the Mean value Theorem for Definite Integrals Theorem: Let f(x) be continuous on
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the Mean value Theorem for Definite Integrals Theorem: Let f(x) be continuous on the closed interval [a,b]. Then there exist x in(a,b) so that 1)Since f(x) is continuous on[a,b], then by definition: is the (or )value for f on the closed interval [a,b]. Since f is also continuous on [a,b], by the Value Theorem (from Calculus I), there exist x1 and x1 in [a,b] where : Thus : min leq f(x) leq Max Thus: By the Dominance (or Preservation of Inequality) Theorem, then: since f(x1)=min and f(x2)=Max, and based on the above inequality, the number k= is between f(x2) and f(x1), then by the: Limit Existence Theorem Extreme Value Theorem Average (or Mean) Value Theorem Second Fundamental Theorem of Calculus Mean Value Theorem for Derivatives Intermediate Value Theorem Average (or Extreme) Value Theorem Squeeze (or Pinching or Sandwich) Theorem There exists a c in [x1, x2] subset [a, b] where: Therefore:Explanation / Answer
average or mean value for f on the curve.
extreme value theorem.
(b-a) min <= .......... <= (b-a) max.
thus min <= (....../(b-a)) <= max i.e. the answer for this block is (b-a).
k = 1/(b-a)
The answers are for:
block wise=> average, mean, extreme, min (b-a), max (b-a), (b-a), (b-a).
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