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Question 694

The flow rate of a lubricant, QQQ, in litres per hour, through a high-precision valve is modeled by the function Q(t)=3t3t2+1Q(t) = \frac{3t^3}{t^2+1}Q(t)=t2+13t3​, where ttt is the time in hours since the valve was opened for 0≤t≤40 \le t \le 40≤t≤4.

An engineer is attempting to approximate the total volume of lubricant released, V=∫04Q(t) dtV = \int_{0}^{4} Q(t) \, dtV=∫04​Q(t)dt, using the trapezium rule by splitting the interval into nnn equal strips.

a.

When n=4n = 4n=4: (i) State the number of ordinates that the engineer uses. (ii) Calculate the approximation for the total volume VVV using this method. Give your answer correct to two decimal places.

[4]
b.

Show that the exact volume of lubricant released is 24−32ln⁡1724 - \frac{3}{2}\ln 1724−23​ln17 litres. Fully justify your answer.

[4]
c.

Explain what would happen to the engineer's approximation in part (a)(ii) as n→∞n \to \inftyn→∞.

[1]
Markscheme

Integration Questions

  1. A Level
  2. /Maths
  3. /Integration

864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.

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