4.7 Properties of Definite Integrals

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In this section we use properties of definite integrals to compute and interpret them.

Properties of Definite Integrals

We begin with some basic properties of definite integrals- many of which are familiar from our study of derivatives and their basic properties.

If $f(x)$ and $g(x)$ are continuous on the interval $[a,b]$ then

(a): $\displaystyle {\int _a^b \Big [f(x) \pm g(x) \Big ] \ dx = \int _a^b f(x) \ dx \pm \int _a^b g(x) \ dx}$
(b): $\displaystyle {\int _a^b cf(x) \ dx = c\int _a^b f(x) \ dx}$
(c): $\displaystyle {\int _a^a f(x) \ dx = 0}$
(d): $\displaystyle {\int _b^a f(x) \ dx = -\int _a^b f(x) \ dx}$

and if $f(x) \leq g(x)$ on $[a,b]$ , then

(e): $\displaystyle {\int _a^b f(x) \ dx \leq \int _a^b g(x) \ dx.}$

Recall that if $f(x) \geq 0$ on an interval $[a, b]$ then the definite integral, $\int _a^b f(x) \ dx$ , gives the area under the curve and if $f(x) \leq 0$ on an interval $[a, b]$ then the definite integral, $\int _a^b f(x) \ dx$ , gives -1 times the area above the curve. We now consider the situation where the integrand $f(x)$ changes sign on the interval $[a,b]$ . The key to handling this situation is to use the following property.

Additivity If $f(x)$ is continuous on the interval $[a,b]$ and $c$ is a number between $a$ and $b$ , i.e., $a<c<b$ then $\int _a^b f(x) \ dx = \int _a^c f(x) \ dx + \int _c^b f(x) \ dx.$

We apply this proposition in the following examples.

Use the figure below to determine the value of the definite integral $\displaystyle {\int _a^b f(x) \ dx}$ .

By additivity, we can write $\int _a^b f(x) \ dx = \int _a^c f(x) \ dx + \int _c^b f(x) \ dx.$ Next, we can determine the definite integrals on the right hand side of the above equation by relating them to area. On the interval $[a,c], f(x) \geq 0$ and so $\int _a^c f(x) \ dx = \text {area under the curve} = 4.$ On the interval $[c,b], f(x) \leq 0$ and so $\int _c^b f(x) \ dx = (-1)\cdot \text {area above the curve} = -6.$ Now, by the additivity of the definite integral, $\int _a^b f(x) \ dx = 4 + (-6) = -2.$

Use the figure below to determine the definite integrals.

$\int _a^c f(x) = \answer {5}.$ $\int _c^b f(x) = \answer {-3}.$ $\int _a^b f(x) = \answer {2}.$

Use the figure below to determine the definite integrals.

$\int _a^c f(x) = \answer {6}.$ $\int _c^d f(x) = \answer {-5}.$ $\int _d^b f(x) = \answer {4}.$ $\int _a^d f(x) = \answer {1}.$ $\int _c^b f(x) = \answer {-1}.$ $\int _a^b f(x) = \answer {5}.$

Use geometry and additivity of the definite integral to compute $\displaystyle {\int _0^5 (3-x) \ dx}$ . The graph of $y = 3-x$ is a line with slope $-1$ passing through the points $(0,3)$ and $(5,-2)$ . To compute the integral, we need to determine where $3-x$ is positive and where it is negative. We accomplish this by first finding the $x$ -intercept(s) by solving $3-x = 0$ . The solution is clearly, $x = 3$ . Now, if $x < 3$ , then $3 - x >0$ and if $x >3$ , then $3-x <0$ . We now use the additivity property: $\int _0^5 (3-x) \ dx = \int _0^3 (3-x) \ dx + \int _3^5 (3-x) \ dx.$ We can now take an area approach to evaluate the two integrals on the right-hand side. $\int _0^3 (3-x) \ dx = \text {area of triangle under the curve} = \tfrac 12bh = \tfrac 12(3)(3) = \tfrac 92,$ and $\int _3^5 (3-x) \ dx = (-1)\cdot \text {area of triangle above the curve} = -\tfrac 12(2)(2) = -2$ Thus $\int _0^5 (3-x) \ dx = \tfrac 92 + (-2) = \tfrac 52.$

Use geometry and additivity of the definite integral to compute the definite integral. $\int _0^3 (x-2) \ dx= \answer {-3/2}.$

$y=x-2$ is a line with slope $1$ passing thru $(0,-2), (2,0)$ and $(3,1)$

If $f(x) \geq 0$ then the integral gives the area under the curve

If $f(x) \leq 0$ then the integral gives $(-1)$ times the area above the curve

Since the definite integral is related to area, there are situations where computing these integrals can be facilitated by observing a pattern of symmetry in a region. To aid us in our discussion, we will have to recall the idea of even and odd functions.

Even and Odd Functions $f(x)$ is called even if $f(-x) = f(x)$ and
$f(x)$ is called odd if $f(-x) = -f(x)$ .

The terms even and odd come from the observation that functions of the form $f(x) = x^n$ are even functions if $n$ is even and they are odd functions if $n$ is odd. Furthermore, $\cos (x)$ is an even function, while $\sin (x)$ is an odd function. The graph of an even function is symmetric about the $y$ -axis and the graph of an odd function is symmetric about the origin. The following proposition describes how these symmetries affect the definite integral.

Let $a>0$ and suppose $f(x)$ is continuous on the interval $[-a, a]$ . If $f(x)$ is an even function then $\int _{-a}^a f(x) \ dx = 2 \int _0^a f(x) \ dx,$ and if $f(x)$ is an odd function, then $\int _{-a}^a f(x) \ dx = 0.$

This result is especially important for odd functions, since the calculation of definite integrals in this case becomes trivial. For even functions the advantage of this result less significant. It allows us to work with the number $0$ instead of the negative number $-a$ , thereby simplifying any arithmetic associated with the Fundamental Theorem of Calculus.

Since $x^2$ is an even function, $\int _{-3}^3 x^2 \ dx = 2\int _0^3 x^2 \ dx = \tfrac {2}{3}x^3\Bigg |_0^3 = 18.$

Since $\cos (x)$ is an even function, $\int _{-\pi /2}^{\pi /2} \cos (x) \ dx = 2\int _0^{\pi /2} \cos (x) \ dx = 2\sin (x)\Bigg |_0^{\pi /2} = 2.$

Use the fact that the integrand is even to compute the definite integral. $\int _{-1}^1 \left (x^4 + 5x^2 + 3 \right ) \ dx =\answer {146/15}.$

Note that in the last two examples, plugging in the value zero gave zero. This is not a coincidence. An even function always has an odd anti-derivative and odd functions must pass thru the origin.
Now we look at some examples of odd functions.

The following definite integrals are all equal to zero, since in each case, the integrand is an odd function and the interval of integration has the form $[-a, a]$ :

$\int _{-2}^2 \left (2x^5 - 4x^3 + 6x \right )\ dx, \int _{-1}^1 \left (\sqrt [3] x + \sqrt [5]x\right ) \ dx \; \text {and} \; \int _{-\pi }^\pi \Big [\sin (x) + \sin ^3(x)\Big ] \ dx.$

$\int _{-5}^5 \left (x^5 - 8x^3 +2x \right ) \ dx =\answer {0}.$

The integrand is odd

$\int _{-1}^1 \Big [x\cos (x) - x^2\sin (x) \Big ] \ dx =\answer {0}.$

An odd function times an even function is an odd function

$\int _{-\pi }^\pi \Big [\sin ^3(x) + \sin (x^3)\Big ] \ dx =\answer {0}.$

The composition of two odd functions is odd

In the next example, we combine an even function and an odd function.

The function $x^4$ is even and the function $x^5$ is odd. Therefore, $\int _{-2}^2 (x^4 + x^5) \ dx = \int _{-2}^2 x^4 \ dx + \int _{-2}^2 x^5 \ dx = 2\int _0^2 x^4 \ dx + 0 = \tfrac {2}{5}x^5\Bigg |_0^2 = \frac {64}{5}.$

$\int _{-3}^3 \Big [x^2 - \sqrt [3]x +2\sin (x) \Big ] \ dx =\answer {18}.$

$x^2$ is even and $\sqrt [3]x$ and $2\sin (x)$ are odd

$\int _{-1}^{1} x^4\Big [1 + \sin (x)\Big ] \ dx =\answer {2/5}.$

$x^4$ is even and $x^4\sin (x)$ is odd