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Properties Of The Definite Integral
Calculus I
Calculus II
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By Rick Stoll          

 Calc I

 

[Under Construction]

 

 

 

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.

 

 

 

 

 

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].

 

 

 

 

 

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  .

 

 

 

 

 

For more on the Area Problem, click HERE.

 

 

 

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Last modified: April 30, 2001