, dt = \frac{t^{5/4}}{5/4} = \frac{4}{5} t^{5/4}$

2. Apply the limits from $0$ to $x$: $\int_0^x t^{1/4} \, dt = \frac{4}{5} x^{5/4}$

3. Substitute back into the expression for $C(x)$: $C(x) = 2000 \left( 4 + 8 \cdot \frac{4}{5} x^{5/4} \right)$

4. Simplify the expression: $C(x) = 2000 \left( 4 + \frac{32}{5} x^{5/4} \right) = 2000 \cdot 4 + 2000 \cdot \frac{32}{5} x^{5/4}$ $C(x) = 8000 + 12800 x^{5/4}$

So, the function $C(x)$ is:

$C(x) = 8000 + 12800 x^{5/4}$

Part (b) - Find $C(1)$, $C(5)$, and $C(10)$

Now we can plug in the values of $x = 1$, $x = 5$, and $x = 10$ to find $C(1)$, $C(5)$, and $C(10)$:

1. For $C(1)$: $C(1) = 8000 + 12800 \cdot 1^{5/4} = 8000 + 12800 = 20800$

2. For $C(5)$: $C(5) = 8000 + 12800 \cdot 5^{5/4}$

   Calculating $5^{5/4}$: $5^{5/4} \approx 8.032$ $C(5) \approx 8000 + 12800 \cdot 8.032 = 8000 + 102809.6 \approx 110810$

3. For $C(10)$: $C(10) = 8000 + 12800 \cdot 10^{5/4}$

   Calculating $10^{5/4}$: $10^{5/4} \approx 17.783$ $C(10) \approx 8000 + 12800 \cdot 17.783 = 8000 + 227622.4 \approx 235622$

Summary of Results:

• $C(1) = 20800$
• $C(5) \approx 110810$
• $C(10) \approx 235622$

Would you like further clarification on any of these steps or additional information?

Here are some related questions for further practice:

1. What is the value of $C(2)$ using the same function?
2. How does the function $C(x)$ behave as $x$ approaches infinity?
3. How would the cost change if the exponent in the integral were different, e.g., $t^{1/2}$ instead of $t^{1/4}$?
4. What does the term $x^{5/4}$ imply about the growth rate of the cost?
5. How would you interpret the constant 8000 in the context of the total cost?

Tip: When integrating powers of $t$, remember to adjust the coefficient based on the new exponent after integration.