Average value Average value
The
average value of a continuous function
f f f on
[ a , b ] [a, b] [ a , b ] is
f avg = 1 b − a ∫ a b f ( x ) d x . f_{\text{avg}} = \frac{1}{b - a}\int_{a}^{b} f(x)\,dx. f avg = b − a 1 ∫ a b f ( x ) d x . The Fundamental Theorem page drew this as a rectangle on
[ a , b ] [a, b] [ a , b ] with the same area as the region under
f f f . The Mean Value Theorem for Integrals says
f f f reaches that height at least once on the interval.
Example 1
Let
f ( x ) = x x 2 + 9 f(x) = \frac{x}{\sqrt{x^2 + 9}} f ( x ) = x 2 + 9 x . (a) Find the average value of
f f f on
[ 0 , 4 ] [0, 4] [ 0 , 4 ] . (b) Find the average rate of change of
f f f on
[ 0 , 4 ] [0, 4] [ 0 , 4 ] .
(a) With
u = x 2 + 9 u = x^2 + 9 u = x 2 + 9 , an antiderivative of
f f f is
x 2 + 9 \sqrt{x^2 + 9} x 2 + 9 , so
∫ 0 4 x x 2 + 9 d x = x 2 + 9 ∣ 0 4 = 5 − 3 = 2. \begin{aligned}
\int_{0}^{4} \frac{x}{\sqrt{x^2 + 9}}\,dx &= \sqrt{x^2 + 9}\,\Big|_{0}^{4} \\[4pt]
&= 5 - 3 = 2.
\end{aligned} ∫ 0 4 x 2 + 9 x d x = x 2 + 9 0 4 = 5 − 3 = 2. f avg = 1 4 − 0 ∫ 0 4 f ( x ) d x = 2 4 = 1 2 . f_{\text{avg}} = \frac{1}{4 - 0}\int_{0}^{4} f(x)\,dx = \frac{2}{4} = \frac{1}{2}. f avg = 4 − 0 1 ∫ 0 4 f ( x ) d x = 4 2 = 2 1 .
The rectangle of height 1/2 on [0, 4] has the same area as the region under f. (b) The average rate of change uses only the endpoints:
f ( 4 ) − f ( 0 ) 4 − 0 = 4 5 − 0 4 = 1 5 . \frac{f(4) - f(0)}{4 - 0} = \frac{\frac{4}{5} - 0}{4} = \frac{1}{5}. 4 − 0 f ( 4 ) − f ( 0 ) = 4 5 4 − 0 = 5 1 . The two answers differ because they describe different functions. The average value of
f f f is the average rate of change of its antiderivative
g ( x ) = x 2 + 9 g(x) = \sqrt{x^2 + 9} g ( x ) = x 2 + 9 , since
1 4 ∫ 0 4 g ′ ( x ) d x = g ( 4 ) − g ( 0 ) 4 = 1 2 . \frac{1}{4}\int_{0}^{4} g'(x)\,dx = \frac{g(4) - g(0)}{4} = \frac{1}{2}. 4 1 ∫ 0 4 g ′ ( x ) d x = 4 g ( 4 ) − g ( 0 ) = 2 1 . When the function is a table of measurements, a Riemann or trapezoidal sum estimates the integral. Dividing that by the length of the interval estimates the average.
Example 2
A rooftop solar panel's power output
P ( t ) P(t) P ( t ) , in kilowatts, is measured at times
t t t hours after 8:00 AM.
t t t P ( t ) P(t) P ( t ) 0 0 0 0.6 0.6 0.6 1 1 1 1.4 1.4 1.4 3 3 3 2.9 2.9 2.9 4 4 4 3.1 3.1 3.1 6 6 6 2.2 2.2 2.2 8 8 8 0.8 0.8 0.8
Use a trapezoidal sum with the five subintervals in the table to estimate the average output from 8:00 AM to 4:00 PM.
Each trapezoid is the average of its two readings times its width:
∫ 0 8 P ( t ) d t ≈ 0.6 + 1.4 2 ( 1 ) + 1.4 + 2.9 2 ( 2 ) + 2.9 + 3.1 2 ( 1 ) + 3.1 + 2.2 2 ( 2 ) + 2.2 + 0.8 2 ( 2 ) = 1.0 + 4.3 + 3.0 + 5.3 + 3.0 = 16.6. \begin{aligned}
\int_{0}^{8} P(t)\,dt &\approx \tfrac{0.6 + 1.4}{2}(1) \\
&\quad + \tfrac{1.4 + 2.9}{2}(2) \\
&\quad + \tfrac{2.9 + 3.1}{2}(1) \\
&\quad + \tfrac{3.1 + 2.2}{2}(2) \\
&\quad + \tfrac{2.2 + 0.8}{2}(2) \\[4pt]
&= 1.0 + 4.3 + 3.0 + 5.3 \\
&\quad + 3.0 \\
&= 16.6.
\end{aligned} ∫ 0 8 P ( t ) d t ≈ 2 0.6 + 1.4 ( 1 ) + 2 1.4 + 2.9 ( 2 ) + 2 2.9 + 3.1 ( 1 ) + 2 3.1 + 2.2 ( 2 ) + 2 2.2 + 0.8 ( 2 ) = 1.0 + 4.3 + 3.0 + 5.3 + 3.0 = 16.6. The integral of a power in kilowatts over hours is energy, so the panel produces about 16.6 kilowatt-hours from 8:00 AM to 4:00 PM. The average output is
1 8 − 0 ∫ 0 8 P ( t ) d t ≈ 16.6 8 = 2.075 kilowatts . \begin{aligned}
\frac{1}{8 - 0}\int_{0}^{8} P(t)\,dt &\approx \frac{16.6}{8} \\
&= 2.075 \text{ kilowatts}.
\end{aligned} 8 − 0 1 ∫ 0 8 P ( t ) d t ≈ 8 16.6 = 2.075 kilowatts . Displacement and total distance For a particle on a line, velocity is the rate of change of position. The integral of a rate is the net change of the quantity, so the integral of velocity is the net change in position.
Speed
∣ v ( t ) ∣ |v(t)| ∣ v ( t ) ∣ is never negative, so its integral counts every leg of the trip as positive. To evaluate it, split the interval where
v v v changes sign and integrate
− v -v − v on the pieces where
v < 0 v < 0 v < 0 .
Example 3
A particle's velocity is
v ( t ) = 3 t − t v(t) = 3\sqrt{t} - t v ( t ) = 3 t − t meters per second for
0 ≤ t ≤ 16 0 \le t \le 16 0 ≤ t ≤ 16 seconds, and its position at
t = 0 t = 0 t = 0 is
x = 5 x = 5 x = 5 . (a) Find the displacement and the total distance traveled over
0 ≤ t ≤ 16 0 \le t \le 16 0 ≤ t ≤ 16 . (b) Find the particle's position at
t = 16 t = 16 t = 16 . (c) Find when, on
0 ≤ t ≤ 16 0 \le t \le 16 0 ≤ t ≤ 16 , the particle is farthest from where it started.
(a) Factor
v ( t ) = t ( 3 − t ) v(t) = \sqrt{t}\left(3 - \sqrt{t}\right) v ( t ) = t ( 3 − t ) . It's positive on
( 0 , 9 ) (0, 9) ( 0 , 9 ) and negative on
( 9 , 16 ) (9, 16) ( 9 , 16 ) . An antiderivative is
V ( t ) = 2 t 3 / 2 − t 2 2 V(t) = 2t^{3/2} - \frac{t^2}{2} V ( t ) = 2 t 3/2 − 2 t 2 , with
V ( 0 ) = 0 V(0) = 0 V ( 0 ) = 0 ,
V ( 9 ) = 13.5 V(9) = 13.5 V ( 9 ) = 13.5 , and
V ( 16 ) = 0 V(16) = 0 V ( 16 ) = 0 .
∫ 0 16 v ( t ) d t = V ( 16 ) − V ( 0 ) = 0 meters . \begin{aligned}
\int_{0}^{16} v(t)\,dt &= V(16) - V(0) \\
&= 0 \text{ meters}.
\end{aligned} ∫ 0 16 v ( t ) d t = V ( 16 ) − V ( 0 ) = 0 meters . ∫ 0 16 ∣ v ( t ) ∣ d t = ∫ 0 9 v ( t ) d t − ∫ 9 16 v ( t ) d t = 13.5 − ( 0 − 13.5 ) = 27 meters . \begin{aligned}
\int_{0}^{16} |v(t)|\,dt &= \int_{0}^{9} v(t)\,dt \\
&\qquad - \int_{9}^{16} v(t)\,dt \\[4pt]
&= 13.5 - (0 - 13.5) \\
&= 27 \text{ meters}.
\end{aligned} ∫ 0 16 ∣ v ( t ) ∣ d t = ∫ 0 9 v ( t ) d t − ∫ 9 16 v ( t ) d t = 13.5 − ( 0 − 13.5 ) = 27 meters .
The region above the axis and the region below it both have area 13.5. They cancel in the displacement and add in the distance. (b) x ( 16 ) = 5 + 0 = 5 x(16) = 5 + 0 = 5 x ( 16 ) = 5 + 0 = 5 meters. The particle is back where it started.
(c) The distance from the start is
∣ x ( t ) − 5 ∣ = ∣ V ( t ) ∣ |x(t) - 5| = |V(t)| ∣ x ( t ) − 5∣ = ∣ V ( t ) ∣ . On
[ 0 , 9 ] [0, 9] [ 0 , 9 ] the particle travels right, and on
[ 9 , 16 ] [9, 16] [ 9 , 16 ] it travels left, so
V V V decreases from 13.5 to
V ( 16 ) = 0 V(16) = 0 V ( 16 ) = 0 . So
V V V is largest at
t = 9 t = 9 t = 9 and never negative. The particle is farthest from its start at
t = 9 t = 9 t = 9 , 13.5 meters to the right, at
x = 18.5 x = 18.5 x = 18.5 .
Example 4
The graph shows the velocity
v ( t ) v(t) v ( t ) , in meters per second, of a particle for
0 ≤ t ≤ 9 0 \le t \le 9 0 ≤ t ≤ 9 seconds. It's made of three line segments. The particle is at
x = − 3 x = -3 x = − 3 when
t = 0 t = 0 t = 0 .
The velocity is 4 at t = 0, −2 at t = 3, 0 at t = 7, and 3 at t = 9. (a) Find the position at
t = 9 t = 9 t = 9 and the total distance traveled over
0 ≤ t ≤ 9 0 \le t \le 9 0 ≤ t ≤ 9 . (b) Find the times when the particle is farthest left and farthest right on that interval, and its positions then. (c) Find every time in
0 < t ≤ 9 0 < t \le 9 0 < t ≤ 9 when the particle is back at
x = − 3 x = -3 x = − 3 . (d) Find the average speed over
0 ≤ t ≤ 9 0 \le t \le 9 0 ≤ t ≤ 9 .
(a) The graph crosses the axis at
t = 2 t = 2 t = 2 and
t = 7 t = 7 t = 7 . The three triangles between the graph and the axis have areas
4 4 4 ,
5 5 5 , and
3 3 3 .
Above the axis on (0, 2) and (7, 9), below it on (2, 7). x ( 9 ) = − 3 + ∫ 0 9 v ( t ) d t = − 3 + ( 4 − 5 + 3 ) = − 1 meter . \begin{aligned}
x(9) &= -3 + \int_{0}^{9} v(t)\,dt \\
&= -3 + (4 - 5 + 3) \\
&= -1 \text{ meter}.
\end{aligned} x ( 9 ) = − 3 + ∫ 0 9 v ( t ) d t = − 3 + ( 4 − 5 + 3 ) = − 1 meter . The total distance counts every area as positive:
∫ 0 9 ∣ v ( t ) ∣ d t = ∫ 0 2 v d t − ∫ 2 7 v d t + ∫ 7 9 v d t = 4 + 5 + 3 = 12 meters . \begin{aligned}
\int_{0}^{9} |v(t)|\,dt &= \int_{0}^{2} v\,dt \\
&\quad - \int_{2}^{7} v\,dt + \int_{7}^{9} v\,dt \\
&= 4 + 5 + 3 \\
&= 12 \text{ meters}.
\end{aligned} ∫ 0 9 ∣ v ( t ) ∣ d t = ∫ 0 2 v d t − ∫ 2 7 v d t + ∫ 7 9 v d t = 4 + 5 + 3 = 12 meters . (b) The particle changes direction only at
t = 2 t = 2 t = 2 and
t = 7 t = 7 t = 7 , so its extreme positions are among those times and the ends:
x ( 0 ) = − 3 , x ( 2 ) = − 3 + 4 = 1 , x ( 7 ) = 1 − 5 = − 4 , x ( 9 ) = − 4 + 3 = − 1. \begin{aligned}
x(0) &= -3, \\
x(2) &= -3 + 4 = 1, \\
x(7) &= 1 - 5 = -4, \\
x(9) &= -4 + 3 = -1.
\end{aligned} x ( 0 ) x ( 2 ) x ( 7 ) x ( 9 ) = − 3 , = − 3 + 4 = 1 , = 1 − 5 = − 4 , = − 4 + 3 = − 1. The particle is farthest right at
t = 2 t = 2 t = 2 , at
x = 1 x = 1 x = 1 meter, and farthest left at
t = 7 t = 7 t = 7 , at
x = − 4 x = -4 x = − 4 meters.
(c) On
[ 0 , 2 ] [0, 2] [ 0 , 2 ] the particle travels right from
− 3 -3 − 3 to
1 1 1 , and on
[ 2 , 3 ] [2, 3] [ 2 , 3 ] it stays to the right of
0 0 0 , reaching
x = 0 x = 0 x = 0 at
t = 3 t = 3 t = 3 . So it isn't back at
− 3 -3 − 3 before
t = 3 t = 3 t = 3 . On
[ 3 , 7 ] [3, 7] [ 3 , 7 ] ,
v ( t ) = − 2 + 1 2 ( t − 3 ) v(t) = -2 + \frac{1}{2}(t - 3) v ( t ) = − 2 + 2 1 ( t − 3 ) , and with
u = t − 3 u = t - 3 u = t − 3 ,
x ( t ) = 0 + ∫ 3 t v ( s ) d s = − 2 u + u 2 4 . x(t) = 0 + \int_{3}^{t} v(s)\,ds = -2u + \frac{u^2}{4}. x ( t ) = 0 + ∫ 3 t v ( s ) d s = − 2 u + 4 u 2 . Setting this equal to
− 3 -3 − 3 gives
u 2 − 8 u + 12 = 0 u^2 - 8u + 12 = 0 u 2 − 8 u + 12 = 0 , so
u = 2 u = 2 u = 2 or
u = 6 u = 6 u = 6 . Only
u = 2 u = 2 u = 2 lies in
[ 0 , 4 ] [0, 4] [ 0 , 4 ] , which gives
t = 5 t = 5 t = 5 . On
[ 7 , 9 ] [7, 9] [ 7 , 9 ] ,
v ( t ) = 3 2 ( t − 7 ) v(t) = \frac{3}{2}(t - 7) v ( t ) = 2 3 ( t − 7 ) , so
x ( t ) = − 4 + ∫ 7 t v ( s ) d s = − 4 + 3 4 ( t − 7 ) 2 . \begin{aligned}
x(t) &= -4 + \int_{7}^{t} v(s)\,ds \\
&= -4 + \frac{3}{4}(t - 7)^2.
\end{aligned} x ( t ) = − 4 + ∫ 7 t v ( s ) d s = − 4 + 4 3 ( t − 7 ) 2 . Setting this equal to
− 3 -3 − 3 gives
( t − 7 ) 2 = 4 3 (t - 7)^2 = \frac{4}{3} ( t − 7 ) 2 = 3 4 , so
t = 7 + 2 3 ≈ 8.155 t = 7 + \frac{2}{\sqrt{3}} \approx 8.155 t = 7 + 3 2 ≈ 8.155 . The particle is back at
x = − 3 x = -3 x = − 3 at
t = 5 t = 5 t = 5 and at
t ≈ 8.155 t \approx 8.155 t ≈ 8.155 seconds.
(d) The average speed is the average value of
∣ v ∣ |v| ∣ v ∣ :
1 9 − 0 ∫ 0 9 ∣ v ( t ) ∣ d t = 12 9 = 4 3 . \frac{1}{9 - 0}\int_{0}^{9} |v(t)|\,dt = \frac{12}{9} = \frac{4}{3}. 9 − 0 1 ∫ 0 9 ∣ v ( t ) ∣ d t = 9 12 = 3 4 . That's
4 3 \frac{4}{3} 3 4 meters per second.
Velocity from acceleration Acceleration is the rate of change of velocity, so the same reasoning gives
v ( t ) = v ( t 0 ) + ∫ t 0 t a ( s ) d s . v(t) = v(t_0) + \int_{t_0}^{t} a(s)\,ds. v ( t ) = v ( t 0 ) + ∫ t 0 t a ( s ) d s . The integral of
a ( t ) a(t) a ( t ) over an interval is the net change in velocity, in units of velocity.
Example 5
A particle travels along a line with acceleration
a ( t ) = − 12 ( t + 2 ) 2 a(t) = -\frac{12}{(t + 2)^2} a ( t ) = − ( t + 2 ) 2 12 meters per second per second for
0 ≤ t ≤ 6 0 \le t \le 6 0 ≤ t ≤ 6 , and its velocity at
t = 0 t = 0 t = 0 is 4 meters per second. (a) Find the integral of
a ( t ) a(t) a ( t ) from
t = 0 t = 0 t = 0 to
t = 1 t = 1 t = 1 , and explain its meaning in context, with units. (b) Find
v ( t ) v(t) v ( t ) , and find when the particle changes direction. (c) Find the total distance traveled over
0 ≤ t ≤ 6 0 \le t \le 6 0 ≤ t ≤ 6 .
(a) An antiderivative of
a ( t ) a(t) a ( t ) is
12 t + 2 \frac{12}{t + 2} t + 2 12 , so
∫ 0 1 a ( t ) d t = 12 t + 2 ∣ 0 1 = 4 − 6 = − 2. \int_{0}^{1} a(t)\,dt = \frac{12}{t + 2}\,\Big|_{0}^{1} = 4 - 6 = -2. ∫ 0 1 a ( t ) d t = t + 2 12 0 1 = 4 − 6 = − 2. The particle's velocity changes by
− 2 -2 − 2 meters per second from
t = 0 t = 0 t = 0 to
t = 1 t = 1 t = 1 . It decreases from 4 to 2 meters per second.
(b) For any
t t t in
[ 0 , 6 ] [0, 6] [ 0 , 6 ] ,
v ( t ) = 4 + ∫ 0 t a ( s ) d s = 4 + 12 t + 2 − 6 = 12 t + 2 − 2. \begin{aligned}
v(t) &= 4 + \int_{0}^{t} a(s)\,ds \\
&= 4 + \frac{12}{t + 2} - 6 \\
&= \frac{12}{t + 2} - 2.
\end{aligned} v ( t ) = 4 + ∫ 0 t a ( s ) d s = 4 + t + 2 12 − 6 = t + 2 12 − 2. Since
v ′ ( t ) = a ( t ) < 0 v'(t) = a(t) < 0 v ′ ( t ) = a ( t ) < 0 ,
v v v is decreasing, so it's positive for
t < 4 t < 4 t < 4 , zero at
t = 4 t = 4 t = 4 , and negative for
t > 4 t > 4 t > 4 . The velocity changes sign at
t = 4 t = 4 t = 4 , so the particle changes direction there.
(c) An antiderivative of
v v v is
V ( t ) = 12 ln ( t + 2 ) − 2 t V(t) = 12\ln(t + 2) - 2t V ( t ) = 12 ln ( t + 2 ) − 2 t . Split at
t = 4 t = 4 t = 4 :
∫ 0 4 v ( t ) d t = ( 12 ln 6 − 8 ) − 12 ln 2 = 12 ln 3 − 8 , ∫ 4 6 v ( t ) d t = ( 12 ln 8 − 12 ) − ( 12 ln 6 − 8 ) = 12 ln 4 3 − 4. \begin{aligned}
\int_{0}^{4} v(t)\,dt &= (12\ln 6 - 8) - 12\ln 2 \\
&= 12\ln 3 - 8, \\[6pt]
\int_{4}^{6} v(t)\,dt &= (12\ln 8 - 12) \\
&\qquad - (12\ln 6 - 8) \\
&= 12\ln\tfrac{4}{3} - 4.
\end{aligned} ∫ 0 4 v ( t ) d t ∫ 4 6 v ( t ) d t = ( 12 ln 6 − 8 ) − 12 ln 2 = 12 ln 3 − 8 , = ( 12 ln 8 − 12 ) − ( 12 ln 6 − 8 ) = 12 ln 3 4 − 4. The second integral is about
− 0.548 -0.548 − 0.548 , so its absolute value is
4 − 12 ln 4 3 4 - 12\ln\frac{4}{3} 4 − 12 ln 3 4 . The total distance, in meters, is
∫ 0 6 ∣ v ( t ) ∣ d t = ∫ 0 4 v d t − ∫ 4 6 v d t = ( 12 ln 3 − 8 ) + ( 4 − 12 ln 4 3 ) = 12 ln 9 4 − 4 ≈ 5.731 \begin{aligned}
\int_{0}^{6} |v(t)|\,dt &= \int_{0}^{4} v\,dt - \int_{4}^{6} v\,dt \\
&= (12\ln 3 - 8) \\
&\qquad + \left(4 - 12\ln\tfrac{4}{3}\right) \\
&= 12\ln\tfrac{9}{4} - 4 \approx 5.731
\end{aligned} ∫ 0 6 ∣ v ( t ) ∣ d t = ∫ 0 4 v d t − ∫ 4 6 v d t = ( 12 ln 3 − 8 ) + ( 4 − 12 ln 3 4 ) = 12 ln 4 9 − 4 ≈ 5.731
The velocity: about 5.183 meters traveled forward on (0, 4) and about 0.548 meters back on (4, 6).