Accumulation: rate in, amount out Problem set B One question at a time

Find \(\int (3x+2)(x^2-5)\,dx\).

• Expand: \(3 x^{3} + 2 x^{2} - 15 x - 10\)
• Integrate each term: \(\frac{3 x^{4}}{4} + \frac{2 x^{3}}{3} - \frac{15 x^{2}}{2} - 10 x + c\)

Find \(\int s^3\left(2s+\frac{5}{s}\right)\,ds\).

• Expand: \(2 s^{4} + 5 s^{2}\)
• Integrate each term: \(\frac{2 s^{5}}{5} + \frac{5 s^{3}}{3} + c\)

Find \(\int \frac{3x^7+4x}{6x^4}\,dx\).

• Divide each term: \(\frac{x^{3}}{2} + \frac{2}{3 x^{3}}\)
• Integrate each term: \(\frac{x^{4}}{8} - \frac{1}{3 x^{2}} + c\)

Find \(\int \frac{x^5+8}{x^3}\,dx\).

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

Water drips into a basin at a rate of \(0.9t^2\) millilitres per second, where t is the time in seconds. After 2 seconds the basin holds 8 millilitres. Find the volume of water in the basin after 5 seconds.

• From the calculator: \(\int_{2}^{5} 0.9t^2\,dt\) = 35.1
• Volume = 8 + 35.1 = 43.1 millilitres

The rate of growth of the mass m grams of a yeast culture is modelled by \(\dfrac{dm}{dt}=kt+0.2\), where t is the time in days. When \(t=4\) the rate of growth is 1.4 grams per day.

1. (a) Find the value of k.

   • \(4k+0.2=1.4\)
   • \(k=0.3\)

2. (b) The culture has negligible mass when \(t=0\). Find its mass after 6 days.

   • Mass = \(\int_{0}^{6} (0.3t+0.2)\,dt\)
   • From the calculator: \(\int_{0}^{6} (0.3t+0.2)\,dt\) = 6.6

A rain gauge is filling at a rate of \(\dfrac{72}{t^2}\) millilitres per minute, where t is the time in minutes. The gauge is empty when \(t=1\). At what time does the gauge hold 48 millilitres?

• Amount = \(\int_{1}^{T} 72t^{-2}\,dt\)
• Solve for T: \(\int_{1}^{T} 72t^{-2}\,dt=48\)
• T = 3 minutes

A water butt is being filled at a rate of \(\dfrac{56}{t^2}\) litres per minute, where t is the time in minutes. The butt is empty when \(t=1\). Its capacity is 50 litres.

1. (a) Find the volume in the butt when \(t=5\).

   • From the calculator: \(\int_{1}^{5} 56t^{-2}\,dt\) = 44.8

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

   • Try a large upper limit. From the calculator: \(\int_{1}^{100} 56t^{-2}\,dt\) = 55.4, which is more than 50, so the butt overflows
   • Solve for T: \(\int_{1}^{T} 56t^{-2}\,dt=50\)
   • T = 9.33 minutes