5.2 The definite integral  (Page 6/16)

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

Suppose, however, that we have a function $v\left(t\right)$ that gives us the speed of an object at any time t , and we want to find the object’s average speed. The function $v\left(t\right)$ 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 $f\left(x\right)$ be continuous over the interval $\left[a,b\right]$ and let $\left[a,b\right]$ be divided into n subintervals of width $\text{Δ}x=(b-a)\text{/}n.$ Choose a representative $x_i^{*}$ in each subinterval and calculate $f\left(x_i^{*}\right)$ for $i=1,2\text{,…,}n.$ In other words, consider each $f\left(x_i^{*}\right)$ as a sampling of the function over each subinterval. The average value of the function may then be approximated as

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

But we know $\text{Δ}x=\frac{b-a}{n},$ so $n=\frac{b-a}{\text{Δ}x},$ and we get

Following through with the algebra, the numerator is a sum that is represented as ${\displaystyle \sum _{i=1}^{n}f\left(x_i^{*}\right)},$ 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

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

Finding the average value of a linear function

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

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

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

$\begin{array}{cc}{\displaystyle {\int }_0^{5}x}+1dx\hfill & =\frac{1}{2}h\left(a+b\right)\hfill \\ & =\frac{1}{2}·5·\left(1+6\right)\hfill \\ & =\frac{35}{2}.\hfill \end{array}$

Thus the average value of the function is

$\frac{1}{5-0}{\displaystyle {\int }_0^{5}x}+1dx=\frac{1}{5}·\frac{35}{2}=\frac{7}{2}.$

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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
  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\underset{n\to \infty }{{\displaystyle \text{lim}}}{\displaystyle \sum _{i=1}^{n}f\left(x_i^{*}\right)\text{Δ}x}$
• Properties of the Definite Integral
  ${\displaystyle {\int }_a^{a}f\left(x\right)dx=0}$
  ${\displaystyle {\int }_b^{a}f\left(x\right)dx}=\text{−}{\displaystyle {\int }_a^{b}f\left(x\right)dx}$
  ${\displaystyle {\int }_a^b\left[f\left(x\right)+g\left(x\right)\right]dx}={\displaystyle {\int }_a^{b}f\left(x\right)dx}+{\displaystyle {\int }_a^{b}g\left(x\right)dx}$
  ${\displaystyle {\int }_a^b\left[f\left(x\right)-g\left(x\right)\right]dx={\displaystyle {\int }_a^{b}f\left(x\right)dx-{\displaystyle {\int }_a^{b}g\left(x\right)dx}}}$
  ${\displaystyle {\int }_a^{b}cf\left(x\right)dx}=c{\displaystyle {\int }_a^{b}f\left(x\right)}$ for constant c
  ${\displaystyle {\int }_a^{b}f\left(x\right)dx}={\displaystyle {\int }_a^{c}f\left(x\right)dx}+{\displaystyle {\int }_c^{b}f\left(x\right)dx}$