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3.6 Integration (A-level only)

3.6 Integration (A-level only)

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Question 147

The rate of change of the volume of water in a large industrial storage tank, R(t)R(t)R(t) in m3/day\text{m}^3/\text{day}m3/day, is modeled by the function:

R(t)=54t2+2t−14,t>0 R(t) = \frac{54}{t^2} + 2t - 14, \quad t > 0 R(t)=t254​+2t−14,t>0

where t t\,t is the time in days since the start of a maintenance cycle. Using calculus:

a.

Find the set of values of t t\,t for which the rate of change R(t)R(t)R(t) is increasing, giving your answer in the form t>ab3t > a\sqrt[3]{b}t>a3b​ where a a\,a and b b\,b are integers.

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b.

Show that ∫39(54t2+2t−14)dt=0\displaystyle \int_{3}^{9} \left( \frac{54}{t^2} + 2t - 14 \right) dt = 0∫39​(t254​+2t−14)dt=0.

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c.

Given that ∫36(54t2+2t−14)dt=−6\displaystyle \int_{3}^{6} \left( \frac{54}{t^2} + 2t - 14 \right) dt = -6∫36​(t254​+2t−14)dt=−6:

(i) State the value of ∫69(54t2+2t−14)dt\displaystyle \int_{6}^{9} \left( \frac{54}{t^2} + 2t - 14 \right) dt∫69​(t254​+2t−14)dt.

(ii) Find the value of the constant k k\,k such that ∫36(54t2+2t+k)dt=0\displaystyle \int_{3}^{6} \left( \frac{54}{t^2} + 2t + k \right) dt = 0∫36​(t254​+2t+k)dt=0.

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Markscheme

3.6 Integration (A-level only) Questions

  1. A Level
  2. /Maths
  3. /3.6 Integration (A-level only)

363 exam-style questions on CCEA A Level Maths 3.6 Integration (A-level only), covering 3.6.1 Integration (A-level only), 3.6.2 Integration (A-level only), 3.6.3 Integration (A-level only), 3.6.4 Integration (A-level only), 3.6.5 Integration (A-level only), 3.6.6 Integration (A-level only), and 3.6.7 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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