Maths · Calculus · chapter 7 of 13
Class 12 Maths Chapter 7: Integrals case study question with answers
One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
A housing society in Jaipur fills its overhead tank with a pump whose output varies as the motor warms up. The caretaker notes that t minutes after switching on, water enters at the rate r(t) = 3t2 − 12t + 15 litres per minute, for 0 ≤ t ≤ 6. The volume added in the first x minutes is the area function V(x) = ∫0x r(t) dt, and the volume added between two times is the definite integral of r over that interval.
Part 1 · 1 mark
V′(2) equals
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Answer: (a) 3
By the first fundamental theorem V′(x) = r(x), so V′(2) = 12 − 24 + 15 = 3 litres per minute. Option (b) is V(2) = 8 − 24 + 30 = 14, the volume rather than the rate; (c) is r′(2); (d) is r(0).
Part 2 · 1 mark
The volume added in the first 3 minutes, in litres, is
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Answer: (a) 18
V(x) = x3 − 6x2 + 15x, so V(3) = 27 − 54 + 45 = 18 litres.
Part 3 · 2 marks
Find the volume of water added from t = 3 to t = 5.
Answer and marking scheme
∫35 (3t2 − 12t + 15) dt = [t3 − 6t2 + 15t]35 = (125 − 150 + 75) − (27 − 54 + 45) = 50 − 18 = 32 litres.
- anti derivative t^{3} − 6t^{2} + 15t (1)
- 32 litres (1)
Or
Find the average rate of inflow over the first 6 minutes, 16 ∫06 r(t) dt.
Answer and marking scheme
∫06 r(t) dt = 216 − 216 + 90 = 90 litres, so the average rate is 906 = 15 litres per minute.
- ∫_{0}^{6} r(t) dt = 90 (1)
- average 15 litres per minute (1)
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