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Question Number 122469 by benjo_mathlover last updated on 17/Nov/20

Commented by benjo_mathlover last updated on 17/Nov/20

$i^{'} m f o r g o t t h e f o r m u l a t h e a v e r a g e r a t e o f$ $c h a n g e o f t h e f u n c t i o n . i s i t$ $A (x) = \frac{1}{b - a} \underset{a}{\int^{b}} f (x) d x ?$
Commented by benjo_mathlover last updated on 17/Nov/20

$p r o f M j s, m r W, l i b e r t y a n d o t h e r c a n y o u$ $h e l p m e$
Commented by benjo_mathlover last updated on 17/Nov/20

$T h e a v e r a g e r a t e o f c h a n g e o f$ $t h e f u n c t i o n f (x) = \underset{0}{\int^{x}} \sqrt{1 + \cos (t^{2})} d t$ $o v e r t h e i n t e r v a l [1, 3] i s n e a r e s t$ $t o . . .$
Commented by liberty last updated on 17/Nov/20

$average rate of change x_{1} to x_{2}$ $defined by relation A (x) = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}}$ $A (x) = \frac{f (3) - f (1)}{2}$ $f (x) = \underset{0}{\int^{x}} \sqrt{1 + \cos (t^{2})} dt$ $f (x) = \underset{0}{\int^{x}} \sqrt{1 + 2 \cos^{2} (\frac{t^{2}}{2}) - 1} dt$ $f (x) = \sqrt{2} \underset{0}{\int^{x}} \cos (\frac{t^{2}}{2}) dt$ $f (3) = \sqrt{2} \underset{0}{\int^{3}} \cos (\frac{t^{2}}{2}); f (1) = \sqrt{2} \underset{0}{\int^{1}} \cos (\frac{t^{2}}{2}) dt$ $f (3) - f (1) = \sqrt{2} \underset{1}{\int^{3}} \cos (\frac{t^{2}}{2}) dt$
Commented by mr W last updated on 17/Nov/20

$A (x) = \frac{1}{b - a} \underset{a}{\int^{b}} f (x) d x$ $i s t h e a v e r a g e v a l u e o f f u n c t i o n f (x)$ $i n [a, b] .$
	Answered by liberty last updated on 18/Nov/20

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