Unit mixed practice: areas, rates and accumulation Problem set A Worked solutions

Find the area enclosed between the x-axis and the graph of \(y=-x^2 + 7x - 10\).

• \(y=-(x-2)(x-5)\), so the curve meets the x-axis at \(x=2\) and \(x=5\)
• \(y>0\) between the roots: \(A=\int_{2}^{5}(-x^2 + 7x - 10)\,dx\)
• From the calculator: \(A=9/2 = 4.50\)

Find the area enclosed by the part of the graph of \(y=16-x^2\) for \(x>0\), the x-axis and the y-axis.

• The curve meets the x-axis where \(16-x^2=0\): \(x=4\) (take the positive root)
• \(A=\int_0^{4}(16-x^2)\,dx\)
• From the calculator: \(A=128/3 = 42.7\)

The diagram shows the graphs of \(y=6x-x^2\) and \(y=2x\). They meet at the origin and at the point P.

1. (a) Find the coordinates of P.

   • \(6x-x^2=2x\), so \(4x-x^2=0\) and \(x(4-x)=0\)
   • The non-zero solution is \(x=4\); \(y=2\times4=8\)

2. (b) Find the shaded area.

   • Between the graphs the curve is the upper one: \(A=\int_0^{4}((6x-x^2)-2x)\,dx\)
   • From the calculator: \(A=32/3 = 10.7\)

The function f satisfies \(f'(x)=x^2 + 2x\) and \(f(0)=5\). Find \(\int_0^{3} f(x)\,dx\).

• \(f(x)=\frac{x^3}{3} + x^2+c\)
• \(f(0)=5\) gives \(c=5\), so \(f(x)=\frac{x^3}{3} + x^2 + 5\)
• From the calculator: \(\int_0^{3}(\frac{x^3}{3} + x^2 + 5)\,dx=123/4 = 30.8\)

Given that \(\int f(x)\,dx=3\left(x^4 - \frac{1}{x}+c\right)\), find an expression for \(f(x)\).

• Differentiating an antiderivative gives the function: \(f(x)=\frac{d}{dx}\left[3\left(x^4 - \frac{1}{x}+c\right)\right]\)
• \(f(x)=3\left(4x^3 + \frac{1}{x^2}\right)=12x^3 + \frac{3}{x^2}\)

The diagram shows the graphs of \(y=\frac{12}{x}\) and \(y=10-x\). They meet at the points P and Q.

1. (a) Use your calculator to find the x-coordinates of P and Q.

   • Solve \(\frac{12}{x}=10-x\) on the calculator (or \(x^2-10x+12=0\))
   • \(x=1.394\) and \(x=8.606\)

2. (b) Find the shaded area.

   • The line is above the curve between P and Q: \(A=\int_{1.394}^{8.606}\left((10-x)-\frac{12}{x}\right)dx\)
   • From the calculator: \(A=14.2\)

A 3D printer lays down plastic at a rate of \(6+kt^2\) grams per minute, where t is the time in minutes after it starts. After 4 minutes the rate is 14 grams per minute. How much plastic is laid down in the first 4 minutes?

• At \(t=4\): \(6+k\times4^2=14\), so \(16k=8\) and \(k=0.5\)
• Amount \(=\int_0^{4}(6+0.5t^2)\,dt\)
• From the calculator: 34.7 grams

For the first 30 seconds after the gate opens, grain leaves a hopper at a rate directly proportional to the time since the gate opened. After 5 seconds the rate is 15 kilograms per second. How much grain leaves the hopper in the first 30 seconds?

• Rate proportional to time: \(r=ct\). At \(t=5\), \(r=15\), so \(c=3\)
• Total \(=\int_0^{30}3t\,dt\)
• From the calculator: 1350 kg

A model suggests that water drips into a collecting tray at a rate of \(\frac{80}{t^3}\) millilitres per second, where t is the time in seconds after the tap was turned on. When \(t=1\) there are 6 millilitres in the tray.

1. (a) Find the amount of water in the tray 4 seconds after the tap was turned on.

   • Amount \(=6+\int_{1}^{4}\frac{80}{t^3}\,dt\)
   • From the calculator: \(6+37.5\), so 43.5 millilitres

2. (b) Use technology to estimate the amount of water that will eventually be in the tray.

   • Use a large upper limit: \(6+\int_{1}^{1000}\frac{80}{t^3}\,dt=46.000\)
   • The total settles at about 46.0 millilitres

3. (c) Give one criticism of this model.

   • The rate is not defined at \(t=0\) and is very large just after it: the model cannot be used from the start.

Rainwater drains from a flooded cellar at a rate of \(r=\frac{t^2}{40}\), where r is measured in cubic metres per minute and t is the time in minutes after the pump is switched on. The graph is shown, shaded from \(t=0\) to \(t=45\).

1. (a) Write down an integral that represents the area of the shaded region.

   • The area under the rate curve from 0 to 45 is \(\int_0^{45}\frac{t^2}{40}\,dt\)

2. (b) Find the area of the shaded region.

   • From the calculator: \(6075/8 = 759\) (that is \(45^3\div120\))

3. (c) Interpret the meaning of the shaded area in the context of this question.

   • The total volume of water drained from the cellar in the first 45 minutes: 759 m³.

4. (d) Find the number of minutes it takes to drain 400 m³ of water.

   • \(\int_0^m\frac{t^2}{40}\,dt=400\), so \(\frac{m^3}{120}=400\)
   • Solve for m: \(m=36.3\) minutes

Find the area enclosed by the curve \(y=2x^2\), the y-axis and the line \(y=8\), for \(x\ge0\).

• The curve meets \(y=8\) where \(2x^2=8\): \(x=2\)
• Area = rectangle minus area under the curve: \(A=\int_0^{2}(8-2x^2)\,dx\)
• From the calculator: \(A=32/3 = 10.7\)