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Raep commented at 2008-12-22 09:13:05 » #24996
the area of the interval lies between the area of ([a->b] x [f(a)]) {meaning the red Box} and the area of ([a->b] x [F(b)] {being the green box}.
thus the area below f(x) = (the blue line) up to the x-axis, inbetween a and b is somewhere between the Area of the green and red box.
however, i would devide it into a few more intervals, to make it more accurate.
...
- oh yeah, btw, in the background there are two lesbian chicks having fun.
25 Points Flag
the area of the interval lies between the area of ([a->b] x [f(a)]) {meaning the red Box} and the area of ([a->b] x [F(b)] {being the green box}.
thus the area below f(x) = (the blue line) up to the x-axis, inbetween a and b is somewhere between the Area of the green and red box.
however, i would devide it into a few more intervals, to make it more accurate.
...
- oh yeah, btw, in the background there are two lesbian chicks having fun.
25 Points Flag
mathematician commented at 2013-07-09 02:17:42 » #1356319
This is not the squeeze theorem. Not even close. This is a consequence of the mean value theorem. For any function, the integral from a to b, then divided by (b-a) is the average value of the function on that interval. Thus, if f is strictly increasing, the average value of that function must be between the endpoints. The inequality is not strict if we have simple monotonicity.
7 Points Flag
This is not the squeeze theorem. Not even close. This is a consequence of the mean value theorem. For any function, the integral from a to b, then divided by (b-a) is the average value of the function on that interval. Thus, if f is strictly increasing, the average value of that function must be between the endpoints. The inequality is not strict if we have simple monotonicity.
7 Points Flag
mathematician commented at 2013-07-09 02:22:06 » #1356323
Must be between the function values of the endpoints* i meant to say. so f(a) < average value < f(b)
2 Points Flag
Must be between the function values of the endpoints* i meant to say. so f(a) < average value < f(b)
2 Points Flag
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