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Some Basic Properties of Integrals

April 5th, 2009 |

Author: Merrin

There are a few properties of integrals that we should discuss before we go calculate them for specific functions. Most of these rules are based on the properties of summation mainly because integrals are just sums. Because summation is linear, integration is too.

$\displaystyle{\sum _{k=1}^{N}[f(n) +g(n)]=\sum _{k=1}^{N}f(n) +\sum _{k=1}^{N}g(n) } \\ {\displaystyle \int _{a}^{b}[f(x)+g(x)]dx =\displaystyle \int _{a}^{b}f(x)dx+ \displaystyle \int _{a}^{b}g(x)dx }$

Constants multiplying functions can be brought out of the integral since the same is true for summations. For example, if

$\displaystyle \int_0^1 x dx = 1/2$

then

$\displaystyle \int_0^1 3x dx = 3/2$ .

$\displaystyle{\sum _{k=1}^{N}cf(k)=c \sum _{k=1}^{N}f(k) } \\ {\displaystyle \int _{a}^{b}cf(x)dx=c\displaystyle \int _{a}^{b}f(x)dx }$

For summation, the order in which terms are added is irrelevant. Therefore, splitting an integral into separate intervals or rejoining intervals is possible.

$\displaystyle {\sum _{k=1}^{N}f(k) =\sum _{k=1}^{n}f(k)+\sum _{k=n+1}^{N}f(k) } \\ {\displaystyle \int _{a}^{c}f(x)dx=\displaystyle \int _{a}^{b}f(x)dx+ \displaystyle \int _{b}^{c}f(x)dx }$

The integral with the same limits of integration is zero because it is an area of zero width.

$\displaystyle \int _{a}^{a}f(x)dx=0$

Reversing the direction of the integration requires multiplication by a minus sign. From the definition of the Riemann integral we have

$\displaystyle \int _{a}^{b}f(x)dx=\mathop{\lim }\limits_{N\to \infty } \sum _{k=0}^{N-1}f(a+k\Delta x)\Delta x\qquad \Delta x=\cfrac{b-a}{N}$

If “a” is then larger than b then Delta x is negative and we see where the minus sign comes from
$\displaystyle \int _{a}^{b}f(x)dx=-\displaystyle \int _{b}^{a}f(x)dx$
Another way in which minus signs can enter is if the function is negative. In that case, the area between the function and the x axis is negative. We can speak of signed area which can be either positive or negative. Integrating the absolute value of f(x) from left to right is positive area, and integrating the absolute value of f(x) from right to left is negative area.

If g(x) is an even function, g(x)=g(-x) then
$\displaystyle \int _{-a}^{a}g(x)dx=2\displaystyle \int _{0}^{a}g(x)dx$
This is because the total area is the sum of two pieces which are a reflection of one another about the y axis.

If h(x) is an odd function then h(-x)=-h(x) then
$\displaystyle \int _{-a}^{a}h(x)dx=0$
This is because the area is the sum of two pieces which are reflections about the origin. One of the sides therefore will be a contribution of the negative area of the other. These two areas cancel each other out.

Another basic property of integrals is that if a function f(x)
is always greater than another function g(x), f(x)>= g(x), or is said to dominate, then the integral of f(x) will be greater than the integral of g(x).
$\displaystyle \int _{a}^{b}f(x)dx \ge \displaystyle \int _{a}^{b}g(x)dx$ This can be understood geometrically as the area under the curve will be bigger
if the function encloses a larger area. This is a strict inequality so there
are no points where f(x) dips below g(x).

We have listed some of the basic properties of integration, linearity, division of the integral into parts, even-odd symmetry, signed area, inversion of integration limits, and dominating functions. Some of these properties can be used to solve a problem in one fell swoop without actually needing to know how to do the integral. And these integration properties are essential in algebraic manipulations.

Example 1. Evaluate the following integral
$I = \displaystyle \int_{-1}^{1}x\text{sin}\,\,(x^2)e^{x^2} dx$
Solution 1. The function is odd so I=0 because the integration interval is symmetric about zero.

Example 2. Which integral is bigger
$I_1=\displaystyle \int_{x=1}^{10}\left|\text{sin}\,\, (e^{x^2})\right| dx \qquad I_2 =\displaystyle \int_{x=1}^{10}e^x dx$

Solution 2. since the integrand of the first integral is always less than the integrand of the second integral, the second integral is bigger without having to actually compute the integrals.

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