1. ## Definite Integrals

explain whe the function f(x) = x2 - 4/x-2 is not continuous on [0,3] what kind of discontinuity occurs?
use areas to show that integral 0,3 x2-4/x-2 dx = 10.5
use areas to show that integral 0,5 int(x) dx = 10.

how do i do these i really need some help on this my math teacher dosen't really teach.

explain whe the function f(x) = x2 - 4/x-2 is not continuous on [0,3] what kind of discontinuity occurs?
use areas to show that integral 0,3 x2-4/x-2 dx = 10.5
use areas to show that integral 0,5 int(x) dx = 10.

how do i do these i really need some help on this my math teacher dosen't really teach.
Have you graphed this? On the surface it looks like there might be a vertical asymptote at x = 2, but there is not:
$\displaystyle f(x) = \frac{x^2 - 4}{x - 2} = x + 2;~x \neq 2$

This has the aspects of a "hole" in the function.

(By the way, please learn to use parentheses in your expressions. The function "x^2 - 4/x - 2" is really:
$\displaystyle f(x) = x^2 - \frac{4}{x} - 2$
and I know that this isn't what you meant.)

-Dan

3. how do you do that like put in the equations

sorry see what you mean about the equation how would i do this and show areas im pretty sure the type of discontinuity is a removable discontinuity but how do i use areas to shw that the last 2 equal that.

$\displaystyle \int_{0}^{3} \frac{x^2-4}{x-2}, dx$= 10.5

$\displaystyle \int_{0}^{5} int(x), dx = 10$

4. ## La Tex

how do you do that like put in the equations

Here is a link Helpisplaying a formula - Wikipedia, the free encyclopedia

good luck.

$\displaystyle \int_{0}^{3} \frac{x^2-4}{x-2}, dx$= 10.5
$\displaystyle \int_{0}^{5} int(x), dx = 10$