The marginal cost is $MC = 0.6x^{2} - 4x + 10$ rupees, and the fixed cost is ₹120. What is the total cost function?

Explanation

1. 01

   $\int (0.6x^{2} - 4x + 10)\,dx = 0.2x^{3} - 2x^{2} + 10x + K$.

2. 02

   With $K = 120$, the fixed cost: $C(x) = 0.2x^{3} - 2x^{2} + 10x + 120$.