To find the total profit from the sale of the first 60 tickets, given the marginal-profit function $P'(x) = 7x - 1037$, we need to compute the total profit $P(x)$ by integrating $P'(x)$ over the interval from 0 to 60.

Steps:

1. Set up the integral: $P(60) = \int_0^{60} P'(x) \, dx = \int_0^{60} (7x - 1037) \, dx$

2. Integrate $7x - 1037$:

   • The integral of $7x$ is $\frac{7x^2}{2}$.
   • The integral of $-1037$ is $-1037x$.
   • So: $\int (7x - 1037) \, dx = \frac{7x^2}{2} - 1037x + C$

3. Apply limits of integration: Evaluate $\frac{7x^2}{2} - 1037x$ from 0 to 60: [ P(60) = \left[ \frac{7(60)^2}{2} - 1037(60) \right] - \left[ \frac{7(0)^2}{2} - 1037(0) \right] ] The second term evaluates to 0 since $x = 0$.

4. Compute: $P(60) = \frac{7(3600)}{2} - 1037(60)$ $P(60) = \frac{25200}{2} - 62220$ $P(60) = 12600 - 62220 = -49620$

Thus, the total profit is $-49,620$ dollars.

Final Answer:

$\text{The total profit is } -\$49,620.00$

Let me know if you'd like a detailed breakdown or further clarification!

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Related Questions:

1. What does the negative profit indicate in this context?
2. How would the profit change if fixed costs were included?
3. What is the marginal profit when 50 tickets are sold?
4. How many tickets must be sold for the profit to break even?
5. What is the role of the integration constant $C$ in this context?

Tip:

Always check the limits of integration and ensure proper substitution to avoid errors in definite integrals.