Q.The marginal cost of producing $x$ units is $MC=3x^2-4x+10$, and the fixed cost is ₹500. Find the total cost function and the cost of producing 5 units.

Integrating the marginal cost and pinning the constant using the fixed cost recovers the total cost function, exactly as before.

$C(x)=\int(3x^2-4x+10)dx=x^3-2x^2+10x+K$; $C(0)=500\Rightarrow K=500$. $C(5)=125-50+50+500=625$.

Integrating the marginal cost

Pinning $K$ using the fixed cost

Cost of producing 5 units

Check (independent recomputation, term by term): $125-50=75$; $75+50=125$; $125+500=625$ — confirmed.

Integrating $-4x$ incorrectly as $-4x^2$ instead of the correct $-2x^2$ (misapplying the power rule's division by the new exponent) — always divide by $n+1$, not merely raise the power.