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

Short Answer

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