Accumulation: rate in, amount out Problem set A Worked solutions

Find \(\int (4x+3)(x^2-2)\,dx\).

• Expand: \(4 x^{3} + 3 x^{2} - 8 x - 6\)
• Integrate each term: \(x^{4} + x^{3} - 4 x^{2} - 6 x + c\)

Find \(\int t^2\left(3t-\frac{4}{t}\right)\,dt\).

• Expand: \(3 t^{3} - 4 t\)
• Integrate each term: \(\frac{3 t^{4}}{4} - 2 t^{2} + c\)

Find \(\int \frac{2x^5-3x}{4x^3}\,dx\).

• Divide each term: \(\frac{x^{2}}{2} - \frac{3}{4 x^{2}}\)
• Integrate each term: \(\frac{x^{3}}{6} + \frac{3}{4 x} + c\)

Find \(\int \frac{x^4-6}{x^2}\,dx\).

• Divide each term: \(x^{2} - \frac{6}{x^{2}}\)
• Integrate each term: \(\frac{x^{3}}{3} + \frac{6}{x} + c\)

Rainwater flows into a barrel at a rate of \(0.6t^2\) litres per minute, where t is the time in minutes. After 1 minute the barrel holds 12 litres. Find the volume of water in the barrel after 4 minutes.

• From the calculator: \(\int_{1}^{4} 0.6t^2\,dt\) = 12.6
• Volume = 12 + 12.6 = 24.6 litres

The rate of growth of the mass m grams of a patch of algae is modelled by \(\dfrac{dm}{dt}=kt+0.4\), where t is the time in weeks. When \(t=3\) the rate of growth is 1.6 grams per week.

1. (a) Find the value of k.

   • \(3k+0.4=1.6\)
   • \(k=0.4\)

2. (b) The patch has negligible mass when \(t=0\). Find its mass after 5 weeks.

   • Mass = \(\int_{0}^{5} (0.4t+0.4)\,dt\)
   • From the calculator: \(\int_{0}^{5} (0.4t+0.4)\,dt\) = 7

A fuel tank is filling at a rate of \(\dfrac{80}{t^2}\) litres per minute, where t is the time in minutes. The tank is empty when \(t=1\). At what time does the tank hold 60 litres?

• Amount = \(\int_{1}^{T} 80t^{-2}\,dt\)
• Solve for T: \(\int_{1}^{T} 80t^{-2}\,dt=60\)
• T = 4 minutes

A pond is being filled at a rate of \(\dfrac{70}{t^2}\) litres per minute, where t is the time in minutes. The pond is empty when \(t=1\). Its capacity is 65 litres.

1. (a) Find the volume in the pond when \(t=10\).

   • From the calculator: \(\int_{1}^{10} 70t^{-2}\,dt\) = 63

2. (b) Use technology to show that the pond will overflow, and find the time when it does.

   • Try a large upper limit. From the calculator: \(\int_{1}^{100} 70t^{-2}\,dt\) = 69.3, which is more than 65, so the pond overflows
   • Solve for T: \(\int_{1}^{T} 70t^{-2}\,dt=65\)
   • T = 14 minutes