Properties of the Definite Integral


The following properties are easy to check:

Theorem. If f (x) and g(x) are defined and continuous on [a, b], except maybe at a finite number of points, then we have the following linearity principle for the integral:

(i)
$\displaystyle \int_{a}^{b}$$\displaystyle \Big($f (x) + g(x)$\displaystyle \Big)$dx = $\displaystyle \int_{a}^{b}$f (x) dx + $\displaystyle \int_{a}^{b}$g(x) dx;
(ii)
$\displaystyle \int_{a}^{b}$$\displaystyle \alpha$f (x) dx = $\displaystyle \alpha$$\displaystyle \int_{a}^{b}$f (x) dx, for any arbitrary number $ \alpha$.

The next results are very useful in many problems.

Theorem. If f (x) is defined and continuous on [a, b], except maybe at a finite number of points, then we have

(i)
$\displaystyle \int_{c}^{c}$f (x) dx = 0;
(ii)
$\displaystyle \int_{a}^{b}$f (x) dx = $\displaystyle \int_{a}^{c}$f (x) dx + $\displaystyle \int_{c}^{b}$f (x) dx;
(iii)
$\displaystyle \int_{b}^{a}$f (x) dx = - $\displaystyle \int_{a}^{b}$f (x) dx;
for any arbitrary numbers a and b, and any c $ \in$ [a, b].

The property (ii) can be easily illustrated by the following picture:

Remark. It is easy to see from the definition of lower and upper sums that if f (x) is positive then $ \int_{a}^{b}$f (x) dx $ \geq$ 0. This implies the following

If f (x) $ \leq$ g(x) for x $ \in$ [a, b]     $\displaystyle \Rightarrow$    $\displaystyle \int_{a}^{b}$f (x) dx $\displaystyle \leq$ $\displaystyle \int_{a}^{b}$g(x) dx  .

Example. We have

$\displaystyle \int_{0}^{1}$(x2 - 2x)dx = $\displaystyle \int_{0}^{1}$x2 dx - 2$\displaystyle \int_{0}^{1}$x dx  .

We have seen previously that

$\displaystyle \int_{0}^{1}$x2 dx = $\displaystyle {\textstyle\frac{1}{3}}$    and    $\displaystyle \int_{0}^{1}$x dx = $\displaystyle {\textstyle\frac{1}{2}}$   .

Hence

$\displaystyle \int_{0}^{1}$(x2 - 2x)dx = $\displaystyle {\textstyle\frac{1}{3}}$ - 2$\displaystyle {\textstyle\frac{1}{2}}$ = - $\displaystyle {\textstyle\frac{2}{3}}$   .

Exercise 1. Given that

$\displaystyle \int_{1}^{2}$f (x)dx = 2  ,    $\displaystyle \int_{1}^{4}$f (x)dx = - 1  ,

find $\displaystyle \int_{2}^{4}$f (x)dx.

Answer.

Exercise 2. Let f (x) be defined and continuous on [a, b]. Assume that f (x) is positive. Show that the function

F(x) = $\displaystyle \int_{a}^{x}$f (t)dt

is increasing on [a, b].

Answer.

Exercise 3. Let f (x) and g(x) be two functions defined and continuous on [a, b]. Show that

$\displaystyle \left(\vphantom{\int_a^b f(x)g(x)dx }\right.$$\displaystyle \int_{a}^{b}$f (x)g(x)dx$\displaystyle \left.\vphantom{\int_a^b f(x)g(x)dx }\right)^{2}_{}$ $\displaystyle \leq$ $\displaystyle \int_{a}^{b}$$\displaystyle \Big($f (x)$\displaystyle \Big)^{2}_{}$dx . $\displaystyle \int_{a}^{b}$$\displaystyle \Big($g(x)$\displaystyle \Big)^{2}_{}$dx  .

Answer.

For more on the Area Problem, click HERE.


[Back] [Next]
[Trigonometry] [Calculus]
[Geometry] [Algebra] [Differential Equations]
[Complex Variables] [Matrix Algebra]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.


Mohamed A. Khamsi
Helmut Knaust

Copyright © 1999-2024 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour