Calculus BC · Unit 8
Divide the definite integral by the width. The integral alone is not the average.
Average value
\(f_{\text{avg}}=\dfrac{1}{b-a}\int_a^b f(x)\,dx\)
Divide by the width. Do not stop at the integral.
The integral is the average times the width
\(\int_a^b f(x)\,dx = f_{\text{avg}}\,(b-a)\)
A continuous function hits its average
If \(f\) is continuous on \([a,b]\), some \(c\) in \([a,b]\) has \(f(c)=f_{\text{avg}}\).
Average velocity is displacement over time
\(\dfrac{s(b)-s(a)}{b-a}\) equals the average value of velocity.
The antiderivative is short, and the width is obvious.
\(\dfrac{1}{3-0}\int_0^3 x^2\,dx\)
\(\dfrac{1}{3}\left[\dfrac{x^3}{3}\right]_0^3\)
\(\dfrac{1}{3}\cdot 9\)
\(3\)
Answer: \(3\)
\(\dfrac{1}{2}\int_1^3 6x\,dx\)
\(\dfrac{1}{2}\left[3x^2\right]_1^3\)
\(\dfrac{1}{2}(27-3)\)
\(12\)
Answer: \(12\)
A constant function’s average is the constant.
\(4\)
Answer: \(4\)
\(\dfrac{1}{4}\int_0^4 (x+1)\,dx\)
\(\dfrac{1}{4}\left[\dfrac{x^2}{2}+x\right]_0^4\)
\(\dfrac{1}{4}(8+4)\)
\(3\)
Answer: \(3\)
\(\int_1^5 f = 6\cdot(5-1)\)
\(24\)
Answer: \(24\)
Divide the integral by the width 4.
\((f(4)-f(0))/4\) is the average rate of change, not the average value, unless \(f\) is a position function and you wanted average velocity.
Answer: A. \(\dfrac{1}{4}\int_0^4 f(x)\,dx\)
A trig integral, a point where the function equals its average, or a table.
\(\dfrac{1}{\pi}\int_0^{\pi}\sin x\,dx\)
\(\dfrac{1}{\pi}\left[-\cos x\right]_0^{\pi}\)
\(\dfrac{1}{\pi}\left(1-(-1)\right)\)
\(\dfrac{2}{\pi}\)
Answer: \(\dfrac{2}{\pi}\)
\(c^2=3\)
\(c=\sqrt{3}\)
\(\sqrt{3}\) is between 0 and 3.
Answer: \(c=\sqrt{3}\)
\(\dfrac{s(4)-s(0)}{4-0}\)
\(\dfrac{16-0}{4}\)
\(4\)
The units are feet per second.
Answer: 4 ft/s
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| f(x) | 1 | 5 | 3 | 2 |
\(\dfrac{2}{2}(1+2\cdot 5+2\cdot 3+2)\)
\(1+10+6+2\)
\(19\)
Divide by the width 6.
\(\dfrac{19}{6}\)
Answer: \(\dfrac{19}{6}\)
\(\dfrac{1}{2}\int_0^2 e^x\,dx\)
\(\dfrac{1}{2}(e^2-1)\)
Answer: \(\dfrac{e^2-1}{2}\)
Average of \(x\): \(\int_0^1 x\,dx=\dfrac{1}{2}\).
Average of \(x^2\): \(\int_0^1 x^2\,dx=\dfrac{1}{3}\).
The width is 1, so those integrals are the averages.
\(\dfrac{1}{2}>\dfrac{1}{3}\)
Answer: \(x\) has the larger average
A root, a signed function, or an average that is not a sample value from a table.
\(\dfrac{1}{4}\int_0^4 x^{1/2}\,dx\)
\(\dfrac{1}{4}\left[\dfrac{2}{3}x^{3/2}\right]_0^4\)
\(4^{3/2}=8\)
\(\dfrac{1}{4}\cdot \dfrac{2}{3}\cdot 8\)
\(\dfrac{4}{3}\)
Answer: \(\dfrac{4}{3}\)
\(\dfrac{1}{4}\int_0^4 (x-2)\,dx\)
\(\dfrac{1}{4}\left[\dfrac{x^2}{2}-2x\right]_0^4\)
\(\dfrac{1}{4}(8-8)\)
\(0\)
The areas above and below \(y=0\) cancel. The average height is 0 even though the function is not always 0.
Answer: \(0\)
The average-value theorem for integrals applies to a continuous function.
Some output on the interval equals the average value.
Answer: yes, some \(c\) has \(f(c)=-3\)
\(\int_0^4 r(t)\,dt = 7\cdot 4\)
\(28\)
Answer: 28 gallons
Split at \(x=1\).
\(\int_0^1 (1-x)\,dx=\dfrac{1}{2}\)
\(\int_1^3 (x-1)\,dx=2\)
The integral is \(\dfrac{5}{2}\).
\(\dfrac{1}{3}\cdot \dfrac{5}{2}\)
\(\dfrac{5}{6}\)
Answer: \(\dfrac{5}{6}\)
\(\int_0^2 f=5\cdot 2=10\)
\(\int_2^8 f=1\cdot 6=6\)
\(\int_0^8 f=16\)
\(\dfrac{16}{8}=2\)
A plain mean of 5 and 1 would ignore the different widths.
Answer: \(2\)
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