- #1
Sirsh
- 267
- 10
Hi all, just wondering if someone could explain to me about define integrals.
Say i have F(x)=0.5x2-2 and F(x)=x3+x2-6x and i want to find the area of the regions which is satisfied from -2 to 2.
so [tex]\int[/tex]0.5x2-2 from -2 to 2. [tex]\int[/tex]x3+x2-6x from -2 to 2.
Now with the cubic, from -2 to 0 is supposedly positive and from 0 to 2 it's negative, is that true i thought the it doesn't matter which region the area is in it'll always be counted as positive.
so, [tex]\int[/tex]x3+x2-6x from -2 to 0. = [tex]\int[/tex](-2)3+(-2)2-6(-2) - [tex]\int[/tex](0)3+(0)2-6(0) = 8-0 = 8units2
[tex]\int[/tex]x3+x2-6x from 0 to 2. = [tex]\int[/tex](0)3+(0)2-6(0) - [tex]\int[/tex](2)3+(2)2-6(2) = 0- 0 = 0 units2
Therefore, Area from (-2 to 0)+ Area from (0 to 2) = 8+0 = 8units2
But using my calculator doing the same steps (except in the graphing section) doing -2 to 0 and adding it to 0 to 2's value i get 16units2( 10.667+5.333)
I don't know what's going on..
Say i have F(x)=0.5x2-2 and F(x)=x3+x2-6x and i want to find the area of the regions which is satisfied from -2 to 2.
so [tex]\int[/tex]0.5x2-2 from -2 to 2. [tex]\int[/tex]x3+x2-6x from -2 to 2.
Now with the cubic, from -2 to 0 is supposedly positive and from 0 to 2 it's negative, is that true i thought the it doesn't matter which region the area is in it'll always be counted as positive.
so, [tex]\int[/tex]x3+x2-6x from -2 to 0. = [tex]\int[/tex](-2)3+(-2)2-6(-2) - [tex]\int[/tex](0)3+(0)2-6(0) = 8-0 = 8units2
[tex]\int[/tex]x3+x2-6x from 0 to 2. = [tex]\int[/tex](0)3+(0)2-6(0) - [tex]\int[/tex](2)3+(2)2-6(2) = 0- 0 = 0 units2
Therefore, Area from (-2 to 0)+ Area from (0 to 2) = 8+0 = 8units2
But using my calculator doing the same steps (except in the graphing section) doing -2 to 0 and adding it to 0 to 2's value i get 16units2( 10.667+5.333)
I don't know what's going on..