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Let F(x)=5x5t2dtF ( x ) = \int _ { 5 } ^ { x } 5 t ^ { 2 } d t

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Let F(x)=5x5t2dtF ( x ) = \int _ { 5 } ^ { x } 5 t ^ { 2 } d t .
a. Use Part 1 of the Fundamental Theorem of Calculus to find F(x)F ^ { \prime } ( x ) .
b. Use Part 2 of the Fundamental Theorem of Calculus to integrate 5x5t2dt\int _ { 5 } ^ { x } 5 t ^ { 2 } d t to obtain an alternative expression for F(x)F ( x ) .
c. Differentiate the expression for F(x)F ( x ) found in part (b). The Fundamental Theorem of Calculus, Part 1 If ff is continuous on [a,b][ a , b ] , then the function FF defined by F(x)=axf(t)dtaxbF ( x ) = \int _ { a } ^ { x } f ( t ) d t \quad a \leq x \leq b is differentiable on (a,b)( a , b ) , and F(x)=ddxaxf(t)dt=f(x)F ^ { \prime } ( x ) = \frac { d } { d x } \int _ { a } ^ { x } f ( t ) d t = f ( x ) The Fundamental Theorem of Calculus, Part 2
If ff is continuous on [a,b][ a , b ] , then abf(x)dx=F(b)F(a)\int _ { a } ^ { b } f ( x ) d x = F ( b ) - F ( a ) where FF is any antiderivative of ff , that is, Ft=fF ^ { t } = f .

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