Definite Integrals

[Sirsh]

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 \int0.5x2-2 from -2 to 2. \intx3+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, \intx3+x2-6x from -2 to 0. = \int(-2)3+(-2)2-6(-2) - \int(0)3+(0)2-6(0) = 8-0 = 8units2


\intx3+x2-6x from 0 to 2. = \int(0)3+(0)2-6(0) - \int(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 whats going on..

[Sirsh]

I only did the one area for the 2nd function, ignore the 1st function.

[pbandjay]

A definite integral is positive if the area is above the y-axis and negative if the area is below the y-axis. If you integrate over an interval that has both, the result will then not represent the complete physical area.

Looking at a graph of your cubic function: http://www.wolframalpha.com/input/?i=x^3%2Bx^2-6x

The area under the curve from -2 to 0 is larger than the area "above" the curve from 0 to 2. Since the positive area above (-2,0) is larger than the negative area below (0,2), the total area will be positive.

When I did these calculations, I got an area of 32/3 on (-2,0) and -16/3 on (0,2). So the definite integral on (-2,2) would be 32/3 - 16/3 = 16/3.

The total physical area would be 32/3 + 16/3 = 48/3 = 16. I had to use the absolute value of the areas to calculate the total physical area. I hope this clears things up!

[HallsofIvy]

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 \int0.5x2-2 from -2 to 2. \intx3+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, \intx3+x2-6x from -2 to 0. = \int(-2)3+(-2)2-6(-2) - \int(0)3+(0)2-6(0) = 8-0 = 8units2
You haven't integrated anything, you have just evaluated the function itself at -2 and 0 and subtracted. The integral is
\int_{-2}^0 x^3+ x^2- 6x dx= \left[\frac{1}{4}x^4+ \frac{1}{3}x^3- 3x^2\right]_{-2}^0
= 0- \left(\frac{1}{4}(16)+ \frac{1}{3}(-8)- 3(4)\right)= \frac{32}{3}



\intx3+x2-6x from 0 to 2. = \int(0)3+(0)2-6(0) - \int(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 whats going on..
That's because your calculator integrated and you didn't!

[Sirsh]

With both did you do F(b) - F(a) ? also, do you know anything about these type of intergrals.

\intf(x)dx+|\intf(x)dx|
|\intf(x)dx|+\intf(x)dx
|\intf(x)dx+\intf(x)dx|

I'm unsure what these | | parts do when theyre associated with the integral. thanks alot.

[HallsofIvy]

With both did you do F(b) - F(a) ? also, do you know anything about these type of intergrals.

\intf(x)dx+|\intf(x)dx|
|\intf(x)dx|+\intf(x)dx
|\intf(x)dx+\intf(x)dx|

I'm unsure what these | | parts do when theyre associated with the integral. thanks alot.
\left|\int f(x)dx\right|
means "first take the integral of the function then take the absolute value of that number".

\int |f(x)|dx
means "first take the absolute value of the function, then integrate".

[Sirsh]

Wow, i'm so retarded I forgot to actually intergrate the function. how embarrassing..

[Sirsh]

\left|\int f(x)dx\right|
means "first take the integral of the function then take the absolute value of that number".

\int |f(x)|dx
means "first take the absolute value of the function, then integrate".

Thanks alot HallsofIvy, I don't really understand. could you please do an example of both?