5.3 Integration & Anti-derivatives The Fundamental Theorem of Calculus Question #4 (Medium)

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4 Feb 2013
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5.3 Integration & Anti-derivatives
The Fundamental Theorem of Calculus
Question #4 (Medium): Finding the Derivative of the Function
Strategy
Notice that the Fundamental Theorem of Calculus Part 1 has 
, where the integral goes
from a numeric value of to a variable . This means that only one endpoint of the interval is a variable.
Two complex variations to this are:
1) One end containing a more complex function instead a simple : Then use Leibniz notation and
substitute a simpler variable. Apply the Chain Rule in conjunction with the Fundamental
Theorem of Calculus Part 1 to find the derivative of the function.
2) Both ends of the integral contain some kind of variable expression: Then split the integral into
two with usually as an easy dividing point, then proceed.





.
Sample Question
Find the derivative of the function.
  

Solution
Recall Leibniz notation of the Fundamental Theorem of Calculus Part 1
 
 .
Here only one end of the interval contains a variable, so Chain Rule is needed.
Let   , then   . Rearranging gives 
  .
Then
   

   

   
   
Therefore, the derivative of the function is:   

is  .
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