Today’s Agenda • Upcoming Homework • Section 5.3: Evaluating Definite Integrals • Quiz Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 1 / 8
Upcoming Homework • Written HW M: Section 5.1, #4. Section 5.2, #10,16,18,30,34,44. Due 11/30/2015. • WeBWorK HW 24: Section 5.3, due 12/2/2015. • WeBWorK HW 25: Section 5.4, due 12/4/2015. • Written HW N: Section 5.3, #54,56,60,62. Section 5.4, #2,14,20,22,26. Due 12/4/2015. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 2 / 8
Section 5.3 As we saw in Sections 5.1 and 5.2, evaluating definite integrals using the limit-sum definition is a long process. As it turns out, there is a much faster way to evaluate definite integrals, and the proof that it works is quite simple. Evaluation Theorem (Fundamental Theorem of Calculus, part 1) If f is continuous on the interval [ a , b ], then � b f ( x ) dx = F ( b ) − F ( a ) , a where F is any antiderivative of f , that is, F ′ ( x ) = f ( x ). Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 3 / 8
Section 5.3 First, we claim that if a = x 0 < x 1 < x 2 < · · · < x n − 1 < x n = b is a partition of the interval [ a , b ], then n � ( F ( x i ) − F ( x i − 1 )) = F ( b ) − F ( a ) . i =1 Observe that n � ( F ( x i ) − F ( x i − 1 )) i =1 = ( F ( x 1 ) − F ( x 0 )) + ( F ( x 2 ) − F ( x 1 )) + ( F ( x 3 ) − F ( x 2 )) + · · · + ( F ( x n − 1 ) − F ( x n − 2 )) + ( F ( x n ) − F ( x n − 1 )) = F ( x n ) − F ( x 0 ) = F ( b ) − F ( a ) . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 4 / 8
Section 5.3 Now we claim that � n i =1 ( F ( x i ) − F ( x i − 1 )) = � n i =1 f ( x ∗ i ) · ∆ x , where x ∗ i is some number between x i − 1 and x i (i.e., x ∗ i ∈ [ x i − 1 , x i ]). From the Mean Value Theorem, we know that there is some value c ∈ [ x i − 1 , x i ] such that F ( x i ) − F ( x i − 1 ) = F ′ ( c ) = f ( c ) . x i − x i − 1 Letting c = x ∗ i , we have that F ( x i ) − F ( x i − 1 ) = ( f ( x ∗ i ))( x i − x i − 1 ) = f ( x ∗ i ) · ∆ x . Therefore, n n � � ( F ( x i ) − F ( x i − 1 )) = f ( x ∗ i ) · ∆ x . i =1 i =1 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 5 / 8
Section 5.3 Combining the results from the previous two slides, we have that n n � � f ( x ∗ F ( b ) − F ( a ) = ( F ( x i ) − F ( x i − 1 )) = i ) · ∆ x . i =1 i =1 Taking the limit of both the left and right sides of this equality, we have that n � f ( x ∗ n →∞ F ( b ) − F ( a ) = lim lim i ) · ∆ x n →∞ i =1 � b ⇒ F ( b ) − F ( a ) = f ( x ) dx , ⇐ a which proves the Evaluation Theorem. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 6 / 8
Section 5.3 The Evaluation Theorem suggests a way of writing indefinite integrals , which are simply families of antiderivatives. An indefinite integral is written the same way as a definite integral, only without limits of integration: � f ( x ) dx . If F ( x ) is an antiderivative of f ( x ), then we write: � f ( x ) dx = F ( x ) + c . Notice that definite integrals represent an actual number that we can calculate, whereas indefinite integrals represent a family of functions that differ by only a constant. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 7 / 8
Section 5.3 Example Problems Evaluate the following definite integrals using the Evaluation Theorem: 1 � 3 4 � 2 − 2 x 2 − 3 dx x 2 − 2 x dx 1 2 � 2 5 � 1 0 (2 x − 3)(4 x 2 + 1) dx 0 x 10 + 10 x dx 3 � π 0 5 e x + 3 sin x dx 6 � √ 3 / 2 1 1 − x 2 dx √ 0 Find the following indefinite integrals: 7 � √ √ x 3 + x 2 dx 3 sin x 9 � 1 − sin 2 x dx 8 � 10 � x 2 − 1 ( x + 4)(2 x + 1) dx x 4 − 1 dx Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 25 November 2015 8 / 8
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