# Du dx by the chain rule e u 2 du dx by ftci e x 4 2 x

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Chapter 0 / Exercise 72
College Algebra and Calculus: An Applied Approach
Larson
Expert Verified
du dx (by the Chain Rule) = e u 2 du dx (by FTCI) = e x 4 · 2 x which is the same answer we calculated by the previous method. In the statement of the Fundamental Theorem of Calculus, the lower limit of the integral was always fixed. That is, it did not vary with x . We can now make our example even more complicated by letting the lower limit of the integral vary as a function of x . Let H ( x ) = Z x 2 cos( x ) e t 2 dt . How would we find H 0 ( x )? We can cleverly use the properties of the integral. In fact, we can write H ( x ) = Z x 2 cos( x ) e t 2 dt = Z 3 cos( x ) e t 2 dt + Z x 2 3 e t 2 dt . Furthermore, we know that Z 3 cos( x ) e t 2 dt = - Z cos( x ) 3 e t 2 dt and this integral is in the form where we can use the Fundamental Theorem. Therefore, we have that H ( x ) = Z x 2 cos( x ) e t 2 dt = Z x 2 3 e t 2 dt - Z cos( x ) 3 e t 2 dt . If we now let H 1 ( x ) = Z cos( x ) 3 e t 2 dt then H ( x ) = G ( x ) - H 1 ( x ) where G ( x ) is defined as before. But then H 0 ( x ) = G 0 ( x ) - H 1 0 ( x ) Calculus 2 (B. Forrest) 2
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Chapter 0 / Exercise 72
College Algebra and Calculus: An Applied Approach
Larson
Expert Verified
Chapter 1: Integration 40 and we already know that G 0 ( x ) = 2 xe x 4 . This means that we only need to find H 1 0 ( x ). To accomplish this we do exactly what we did to find G 0 ( x ). We note that H 1 ( x ) = F (cos( x )) so H 1 0 ( x ) = F 0 (cos( x )) d dx (cos( x )) = - sin( x ) e (cos( x )) 2 Combining all of this together gives us that H 0 ( x ) = G 0 ( x ) - H 1 0 ( x ) = 2 xe x 4 - ( - sin( x ) e (cos( x )) 2 ) = 2 xe x 4 + sin( x ) e (cos( x )) 2 The previous example leads us to an extended version of the Fundamental Theorem of Calculus. THEOREM 6 Extended Version of the Fundamental Theorem of Calculus Assume that f is continuous and that g and h are di erentiable. Let H ( x ) = Z h ( x ) g ( x ) f ( t ) dt . Then H ( x ) is di erentiable and H 0 ( x ) = f ( h ( x )) h 0 ( x ) - f ( g ( x )) g 0 ( x ). 1.6 The Fundamental Theorem of Calculus (Part 2) We have seen that the Fundamental Theorem of Calculus provides us with a simple rule for di erentiating integral functions and so it provides the key link between di erential and integral calculus. However, we will soon see it also provides us with a powerful tool for evaluating integrals . First we must briefly review the topic of antiderivatives from your study of di erential calculus. Calculus 2 (B. Forrest) 2
Section 1.6: The Fundamental Theorem of Calculus (Part 2) 41 1.6.1 Antiderivatives We know a number of techniques for calculating derivatives. In this section, we will review how we can sometimes “undo” di erentiation. That is, given a function f , we will look for a new function F with the property that F 0 ( x ) = f ( x ). DEFINITION Antiderivative Given a function f , an antiderivative is a function F such that F 0 ( x ) = f ( x ) . If F 0 ( x ) = f ( x ) for all x in an interval I , we say that F is an antiderivative for f on I . EXAMPLE 12 Let f ( x ) = x 3 . Let F ( x ) = x 4 4 . Then F 0 ( x ) = 4 x 4 - 1 4 = x 3 = f ( x ) , so F ( x ) = x 4 4 is an antiderivative of f ( x ) = x 3 . While the derivative of a function is always unique, this is not true of antiderivatives.