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Solve the following.
Find the cost of increasing from 100 to 200 units if the marginal cost in Rupees per unit is given…
Find the cost of increasing from 100 to 200 units if the marginal cost in Rupees per unit is given by the function $MC = 0.003x^2 - 0.01x + 2.5$.
Answer
Answer
AIIntegral: $\int_{100}^{200} \left(0.003x^2 - 0.01x + 2.5\right) dx$
Increase in cost $= \int_{100}^{200}(0.003x^2 - 0.01x + 2.5)dx = \left[0.001x^3 - 0.005x^2 + 2.5x\right]_{100}^{200}$
$= (8000 - 200 + 500) - (1000 - 50 + 250) = 8300 - 1200 = 7100$.
Hence the cost of increasing from 100 to 200 units is ₹ $7100$.
Final answer: $7100$ ₹
From ISC 2018 Specimen Mathematics Paper 1, question 19(a).