Why does integration between curves across the x-axis work?

[DylanBurrowes]

Hey guys,

This is a bit of an extra knowledge question, but why does integration between curves across the y-axis work?
I've tried modelling it to see why it works, but just can't get around WHY it does!
e.g. finding the area enclosed by the lines x=-5, x=5 and the curves of y=x+2 and y=x-3.

I know it works, I just really want to know why!!

Thanks!

Edit: Sorry I meant the x-axis!!

[abeybaby]

Because youre integrating WITH RESPECT TO x. You know the little dx at the end of the integral? that means its with respect to x - ie, everything depends on x values, not on y-values. If it helps you think about it, what if you translated those 2 graphs 20 units to the right, wouldnt the area between them be the same? so whether you cross the y-axis or not, your area is still the same

[DylanBurrowes]

Sorry, I meant the x-axis! I typed this up late last night -___-

[Planck's constant]

Because the definite integral is a 'signed' area.
Provided tha f(x) > g(x) in the entire interval in question, the definite integral of f(x)-g(x) is always positive regardless of whether one or both of f(x) and g(x) lies below the x-axis. The process self-corrects.

Examples.
Both f(x) and g(x) above the x-axis : (positive area) - (smaller positive area) = (positive area)
f(x) above, g(x) below the x-axis:      (positive area) - (negative area) = (positive) + (positive) = (positive area)
Both f(x) and g(x) below the x-axis:   (negative area) - (more negative area) = ( positive area)

[DylanBurrowes]

Yep, I get you. Thanks so much!

[lzxnl]

If you ever forget, shift both graphs up by the same amount so that both graphs are wholly above the x-axis. It should be quite clear now why integrating the difference of the functions gives you the required area.