1. ## some how forgot

Hey guys n gals

somehow forgot, but when solving a definite integral say functiong y=x^2 from 0 to 10... how come when you do the integral it doesnt include anything below y=0.... i know why but i just neeed the proof

and when doing area problems under a curve for example sinx {0, 2pi} the value is 0 but the net area is ???

2. Originally Posted by mpl06c
how come when you do the integral it doesnt include anything below y=0.
I have no idea what this means. Please explain more.

3. Originally Posted by mpl06c
Hey guys n gals

somehow forgot, but when solving a definite integral say functiong y=x^2 from 0 to 10... how come when you do the integral it doesnt include anything below y=0.... i know why but i just neeed the proof
Proof of what? That the square of a real number is never negative? Essentially it is that the product of two positive numbers is positive and that the product of two negative numbers is positive: if you are squaring a number, you are multiplying it by itself so you are always multiplying numbers of the same sign.

and when doing area problems under a curve for example sinx {0, 2pi} the value is 0 but the net area is ???
The "area under the curve" normally means the area below the graph and above the x-axis. For x between pi and 2pi, there is no such are because y= sin(x) is below y= 0. I suspect you mean the area bounded by y= sin(x) and y= 0. Because "area" is never negative, you are subtracting "higher value" minus "lower value". From 0 to pi, the higher value is sin(x) but from pi to 2pi, the higher value is 0 so you have to do that as two separate integrals: $\displaystyle \int_0^\pi (sin(x)- 0)dx+ \int_\pi^{2\pi} (0- sin(x)dx$