AI SL · 13.5
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Unit mixed practice: areas, rates and accumulation
13.5.6
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Find the area enclosed between the x-axis and the graph of \(y=-x^2 + 5x - 4\).
Have a go first. Then open the working one line at a time.
\(-x^2+5x-4=0\), so \(x=1\) or \(x=4\)
These are the limits. Between them the curve is above the x-axis.
\(A=\int_1^4(-x^2+5x-4)\,dx\)
The curve is above the axis, so the integral is the area.
From the calculator: \(A=9/2 = 4.50\)
Area is positive. Give 3 significant figures.
The diagram shows the graphs of \(y=8x-x^2\) and \(y=3x\). They meet at the origin and at the point P. (a) Find the coordinates of P. (b) Find the shaded area.
Have a go first. Then open the working one line at a time.
\(8x-x^2=3x\), so \(5x-x^2=0\)
Where the graphs meet, the y-values are equal.
\(x=5\), \(y=3\times5=15\): P = (5, 15)
The other solution, x = 0, is the origin.
\(A=\int_0^5\big((8x-x^2)-3x\big)\,dx\)
The curve is above the line between 0 and 5.
From the calculator: \(A=125/6 = 20.8\)
Give 3 significant figures.
A dye pump warms up and delivers dye at a rate of \(4+kt^2\) millilitres per minute, where t is the time in minutes after it is switched on. After 3 minutes the rate is 13 millilitres per minute. How much dye is delivered in the first 3 minutes?
Have a go first. Then open the working one line at a time.
\(4+k\times3^2=13\), so \(9k=9\) and \(k=1\)
Use the rate you are given to find the constant.
\(V=\int_0^3(4+t^2)\,dt\)
Amount delivered = area under the rate from 0 to 3.
From the calculator: \(V=21.0\) ml
Units: rate in ml per minute, time in minutes, so ml.
For the first 40 seconds of a spray, the rate at which a nozzle emits liquid is directly proportional to the time it has been on. After 5 seconds the rate is 30 grams per second. How much liquid is emitted in the first 40 seconds?
Have a go first. Then open the working one line at a time.
\(r=ct\)
Directly proportional to t means r equals a constant times t.
\(30=c\times5\), so \(c=6\) and \(r=6t\)
Use the rate after 5 seconds.
\(\int_0^{40}6t\,dt\); from the calculator: 4800
The total is in grams.
A model suggests that sand falls into the base of a timer at a rate of \(\frac{60}{t^3}\) grams per second, where t is the time in seconds after the timer was started. When \(t=1\) there are 8 grams of sand in the base. (a) Find the amount of sand in the base when \(t=4\). (b) Use technology to estimate the amount that will eventually fall into the base. (c) Give one criticism of the model.
Have a go first. Then open the working one line at a time.
Amount \(=8+\int_1^4\frac{60}{t^3}\,dt\)
The integral gives the sand that falls from t = 1 to t = 4. Add what was there at t = 1.
From the calculator: \(8+28.125=36.1\) g
3 significant figures.
\(8+\int_1^{1000}\frac{60}{t^3}\,dt=38.000\); try \(10^6\) as well
The value settles, so about 38.0 g will eventually fall into the base.
The rate is very large just after \(t=0\) and is not defined at \(t=0\).
The model only makes sense for \(t\ge1\). Also, a real timer runs out of sand.
Oil is pumped into a storage tank at a rate of \(r=\frac{t^2}{20}\), where r is measured in cubic metres per minute and t is the time in minutes. The graph is shown, shaded from \(t=0\) to \(t=30\). (a) Write down an integral that represents the area of the shaded region. (b) Find the area of the shaded region. (c) Interpret the meaning of the shaded area. (d) Find the number of minutes it takes to pump 200 m³ of oil into the tank.
Have a go first. Then open the working one line at a time.
\(A=\int_0^{30}\frac{t^2}{20}\,dt\)
Area under the rate curve, from t = 0 to t = 30.
From the calculator: \(A=450\) m³
This is \(30^3\div60\).
The total volume of oil pumped into the tank in the first 30 minutes: 450 m³.
Area under a rate is the amount that has flowed.
\(\int_0^m\frac{t^2}{20}\,dt=200\), so \(\frac{m^3}{60}=200\). Solve for m: \(m=22.9\)
Solve for the upper limit: 22.9 minutes.
The function f satisfies \(f'(x)=4x-6x^2\) and \(f(0)=1\). Find \(\int_0^2 f(x)\,dx\).
Have a go first. Then open the working one line at a time.
\(f(x)=2x^2-2x^3+c\)
Raise each power by one and divide by the new power.
\(f(0)=1\) gives \(c=1\), so \(f(x)=2x^2-2x^3+1\)
Put x = 0 in f.
From the calculator: \(\int_0^2(2x^2-2x^3+1)\,dx=-2/3 = -0.667\)
Now integrate f itself, using the new f.
The diagram shows the graphs of \(y=\frac{10}{x}\) and \(y=9-x\). They meet at the points P and Q. (a) Use your calculator to find the x-coordinates of P and Q. (b) Find the shaded area.
Have a go first. Then open the working one line at a time.
\(\frac{10}{x}=9-x\)
Where the graphs meet, the y-values are equal.
From the calculator: \(x=1.298\) and \(x=7.702\)
Keep the unrounded values in the calculator for the limits.
\(A=\int_{1.298}^{7.702}\left((9-x)-\frac{10}{x}\right)dx\)
The line is above the curve between P and Q.
From the calculator: \(A=11.0\)
Give 3 significant figures.