Lesson Overview

Theorem (Mean Value Theorem for Integrals)

If f is a continuous function on the closed, bounded interval [ , ], then there is at least one number  in ( ,  ) for which

      .  

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Example 1: Average Value of a Function

Let f be a function defined on the interval [ ,  ].  Consider a regular partition of [ ,  ] into  subintervals with endpoints

    =  <  < .. <  <  = ,  

that is, for each , 1, .., ,  define  and   

The average of  function values, , , ..., , is

     .  

Observe that in the final two steps, the definition of  has been used to write  and the constant  has been factored out of the sum.

The sum in this last expression is a Riemann sum.  If the number of data points is allowed to increase without bound a definite integral can be identified:

    = .  

This is the motivation for the definition of the average value of a function:

The FunctionAverage command in Maple's Student[Calculus1] package

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Example 2: Hypotheses of MVT Satisfied

The function

>	f2 := piecewise( x<0, 0, x<1, x, exp(1-x) ): f(x) = f2;

is continuous for all real numbers.  In particular, the MVT for Integrals applies on the interval [ -5, 5 ].

>	plot( f2, x=-5..5, axes=framed );

The average value of f on [ -5, 5 ] is

>	f2_avg := FunctionAverage( f2, x=-5..5, output=integral ): f[avg] = f2_avg; `` = value( f2_avg ); `` = evalf( f2_avg );

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When the average value is included in the plot of the function, the shaded area below the average value should be equal to the area under the function.

>	FunctionAverage( f2, x=-5..5, output=plot,                  averageoptions=[color=cyan,filled=true],                  view=[DEFAULT,0..1] );

The graph suggests that there are two points in the interval ( -5, 5 ) where the function value is equal to the average value of the function on [ -5, 5 ].  These two points are

>	s2 := solve( f2=value(f2_avg), {x} ): s2;

or, numerically,

>	evalf( s2 );

These values are consistent with the graph.

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Example 3: Hypotheses of MVT Not Satisfied

The piecewise constant function

>	f3 := piecewise( x<0, -1, x<2, 2, 1 ): f(x) = f3;

has jump discontinuities at  and .  Because f is not continuous at  or , the MVT for Integrals does not apply on the interval [ -3, 3 ].

>	plot( f3, x=-3..3, discont=true );

To understand why the conclusion of the MVT for Integrals is not satisfied, observe that the average value of f on [ -3, 3 ] is

>	f3_avg := FunctionAverage( f3, x=-3..3, output=integral ): f3_avg = value( f3_avg );

Since the function values are -1, 1, and 2, it is clear that there are no points on ( -3, 3 ) where  = .

>	FunctionAverage( f3, x=-3..3, output=plot,                  averageoptions=[color=cyan,filled=true],                  view=[DEFAULT,-1..2] );

>	solve( f3=value(f3_avg), {x} );

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Note that the average value of the function on [ -1, 3 ] is

>	f3_avg2 := FunctionAverage( f3, x=-1..3, output=integral ): f3_avg2 = value( f3_avg2 );

Now, since the average value of the function on [ -1, 3 ] is one of the values of this function, there are points where the function value coincides with the average value on [ -1, 3 ].  Graphically,

>	FunctionAverage( f3, x=-1..3, output=plot,                  averageoptions=[color=cyan,filled=true],                  view=[DEFAULT,-1..2] );

And, algebraically,

>	solve( f3=value(f3_avg2), {x} );

While the MVT for Integrals does not apply on either [ -3, 3 ] or  [ -1, 3 ], it is not a contradiction to find a case where the conclusion holds even when the hypotheses are false.  (In terms of boolean logic:  False implies True is a True statement.)

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Example 4: Human Respiration

The essential life function of breathing is cyclic.  At rest a fit human will complete approximately 10 full respiratory cycles every minute.  The maximum rate of air flow into the lungs is about one-half liter per second.  Based on these assumptions the rate of air flow into the lungs can be modeled by the function

>	num_resp_cycle_per_minute := 10; max_rate_of_air_flow := 1/2;

>	rate := max_rate_of_air_flow          * sin( 2*Pi*x * (num_resp_cycle_per_minute/60) ): f(t) = rate;

The volume (in liters) of air in the lungs at time  (in seconds) is

>	VOL := Int( rate, x=0..t ): vol := value( VOL ): V(t) = VOL; `` = vol;

The relationship between the rate of air flow and the volume of air in the lungs is clearly shown in the following plot.

>	P1 := plot( [vol, eval(rate,x=t)], t=0..30,             title="Volume of Air in Lungs",             legend=["volume (L)","rate of air flow (L/s)"] ): P1;

Observe that the volume is periodic, with period 6 seconds, and that the maximum amount of air in the lungs is  L, or approximately 0.9549 L.  The maximum occurs every time ; the minimum value, 0 L, is attained every time .  (During times of exercise and other stresses respiration increases; the maximum lung capacity is about 3.7 L.)

The average amount of air in the lungs during one complete respiration cycle is

>	avg_vol := FunctionAverage( vol, t=0..6, output=integral ): V[avg] = avg_vol; `` = value( avg_vol ); `` = evalf( avg_vol );

The average amount of air in the lungs during each respiratory cycle (while at rest) is a little under  L.

>	P2 := plot( value(avg_vol), x=0..30,             color=blue, legend="average value of volume (L)" ): P3 := plot( value(avg_vol), x=0..30,             color=cyan, legend="", filled=true ): display( P1,P3 );

Note that the average value is not the maximum rate of air flow.  Numerically the two quantities are close, but the values (and units) are different.

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