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...0 to 1 of g(x) dx, then f(x) <= g(x) for all x in (0,1).

Is the statement false?

Thanks!

2007-10-19 01:03:43 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

If f(x) < g(x) for every x in an interval [a, b], then Int (a to b ) f(x) dx <= Int (a to b) g(x) dx . But, as seen in Mathematician example, Int (a to b ) f(x) dx < Int (a to b) g(x) dx does NOT imply f(x) < g(x) (nor f(x) <- g(x) in (a, b).

2007-10-19 02:15:57 · answer #1 · answered by Steiner 7 · 0 0

False - but dont know a nice proof.

If you think of the integral as the area under each function, then just because the area of g(x) is larger than f(x) does not mean that the function itself necessarily is always larger.

2007-10-19 01:17:02 · answer #2 · answered by Anonymous · 0 0

Flase. Let f(x)=1/4 (constant) and g(x)=x.

2007-10-19 02:05:42 · answer #3 · answered by mathematician 7 · 1 0

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