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89 + 90 + 56 + 78 + 100 + 69 6 = 482 6 80.33 .

Therefore, your average test grade is approximately 80.33, which translates to a B− at most schools.

Suppose, however, that we have a function v ( t ) that gives us the speed of an object at any time t , and we want to find the object’s average speed. The function v ( t ) takes on an infinite number of values, so we can’t use the process just described. Fortunately, we can use a definite integral to find the average value of a function such as this.

Let f ( x ) be continuous over the interval [ a , b ] and let [ a , b ] be divided into n subintervals of width Δ x = ( b a ) / n . Choose a representative x i * in each subinterval and calculate f ( x i * ) for i = 1 , 2 ,…, n . In other words, consider each f ( x i * ) as a sampling of the function over each subinterval. The average value of the function may then be approximated as

f ( x 1 * ) + f ( x 2 * ) + + f ( x n * ) n ,

which is basically the same expression used to calculate the average of discrete values.

But we know Δ x = b a n , so n = b a Δ x , and we get

f ( x 1 * ) + f ( x 2 * ) + + f ( x n * ) n = f ( x 1 * ) + f ( x 2 * ) + + f ( x n * ) ( b a ) Δ x .

Following through with the algebra, the numerator is a sum that is represented as i = 1 n f ( x i * ) , and we are dividing by a fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of the function is given by

i = 1 n f ( x i * ) ( b a ) Δ x = ( Δ x b a ) i = 1 n f ( x i * ) = ( 1 b a ) i = 1 n f ( x i * ) Δ x .

This is a Riemann sum. Then, to get the exact average value, take the limit as n goes to infinity. Thus, the average value of a function is given by

1 b a lim n i = 1 n f ( x i ) Δ x = 1 b a a b f ( x ) d x .

Definition

Let f ( x ) be continuous over the interval [ a , b ] . Then, the average value of the function f ( x ) (or f ave ) on [ a , b ] is given by

f ave = 1 b a a b f ( x ) d x .

Finding the average value of a linear function

Find the average value of f ( x ) = x + 1 over the interval [ 0 , 5 ] .

First, graph the function on the stated interval, as shown in [link] .

A graph in quadrant one showing the shaded area under the function f(x) = x + 1 over [0,5].
The graph shows the area under the function f ( x ) = x + 1 over [ 0 , 5 ] .

The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid A = 1 2 h ( a + b ) , where h represents height, and a and b represent the two parallel sides. Then,

0 5 x + 1 d x = 1 2 h ( a + b ) = 1 2 · 5 · ( 1 + 6 ) = 35 2 .

Thus the average value of the function is

1 5 0 0 5 x + 1 d x = 1 5 · 35 2 = 7 2 .
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Find the average value of f ( x ) = 6 2 x over the interval [ 0 , 3 ] .

3

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Key concepts

  • The definite integral can be used to calculate net signed area, which is the area above the x -axis less the area below the x -axis. Net signed area can be positive, negative, or zero.
  • The component parts of the definite integral are the integrand, the variable of integration, and the limits of integration.
  • Continuous functions on a closed interval are integrable. Functions that are not continuous may still be integrable, depending on the nature of the discontinuities.
  • The properties of definite integrals can be used to evaluate integrals.
  • The area under the curve of many functions can be calculated using geometric formulas.
  • The average value of a function can be calculated using definite integrals.

Key equations

  • Definite Integral
    a b f ( x ) d x = lim n i = 1 n f ( x i * ) Δ x
  • Properties of the Definite Integral
    a a f ( x ) d x = 0
    b a f ( x ) d x = a b f ( x ) d x
    a b [ f ( x ) + g ( x ) ] d x = a b f ( x ) d x + a b g ( x ) d x
    a b [ f ( x ) g ( x ) ] d x = a b f ( x ) d x a b g ( x ) d x
    a b c f ( x ) d x = c a b f ( x ) for constant c
    a b f ( x ) d x = a c f ( x ) d x + c b f ( x ) d x
Practice Key Terms 8

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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