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

Find the area enclosed between the x-axis and the graph of \(y=-x^2 + 9x - 14\).

• \(y=-(x-2)(x-7)\), so the curve meets the x-axis at \(x=2\) and \(x=7\)
• \(y>0\) between the roots: \(A=\int_{2}^{7}(-x^2 + 9x - 14)\,dx\)
• From the calculator: \(A=125/6 = 20.8\)

Find the area enclosed by the part of the graph of \(y=25-x^2\) for \(x>0\), the x-axis and the y-axis.

• The curve meets the x-axis where \(25-x^2=0\): \(x=5\) (take the positive root)
• \(A=\int_0^{5}(25-x^2)\,dx\)
• From the calculator: \(A=250/3 = 83.3\)

The diagram shows the graphs of \(y=10x-x^2\) and \(y=4x\). They meet at the origin and at the point P.

1. (a) Find the coordinates of P.

   • \(10x-x^2=4x\), so \(6x-x^2=0\) and \(x(6-x)=0\)
   • The non-zero solution is \(x=6\); \(y=4\times6=24\)

2. (b) Find the shaded area.

   • Between the graphs the curve is the upper one: \(A=\int_0^{6}((10x-x^2)-4x)\,dx\)
   • From the calculator: \(A=36\)

The function f satisfies \(f'(x)=2x^2 + 3x\) and \(f(0)=4\). Find \(\int_0^{2} f(x)\,dx\).

• \(f(x)=\frac{2x^3}{3} + \frac{3x^2}{2}+c\)
• \(f(0)=4\) gives \(c=4\), so \(f(x)=\frac{2x^3}{3} + \frac{3x^2}{2} + 4\)
• From the calculator: \(\int_0^{2}(\frac{2x^3}{3} + \frac{3x^2}{2} + 4)\,dx=44/3 = 14.7\)

Given that \(\int f(x)\,dx=5\left(x^3 + \frac{2}{x}+c\right)\), find an expression for \(f(x)\).

• Differentiating an antiderivative gives the function: \(f(x)=\frac{d}{dx}\left[5\left(x^3 + \frac{2}{x}+c\right)\right]\)
• \(f(x)=5\left(3x^2 - \frac{2}{x^2}\right)=15x^2 - \frac{10}{x^2}\)

The diagram shows the graphs of \(y=\frac{7}{x}\) and \(y=6-x\). They meet at the points P and Q.

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

   • Solve \(\frac{7}{x}=6-x\) on the calculator (or \(x^2-6x+7=0\))
   • \(x=1.586\) and \(x=4.414\)

2. (b) Find the shaded area.

   • The line is above the curve between P and Q: \(A=\int_{1.586}^{4.414}\left((6-x)-\frac{7}{x}\right)dx\)
   • From the calculator: \(A=1.32\)

A pump fills a tank at a rate of \(3+kt^2\) litres per minute, where t is the time in minutes after it is switched on. After 5 minutes the rate is 8 litres per minute. How much water is pumped in the first 5 minutes?

• At \(t=5\): \(3+k\times5^2=8\), so \(25k=5\) and \(k=0.2\)
• Amount \(=\int_0^{5}(3+0.2t^2)\,dt\)
• From the calculator: 23.3 litres

A fountain is switched on and, for the first 20 seconds, the rate at which water leaves it is directly proportional to the time it has been on. After 8 seconds the rate is 10 litres per second. How much water leaves the fountain in the first 20 seconds?

• Rate proportional to time: \(r=ct\). At \(t=8\), \(r=10\), so \(c=1.25\)
• Total \(=\int_0^{20}1.25t\,dt\)
• From the calculator: 250 litres

A model suggests that ice melts into a dish at a rate of \(\frac{120}{t^3}\) millilitres per minute, where t is the time in minutes after the ice was put out. When \(t=2\) there are 15 millilitres of water in the dish.

1. (a) Find the amount of water in the dish 10 minutes after the ice was put out.

   • Amount \(=15+\int_{2}^{10}\frac{120}{t^3}\,dt\)
   • From the calculator: \(15+14.4\), so 29.4 millilitres

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

   • Use a large upper limit: \(15+\int_{2}^{1000}\frac{120}{t^3}\,dt=30.000\)
   • The total settles at about 30.0 millilitres

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

   • The rate is not defined at \(t=0\) and is very large just after it, so the model fails at the start. (Or: the ice would all melt in a finite time, but the model goes on for ever.)

Grain is loaded into a silo at a rate of \(r=\frac{t^2}{60}\), where r is measured in tonnes per hour and t is the time in hours after loading starts. The graph is shown, shaded from \(t=0\) to \(t=36\).

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

   • The area under the rate curve from 0 to 36 is \(\int_0^{36}\frac{t^2}{60}\,dt\)

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

   • From the calculator: \(1296/5 = 259\) (that is \(36^3\div180\))

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

   • The total mass of grain loaded into the silo in the first 36 hours: 259 tonnes.

4. (d) Find the number of hours it takes to load 100 tonnes of grain.

   • \(\int_0^m\frac{t^2}{60}\,dt=100\), so \(\frac{m^3}{180}=100\)
   • Solve for m: \(m=26.2\) hours

Find the area enclosed by the curve \(y=3x^2\), the y-axis and the line \(y=27\), for \(x\ge0\).

• The curve meets \(y=27\) where \(3x^2=27\): \(x=3\)
• Area = rectangle minus area under the curve: \(A=\int_0^{3}(27-3x^2)\,dx\)
• From the calculator: \(A=54\)

A one-to-one function \(g(x)\) satisfies \(g(0)=0\) and \(g(x)\ge0\). Given that \(\int_0^{b}g(x)\,dx=P\) for \(b>0\), find an expression for \(\int_0^{g(b)}g^{-1}(x)\,dx\).

• The area under \(y=g(x)\) from 0 to b is P
• The rectangle with corners at the origin and \((b,g(b))\) has area \(b\,g(b)\)
• The inverse function swaps the axes: the area under \(y=g^{-1}(x)\) is the rest of the rectangle
• \(\int_0^{g(b)}g^{-1}(x)\,dx=b\,g(b)-P\)