Q. 82 Prove the Mean Value Theorem for... [FREE SOLUTION]

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Chapter 4: Q. 82 (page 388)

Prove the Mean Value Theorem for Integrals by following these steps:
(a) Use the Extreme Value Theorem to argue that f has a maximum value M and a minimum value m on the interval [a, b].
(b) Use an upper sum and a lower sum with one rectangle to argue that $m (b - a) 猢� {鈭�}_{a}^{b} f (x) d x 猢� M (b - a)$ and thus that the average value of f on [a, b] is between m and M.

Short Answer

Expert verified

Hence, proved.

Step by step solution

Step 1. Given Information.

A function f(x) in the interval [a, b] has maximum value M and minimum value m.

Step 2. part (a) Extreme value theorem.

$T h e e x t r e m e v a l u e t h e o r e m s t a t e s t h a t i f a r e a l v a l u e d f u n c t i o n f i s c o n t i n u o u s i n t h e c l o s e d a n d b o u n d e d i n t e r v a l [a, b], t h e n f m u s t a t t a i n i t s m a x i m u m a n d m i n i m u m, e a c h a t l e a s t o n c e . T h e r e f o r e f r o m t h i s t h e o r e m f (x) = M w h i c h i s i t s m a x i m u m v a l u e a n d f (x) = m w h i c h i s i t s m i n i m u m v a l u e i n t h e i n t e r v a l [a, b] a t l e a s t o n c e .$

Step 3. part (b) Proof.

${鈭�}_{a}^{b} f (x) d x = \lim_{n 鈫� 鈭�} {鈭�}_{k = 1}^{n} f (x_{k}) 鈭� x, S o, {鈭�}_{k = 1}^{n} f (x_{k}) 鈭� x 猢� {鈭�}_{k = 1}^{n} M 鈭� x, {鈭�}_{k = 1}^{n} f (x_{k}) 鈭� x 猢� M ({鈭�}_{k = 1}^{n} 鈭� x), S i n c e, {鈭�}_{k = 1}^{n} 鈭� x = b - a, w e g e t, {鈭�}_{k = 1}^{n} f (x_{k}) 鈭� x 猢� M (b - a), T a k i n g l i m i t w e g e t, {鈭�}_{a}^{b} f (x) d x 猢� M (b - a) . S i m i l a r l y i t c a n b e s h o w n t h a t, {鈭�}_{a}^{b} f (x) d x 猢� m (b - a) . T h e r e f o r e, i t i s p r o v e d t h a t, m (b - a) 猢� {鈭�}_{a}^{b} f (x) d x 猢� M (b - a) . H e n c e, p r o v e d .$

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. part (a) Extreme value theorem.

Step 3. part (b) Proof.

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