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## Section5.4Properties of Integration

###### Motivating Questions
• What are some properties of definite integrals, and how can we use these to evaluate given integrals?

• How is the average value of a function on an interval related to the definite integral of that function on the interval?

• How can we find the area between two curves?

### SubsectionSome properties of the definite integral

Regarding the definite integral of a function $f$ over an interval $[a,b]$ as the net signed area bounded by $f$ and the $x$-axis, we discover several standard properties of the definite integral.

For any real number $a$ and the definite integral $\displaystyle \int\limits_a^a f(x) \, dx\text{,}$ it is evident that no area is enclosed because the interval begins and ends with the same point. Hence, we have our first property of definite integrals:

If $f$ is a continuous function and $a$ is a real number, then $\displaystyle \int\limits_a^a f(x) \,dx = 0\text{.}$

Next, we consider the result of subdividing the interval of integration. In Figure5.59, we see that

\begin{gather*} \int\limits_a^b f(x) \, dx = A_1, \ \int\limits_b^c f(x) \, dx = A_2,\\ \text{and }\int\limits_a^c f(x) \, dx = A_1 + A_2\text{,} \end{gather*}

which illustrates the following general rule. Figure5.59The area bounded by $y=f(x)$ on the interval $[a,c]\text{.}$

If $f$ is a continuous function and $a\text{,}$ $b\text{,}$ and $c$ are real numbers, then

\begin{equation*} \int\limits_a^c f(x) \,dx = \int\limits_a^b f(x) \,dx + \int\limits_b^c f(x) \,dx\text{.} \end{equation*}

While this rule is easy to see if $a \lt b \lt c\text{,}$ it in fact holds in general for any values of $a\text{,}$ $b\text{,}$ and $c\text{.}$

###### Example5.60

Consider the continuous piecewise function $f(x)\text{,}$ defined below

\begin{equation*} f(x) = \begin{cases} x^2+3 \amp \text{if } x\le 1\\ 4x \amp \text{if } x\gt 1 \end{cases}. \end{equation*}

Evaluate

\begin{equation*} \int\limits_0^2 f(x)dx. \end{equation*}
\begin{equation*} \int\limits_0^2 f(x)dx=\frac{28}{3}. \end{equation*}
Solution

For a continuous piecewise function, we break into two integrals, splitting at $c=1\text{,}$ the value for which the piecewise function changes:

\begin{equation*} \int\limits_0^2 f(x)dx=\int\limits_0^1 f(x)dx+\int\limits_1^2 f(x) dx\text{.} \end{equation*}

Then plug in the appropriate pieces of the function

\begin{equation*} \int\limits_0^2 f(x)dx=\int\limits_0^1( x^2+3)dx+\int\limits_1^2 4x dx\text{.} \end{equation*}

Then evaluate each integral.

\begin{equation*} \int\limits_0^1( x^2+3)dx+\int\limits_1^2 4x dx =\left. \left( \frac{x^3}{3}+3x\right)\right|_0^1+\left.\left(\frac{4x^2}{2}\right)\right|_1^2 \end{equation*}
\begin{equation*} =\left(\frac{1^3}{3}+3(1)\right)-\left(\frac{0^3}{3}+3(0)\right)+\left(2(2^2)-2(1)^2\right)=\frac{28}{3}\text{.} \end{equation*}

Another property of the definite integral states that if we reverse the order of the limits of integration, we change the sign of the integral's value.

If $f$ is a continuous function and $a$ and $b$ are real numbers, then

\begin{equation*} \int\limits_b^a f(x) \,dx = -\int\limits_a^b f(x) \,dx\text{.} \end{equation*}
###### Example5.61

Show that

\begin{equation*} \int\limits_0^2 2x dx=-\int\limits_2^0 2x dx. \end{equation*}
Solution

To show this property holds we will compute each integral separately:

\begin{equation*} \int\limits_0^2 2x dx=\left. x^2\right|_0^2=2^2-0^2=4 \end{equation*}
\begin{equation*} \int\limits_2^0 2x dx=\left. x^2\right|_2^0=0^2-2^2=-4\text{.} \end{equation*}

Thus

\begin{equation*} \int\limits_0^2 2x dx=4=-(-4)=-\int\limits_2^0 2x dx. \end{equation*}

### SubsectionHow the definite integral is connected to a function's average value

One of the most valuable applications of the definite integral is that it provides a way to discuss the average value of a function, even for a function that takes on infinitely many values. Recall that if we wish to take the average of $n$ numbers $y_1\text{,}$ $y_2\text{,}$ $\ldots\text{,}$ $y_n\text{,}$ we compute

\begin{equation*} AVG = \frac{y_1 + y_2 + \cdots + y_n}{n}\text{.} \end{equation*}

Since integrals arise from Riemann sums in which we add $n$ values of a function, it should not be surprising that evaluating an integral is similar to averaging the output values of a function.

###### Average value of a function

If $f$ is a continuous function on $[a,b]\text{,}$ then its average value on $[a,b]$ is given by the formula

\begin{equation*} f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \cdot \int\limits_a^b f(x) \, dx\text{.} \end{equation*}
###### Example5.62

Find the average value of the function $f(x)=x^3$ on the interval $[1,3]\text{.}$

\begin{equation*} f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \cdot \int\limits_a^b f(x) \, dx\text{.} \end{equation*}

where $a=1$ and $b=3\text{,}$ then evaluate the integral:

\begin{equation*} f_{\operatorname{AVG} [1,3]} = \frac{1}{3-1} \cdot \int\limits_1^3 x^3 \, dx = \frac{1}{2}\left.\left(\frac{x^4}{4}\right)\right|_1^3=\frac{1}{2}\left(\frac{3^4}{4}-\frac{1^4}{4}\right)=10. \end{equation*}
###### Example5.63

Compute the average value of each of the following functions over the given intervals

1. $\displaystyle f(x)=e^x \text{ on the interval } [0,10]$

2. $\displaystyle g(x)=6x^2+3\sqrt{x} \text{ on the interval } [1,4]$

3. $\displaystyle h(x)=\frac{1}{x} \text{ on the interval } [1,e]$

Solution
1. \begin{equation*} f_{\operatorname{AVG} [0,10]} = \frac{1}{10-0} \cdot \int\limits_0^{10} e^x \, dx = \left.\frac{1}{10}\left(e^x\right)\right|_0^{10} \end{equation*}
\begin{equation*} =\frac{1}{10}\left(e^{10}-e^0\right)=\frac{e^{10}-1}{10} \end{equation*}
2. \begin{equation*} g_{\operatorname{AVG} [1,4]} = \frac{1}{4-1} \cdot \int\limits_1^{4} 6x^2+3\sqrt{x} \, dx = \left.\frac{1}{3}\left(6\frac{x^3}{3}+3\frac{x^{3/2}}{3/2}\right)\right|_1^{4} \end{equation*}
\begin{equation*} =\frac{1}{3}\left(2(4)^3+2(4)^{3/2}-(2(1)^3+2(1)^{3/2})\right)=\frac{140}{3} \end{equation*}
3. \begin{equation*} h_{\operatorname{AVG} [1,e]} = \frac{1}{e-1} \cdot \int\limits_1^{e} \frac{1}{x} \, dx = \left.\frac{1}{e-1}\left(\ln(x)\right)\right|_1^{e} \end{equation*}
\begin{equation*} =\frac{1}{e-1}\left(\ln(e)-\ln(1)\right)=\frac{1}{e-1} \end{equation*}
###### Example5.64

A company determines that the total cost, in dollars, to produce $x$ units of a product can be given by the function

\begin{equation*} C(x)=0.4x^2+140x+110. \end{equation*}

Find average cost over for producing the first 200 items.

The average cost is $\19443.30\text{.}$

Solution

Here we want to compute the average value of $C(x)$ over the interval $[0,200]\text{,}$ that is

\begin{align*} C_{\operatorname{AVG} [0,200]} =\mathstrut \amp \frac{1}{200-0} \cdot \int\limits_0^{200} (0.4x^2+140x+110) \, dx \\ =\mathstrut \amp \left.\frac{1}{200}\left(0.4\frac{x^3}{3}+140\frac{x^2}{2}+110x\right)\right|_0^{200} \\ =\mathstrut \amp \frac{1}{200}\left(\frac{0.4}{3}(200)^3+70(200)^2+110(200)\right) \\ =\mathstrut \amp 19443.30 \end{align*}

### SubsectionIntegration and the Area between Two Curves

Integration allows us to find the area between a curve and the $x$-axis. In fact, by visually inspecting the graphs of two different curves we can develop a technique for finding the area between two curves. Consider the graphs of the curves $y=x^2+1$ and $y=x+3$ displayed in Figure5.65. We might ask if there is a way to find the area enclosed by these two curves as is colored in the figure. Figure5.65Graphs of the curves $y=x^2+1$ and $y=x+3\text{.}$

In fact, using the knowledge we have about integration, we can figure out this quantity exactly.

###### Example5.66

What is the exact area enclosed by the curves given by the functions $f(x)=x^2+1$ and $g(x)=x+3\text{?}$

The first step in solving this problem is to determine where the curves intersect. We can find the point of intersection by setting the two curves equal to each other.

\begin{align*} x^2+1 \amp=x+3\\ x^2-x-2 \amp=0\\ (x+1)(x-2) \amp=0 \end{align*}

Therefore, the curves intersect at two $x$-values, $x=-1,2\text{.}$ This means were are concerned with the interval $[-1,2]\text{.}$

We next note that on the interval in question, $[-1,2]\text{,}$ the curve $g(x)=x+3$ is above the curve $f(x)=x^2+1\text{.}$ We also note that $\displaystyle \int\limits_{-1}^{2}(x+3)dx$ would give the area shown in Figure5.67 and $\displaystyle \int\limits_{-1}^{2}(x^2+1)dx$ Figure5.68. Figure5.67Graph of the curve $y=x+3\text{.}$ Figure5.68Graph of the curve $y=x^2+1\text{.}$

By overlaying Figure5.67 and Figure5.68 as in Figure5.69 you can see that Figure5.65 should be given by subtracting $\int_{-1}^{2}(x^2+1)dx$ from $\int_{-1}^{2}(x+3)dx\text{.}$ Figure5.69Graphs of the curves $y=x^2+1$ and $y=x+3\text{.}$

Therefore, the area between the curves $y=x^2+1$ and $y=x+3$ is given by the following quantity:

\begin{align*} \int\limits_{-1}^{2}(x+3)dx-\int\limits_{-1}^{2}(x^2+1)dx \amp=\int\limits_{-1}^{2}\left((x+3)-(x^2+1)\right)dx\\ \mathstrut \amp=\int\limits_{-1}^{2}(-x^2+x+2)dx\\ \mathstrut \amp=\left. -\frac{1}{3}x^3+\frac{1}{2}x^2+2x \right|_{-1}^{2}\\ \mathstrut \amp= -\frac{1}{3}(2)^3+\frac{1}{2}(2)^2+2(2)-\left(-\frac{1}{3}(-1)^3+\frac{1}{2}(-1)^2+2(-1)\right)\\ \mathstrut \amp =-\frac{8}{3}+2+4-\left(\frac{1}{3}+\frac{1}{2}-2\right)\\ \mathstrut \amp= \frac{9}{2} \end{align*}

In general we can find the area enclosed by two curves by integrating the top curve minus the bottom curve.

###### Area between two curves

If $f(x)\ge g(x)$ on the interval $[a,b]\text{,}$ then the area between $f(x)$ and $g(x)$ is

\begin{equation*} \int\limits_a^b \left(\text{top curve-bottom curve}\right) dx =\int\limits_a^b (f(x)-g(x))dx \end{equation*}
###### Example5.70

Find the area of the region enclosed between $f(x)=0.1x^2+4\text{,}$ shown as the red curve in the image below, and $g(x)=x\text{,}$ shown as the blue curve in the figure below, between $x=2\text{,}$ and $x=2\text{.}$ Figure5.71Graphs of the curves $f(x)=0.1x^2+4$ and $g(x)=x\text{.}$

The area between $f(x)$ and $g(x)$ is $16.5333\text{.}$

Solution

Here we are given the interval of integration, $[-2,2]\text{,}$ so we integrate:

\begin{equation*} \int\limits_a^b \left(\text{top curve-bottom curve}\right) dx =\int\limits_{-2}^{2} \left(0.1x^2+4-(x)\right)dx \end{equation*}
\begin{equation*} =\int\limits_{-2}^{2} \left(0.1x^2-x+4\right)dx=\left.\left(0.1\frac{x^3}{3}-\frac{x^2}{2}+4x\right)\right|_{-2}^2 \end{equation*}
\begin{equation*} =\left(0.1\frac{2^3}{3}-\frac{2^2}{2}+4(2)\right)-\left(0.1\frac{(-2)^3}{3}-\frac{(-2)^2}{2}+4(-2)\right)=16.5333 \end{equation*}
###### Example5.72

Find the area of the region enclosed by the two functions $y=4x^2$ and $y=x^2+3\text{,}$ shown below. Figure5.73Graphs of the curves $y=4x^2$ and $y=x^2+3\text{.}$

The area between the curves $y=4x^2$ and $y=x^2+3$ is $4\text{.}$

Solution

To start with need to find when these two curves intersect. We do this by setting the two curves equal to each other and solving.

\begin{equation*} 4x^2=x^2+3 \implies 3x^2=3 \implies x^2=1 \end{equation*}

Thus, these two curves intersect at $x=\pm1\text{.}$ This gives us the bounds of integration. Now we need to determine which curve is the top curve. To do so, plug in a value between the bounds, for example, $x=0\text{.}$ From this we see that $y=x^2+3$ is the top curve. So the area between the curves can be determined by the integral:

\begin{equation*} \int\limits_{-1}^1 \left(x^2+3-4x^2\right)dx=\int\limits_{-1}^1 \left(3-3x^2\right)dx \end{equation*}
\begin{equation*} =\left(3x-\frac{3x^3}{3}\right) \bigg|_{-1}^1=\left(3(1)-(1)^3\right)-\left(3(-1)-(-1)^3\right)=4 \end{equation*}

### SubsectionSummary

• We have explored several properties of definite integrals, such as the integral over the interval $[a,a] \text{,}$ subdividing the interval of integration, and reversing the order of the limits of integration. These properties help us to evaluate integrals of functions that may otherwise be difficult to evaluate, such as piecewise functions.

• One of the most valuable applications of the definite integral is that it provides a way to discuss the average value of a function. Indeed, if $f$ is a continuous function on $[a,b] \text{,}$ then its average value on $[a,b]$ is given by the formula $f_{\operatorname{AVG} [a,b]} = \frac{1}{b-a} \int_a^b f(x) \,dx \text{.}$

• Integration also allows us to find the area between two curves. For example, if $f(x)\ge g(x)$ on the interval $[a,b]\text{,}$ then the area between $f(x)$ and $g(x)$ is $\int\limits_a^b \left(\text{top curve-bottom curve}\right) dx =\int\limits_a^b (f(x)-g(x))dx.$

### SubsectionExercises

Consider the graphs of two functions $f$ and $g$ that are provided in Figure5.74. Each piece of $f$ and $g$ is either part of a straight line or part of a circle. Figure5.74Two functions $f$ and $g\text{.}$
1. Determine the exact value of $\int_0^1 [f(x) + g(x)]\,dx\text{.}$

2. Determine the exact value of $\int_1^4 [2f(x) - 3g(x)] \, dx\text{.}$

3. Find the exact average value of $h(x) = g(x) - f(x)$ on $[0,4]\text{.}$

4. For what constant $c$ does the following equation hold?

\begin{equation*} \int_0^4 c \, dx = \int_0^4 [f(x) + g(x)] \, dx \end{equation*}

Let $f(x) = 3 - x^2$ and $g(x) = 2x^2\text{.}$

1. On the interval $[-1,1]\text{,}$ sketch a labeled graph of $y = f(x)$ and write a definite integral whose value is the exact area bounded by $y = f(x)$ on $[-1,1]\text{.}$

2. On the interval $[-1,1]\text{,}$ sketch a labeled graph of $y = g(x)$ and write a definite integral whose value is the exact area bounded by $y = g(x)$ on $[-1,1]\text{.}$

3. Write an expression involving a difference of definite integrals whose value is the exact area that lies between $y = f(x)$ and $y = g(x)$ on $[-1,1]\text{.}$

4. Explain why your expression in (c) has the same value as the single integral $\int_{-1}^1 [f(x) - g(x)] \, dx\text{.}$

5. Explain why, in general, if $p(x) \ge q(x)$ for all $x$ in $[a,b]\text{,}$ the exact area between $y = p(x)$ and $y = q(x)$ is given by

\begin{equation*} \int_a^b [p(x) - q(x)] \, dx\text{.} \end{equation*}