Week2_Section_FTC_and_its_example

Week2_Section_FTC_and_its_example - d dt R t 4 √ udu = d...

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Math 191 The Fundamental Theorem of Calculus via the Chain Rule Let f ( x ) be a continuous function on [ a, b ]. Then by the fundamental theorem of calculus, F ( x ) = R x a f ( t ) dt is continuous on [ a, b ], is differentiable on [ a, b ], and F 0 ( x ) = dF ( x ) dx = d R x a f ( t ) dt dx = f ( x ) . (1) Now consider g ( x ) instead of x as an upper bound of the integral above. Then to find F 0 ( g ( x )), we need to use the chain rule to get F 0 ( g ( x )) = dF ( g ( x )) dx (2) = dF ( g ( x )) dg ( x ) · dg ( x ) dx (3) = d R g ( x ) a f ( t ) dt dg ( x ) · g 0 ( x ) (4) = f ( g ( x )) · g 0 ( x ) , (5) where (3) is from the chain rule for the mapping x g ( x ) F ( g ( x )), and (5) from (1) with g ( x ) instead of x , i.e. the fundamental theorem of calculus. 1 Find the derivative d dt R t 4 0 udu by evaluating the integral and differentiating the result,
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Unformatted text preview: d dt R t 4 √ udu = d dt ± 2 3 u 3 / 2 | t 4 ² = d dt ( 2 3 t 6 ) = 4 t 5 2 Find the derivative d dt R t 4 √ udu by differentiating the integral directly. Since the upper bound is not simply x , We use the fundamental theorem of calculus combining with the chain rule above. Let f ( u ) = √ u and g ( t ) = t 4 . Then d dt Z t 4 √ udu = d dt Z g ( t ) f ( u ) du (6) = f ( g ( t )) · g ( t ) (7) = √ t 4 · d ( t 4 ) dt (8) = t 2 · 4 t 3 (9) = 4 t 5 (10) 1...
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