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Prove the Mean Value Theorem for Integrals by applying the Mean Value Theorem for derivatives (see Section 4.2) to the function $ F(x) = \displaystyle \int_{a}^x f(t) dt $.

$f(c)(b-a)$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 5

Average Value of a Function

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okay, The mean value. Their first off we know efforts continuously closed. Interval A to B and F is also different. Chewable on the open interval. A deadbeat. Therefore, we know there's a number seeing a comma be such that the derivative of prime F C is F A B minus F of a over B minus A. Therefore, we can write this out as half primacy is F A B minus off of a over B minus. And remember the fundamental theme of calculus. It means that we can essentially break up into girls like this. I remember when the integral is from a This means it's just zero, which means we can disregard it. In this case, bring us back to FC is one over B minus eight times the integral from A to B after vax d x. Thus we have now proven the mean value the arm

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