MATH 12002 - CALCULUS I 4.3: The Evaluation Theorem Professor - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 4.3: The Evaluation Theorem Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 9

Evaluation Theorem Evaluation Theorem If y = f ( x ) is a continuous function on [ a , b ] , then � b f ( x ) dx = F ( b ) − F ( a ) , a where F is any antiderivative of f . D.L. White (Kent State University) 2 / 9

Evaluation Theorem Notes: The Evaluation Theorem is actually Part 2 of the Fundamental Theorem of Calculus (see text, page 237). This generalizes the solution to the distance problem: � b a v ( t ) dt = displacement = s ( b ) − s ( a ) . See pages 224–225 of the text for a proof of the Evaluation Theorem. If G is another antiderivative of F , then we know G ( x ) = F ( x ) + C for a constant C . Hence G ( b ) − G ( a ) = ( F ( b ) + C ) − ( F ( a ) + C ) = F ( b ) + C − F ( a ) − C = F ( b ) − F ( a ) , and so it does not matter what antiderivative we choose. For convenience, we will use the notation F ( x ) | b a = F ( b ) − F ( a ) . D.L. White (Kent State University) 3 / 9

Evaluation Theorem Because of this relationship between the antiderivative and the definite integral, we use the notation � f ( x ) dx for the general antiderivative of f , � and we call f ( x ) dx the indefinite integral of f . So for example, � x 2 dx = 1 3 x 3 + C and � 2 x + cos x dx = x 2 + sin x + C . � b Note that a f ( x ) dx is a NUMBER, � whereas f ( x ) dx is a family of FUNCTIONS. D.L. White (Kent State University) 4 / 9

Evaluation Theorem We can translate our properties of antiderivatives into properties of indefinite integrals: 1 � n + 1 x n +1 + C ( n � = − 1) x n dx = � sin x dx = − cos x + C � cos x dx = sin x + C � � kf ( x ) dx = k f ( x ) dx � � � [ f ( x ) ± g ( x )] dx = f ( x ) dx ± g ( x ) dx D.L. White (Kent State University) 5 / 9

Examples � 3 − 1 (6 x 2 + 5) dx . 1 Evaluate � 3 (6 x 2 + 5) dx (2 x 3 + 5 x ) � 3 � = − 1 − 1 [2(3 3 ) + 5(3)] − [2( − 1) 3 + 5( − 1)] = = (54 + 15) − ( − 2 − 5) = 69 − ( − 7) = 76 . � π 2 Evaluate 0 sin x dx . � π − cos x | π sin x dx = 0 0 = ( − cos π ) − ( − cos 0) = − ( − 1) − ( − 1) = 1 + 1 = 2 . D.L. White (Kent State University) 6 / 9

Examples 1 ( x − 2 + 2 √ x − 4 x ) dx . � 4 3 Evaluate � 4 � 4 ( x − 2 + 2 √ x − 4 x ) dx ( x − 2 + 2 x 1 / 2 − 4 x ) dx = 1 1 4 3 x 3 / 2 − 2 x 2 �� � − x − 1 + 4 = � � 1 [ − 4 − 1 + 4 3 (4 3 / 2 ) − 2(4 2 )] − = [ − 1 − 1 + 4 3 (1 3 / 2 ) − 2(1 2 )] [ − 1 4 + 4 3 (8) − 32] − [ − 1 + 4 = 3 − 2] − 1 4 + 32 3 − 32 + 1 − 4 = 3 + 2 − 239 = 12 ≈ − 19 . 9 . D.L. White (Kent State University) 7 / 9

Examples � 1 0 ( x 2 + 1)( x 3 + 5) dx . 4 Evaluate � 1 � 1 ( x 2 + 1)( x 3 + 5) dx ( x 5 + x 3 + 5 x 2 + 5) dx = 0 0 � 1 6 x 6 + 1 4 x 4 + 5 3 x 3 + 5 x � 1 �� = 0 � 1 6 (1 6 ) + 1 4 (1 4 ) + 5 3 (1 3 ) + 5(1) � = − � 1 6 (0 6 ) + 1 4 (0 4 ) + 5 3 (0 3 ) + 5(0) � � 1 6 + 1 4 + 5 � = 3 + 5 − (0) 85 = 12 ≈ 7 . 08 . D.L. White (Kent State University) 8 / 9

Examples Note that the Evaluation Theorem is valid only if f is continuous on [ a , b ]: � 1 5 Consider x 4 dx ; f ( x ) = 1 1 x 4 is not continuous at x = 0. − 1 1 Since x 4 is always positive where it is defined, � 1 1 if x 4 dx exists, it has to be positive. − 1 x 4 = x − 4 has antiderivative − 1 3 x − 3 = − 1 1 However, 3 x 3 , and 1 � �� � � � � − 1 1 1 � = − − − � 3 x 3 3(1 3 ) 3( − 1) 3 � − 1 � − 1 � � − 1 � = − 3 − 3 � − 1 � � 1 � = − 2 = − 3 , 3 3 and so the formula from the Evaluation Theorem fails. D.L. White (Kent State University) 9 / 9