Section 14.7 d Definite Integrals i E 2 Lectures a l l u d b Dr. Abdulla Eid A . College of Science r D MATHS 104: Mathematics for Business II Dr. Abdulla Eid (University of Bahrain) Integrals 1 / 15
Definite Integral Recall: The integral is used to find area under the curve over an interval [ a , b ] d i Idea: To cover the area by as many rectangles as possible and then we will E get better and better estimate if we increase the number of rectangles. a l l u Question: When will we get an exact estimate for the area? d b Answer: When the number of rectangle → ∞ . In that case, we write the A area by . � b r D Area = a f ( x ) dx This integral is called definite integral. The number a and b are called the lower limit and upper limit of integration respectively. Dr. Abdulla Eid (University of Bahrain) Integrals 2 / 15
The Fundamental Theorem of Calculus Question: How to evaluate the definite integral? d Theorem 1 i E If f is continuous on the interval [ a , b ] and F is the anti-derivative of f , a then l l u b � b d a f ( x ) dx = b F ( x ) = F ( b ) − F ( a ) A ���� antiderivative a . r D 1 Definite integral � b a f ( x ) dx gives a number represents the area. 2 Indefinite integral � f ( x ) dx gives a function . Dr. Abdulla Eid (University of Bahrain) Integrals 3 / 15
Example 2 Find � 2 − 1 ( x 3 − 6 x ) dx . d i E Solution: 1 a l l u � 2 � 2 d � 1 � 2 − 1 ( x 3 − 6 x ) dx = − 1 ( x 3 − 6 x ) dx = 4 x 4 − 3 x 2 b A − 1 � 1 � � 1 � = − 21 . 4 ( 2 ) 4 − 3 ( 2 ) 2 4 ( − 1 ) 4 − 3 ( − 1 ) 2 r − D 4 1 Direct evaluation Dr. Abdulla Eid (University of Bahrain) Integrals 4 / 15
Example 3 1 6 √ x dx . Find � 9 d i E Solution: 2 a l l u d � 9 � 9 � � 9 1 6 √ x dx = 62 b 1 3 2 dx = 1 6 x 3 x 2 A 1 � � � � . 3 3 4 ( 9 ) − 4 ( 1 ) = 104 r 2 2 D 2 Direct evaluation Dr. Abdulla Eid (University of Bahrain) Integrals 5 / 15
Exercise 4 Find � 1 − 1 ( x + 1 ) 2 dx . d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Integrals 6 / 15
Example 5 Find � 2 x 5 + 3 x 3 dx . x 4 d 1 i E Solution: 3 a l l u x 5 + 3 x 3 � 2 � 2 d � 1 � 2 1 x + 3 2 x 2 + 3 ln | x | b dx = x dx = x 4 A 1 1 � 1 � � 1 � = 3 . 2 ( 2 ) 2 + 3 ln 2 2 ( 1 ) 2 + 3 ln 1 r − 2 + 3 ln 2 D 3 Direct evaluation Dr. Abdulla Eid (University of Bahrain) Integrals 7 / 15
Example 6 (Substitution and definite integrals) Find � 1 0 x 2 e − x 3 dx Solution: 4 Since this is not a basic integral, we are looking for a good substitution. We are looking for an inner function with almost the d derivative is somewhere in the integral. Let i E a l u = − x 3 l u d du du = − 3 x 2 dx → dx = b − 3 x 2 A if x = 0,then u = 0 . r D if x = 1,then u = − 1 � − 1 � − 1 � 1 0 x 2 e − x 3 dx = du 1 e u du x 2 e u − 3 x 2 = − 3 0 0 � − 1 � − 1 � − 1 � � − 1 � = − 1 3 e − 1 + 1 3 e − 1 3 e 0 3 e u = = = − 3 0 4 Substitution and then evaluation Dr. Abdulla Eid (University of Bahrain) Integrals 8 / 15
Example 7 (Substitution and definite integrals) Find � 4 − 5 x √ x 2 + 9 dx 0 Solution: 5 Since this is not a basic integral, we are looking for a good substitution. We are looking for an inner function with almost the d derivative is somewhere in the integral. Let i E a l u = x 2 + 9 l u d du = 2 x dx → dx = du b A 2 x if x = 0,then u = 9 . r D if x = 4,then u = 25 � 4 � 25 � 25 − 5 x − 5 x 2 x = − 5 1 du √ dx = √ u √ u du x 2 + 9 2 0 9 9 � 25 = − 5 � � 25 ( u ) − 1 1 2 du = − 5 u 2 2 9 9 � � � � 1 1 = − 5 ( 25 ) − − 5 ( 9 ) = − 10 2 2 Dr. Abdulla Eid (University of Bahrain) Integrals 9 / 15
Example 8 (Old Final Exam Question) If � 3 a ( 3 x 2 + 2 x ) dx = 36, then find the value of a . d Solution: 6 i E a � 2 � 3 l a ( 3 x 2 + 2 x ) dx = a ( 3 x 2 + 2 x ) dx = l x 3 + x 2 � 3 � u 36 = d a b � a 3 + a 2 � 36 = ( 36 ) − A 36 = − a 3 − a 2 + 36 . r 0 = − a 3 − a 2 D 0 = − a 2 ( a + 1 ) a = 0 or a = − 1 6 Finding limit of integration Dr. Abdulla Eid (University of Bahrain) Integrals 10 / 15
Example 9 (Old Final Exam Question) If � 2 a ( x + 1 ) 2 dx = 9, then find the value of a . Solution: 7 d i E � 2 � 2 � 1 � 2 a a ( x 2 + 2 x + 1 ) dx = 3 x 3 + x 2 + x a ( x + 1 ) 2 dx = 9 = l l u a d � 26 � � 1 � 3 a 3 + a 2 + a b 9 = − A 3 9 = − 1 3 a 3 − a 2 − a + 26 . r D 3 0 = − 1 3 a 3 − a 2 − a + − 2 3 a = − 2 7 Finding limit of integration Dr. Abdulla Eid (University of Bahrain) Integrals 11 / 15
Properties of Integration Recall: Definite integrals compute the area under the curve, i.e., � b Area = a f ( x ) dx d i E 1 � b a [ c · f ( x )] dx = c · � b a f ( x ) dx . a l l 2 � b a [ f ( x ) + g ( x )] dx = � b a f ( x ) dx + � b u a g ( x ) dx . d b 3 � a A a f ( x ) dx = 0. . r 4 � b a f ( x ) dx = − � a D b f ( x ) dx . 5 � b a f ( x ) dx = � c a f ( x ) dx + � b c g ( x ) dx . 6 If f ( x ) ≤ g ( x ) on [ a , b ] , then � b a f ( x ) dx ≤ � b a g ( x ) dx . Dr. Abdulla Eid (University of Bahrain) Integrals 12 / 15
Example 10 (Old Final Exam Question) If � 2 0 f ( x ) dx = 3, � 2 0 g ( x ) dx = 2, then find � 2 d 0 [ 4 f ( x ) + g ( x )] dx . i E Solution: 8 a l l � 2 � 2 u � 2 d 0 [ 4 f ( x ) + g ( x )] = 4 0 [ f ( x ) dx ] + 0 [ g ( x )] dx b A = 4 ( 3 ) + 2 . r = 14 D 8 Properties of integral Dr. Abdulla Eid (University of Bahrain) Integrals 13 / 15
Exercise 11 If � 5 1 f ( x ) dx = 3, � 3 1 f ( x ) dx = 1, and � 3 1 h ( x ) dx = 5 then find a 1 � 5 1 − 2 f ( x ) dx . 2 � 3 1 [ f ( x ) + h ( x )] dx . d 3 � 3 i 1 [ 2 f ( x ) − 5 h ( x )] dx . E 4 � 1 a 5 f ( x ) dx . l l 5 � 5 u 3 f ( x ) dx . d b 6 � 1 3 [ h ( x ) − f ( x )] dx . A 7 � 3 . 3 [ h ( x ) − f ( x )] dx . r D a Properties of integral Dr. Abdulla Eid (University of Bahrain) Integrals 14 / 15
Example 12 (Old Final Exam Question) Given 4 x + 2, x < 2 3 x 2 − 2, f ( x ) = 2 ≤ x < 6 d i 106, x ≥ 6 E a Evaluate � 4 l 0 f ( x ) dx l u d b Solution: 9 A � 4 � 2 � 4 0 f ( x ) dx = . 0 f ( x ) dx + 2 f ( x ) dx r D � 2 � 4 2 3 x 2 − 2 dx = 0 4 x + 2 dx + 2 x 2 + 2 x x 3 − 2 x � � 2 � � 4 = 0 + 2 = 16 + 56 − 4 = 68 9 Properties of integration Dr. Abdulla Eid (University of Bahrain) Integrals 15 / 15
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