- #1

Sirsh

- 267

- 10

Say i have F(x)=0.5x

^{2}-2 and F(x)=x

^{3}+x

^{2}-6x and i want to find the area of the regions which is satisfied from -2 to 2.

so [tex]\int[/tex]0.5x

^{2}-2 from -2 to 2. [tex]\int[/tex]x

^{3}+x

^{2}-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]x

^{3}+x

^{2}-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 = 8units

^{2}

[tex]\int[/tex]x

^{3}+x

^{2}-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 units

^{2}

Therefore, Area from (-2 to 0)+ Area from (0 to 2) = 8+0 = 8units

^{2}

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 16units

^{2}( 10.667+5.333)

I don't know whats going on..