Section 5.4 Fundmentl Theorem of Clculus 2 Lectures College of Science MATHS : Clculus (University of Bhrin) Integrls / 24
Definite Integrl Recll: The integrl is used to find re under the curve over n intervl [, b] Ide: To cover the re by s mny rectngles s possible nd then we will get better nd better estimte if we increse the number of rectngles. Question: When will we get n exct estimte for the re? Answer: When the number of rectngle. In tht cse, we write the re by Are = b f (x) dx This integrl is clled definite integrl. The number nd b re clled the lower limit nd upper limit of integrtion respectively. (University of Bhrin) Integrls 2 / 24
The Fundmentl Theorem of Clculus, Prt Question: Wht is the reltion between definite integrl (finding the re) nd the indefinite integrl (finding the nti-derivtive)? Answer: The Fundmentl Theorem of Clculus. One of the gret chievement of the humn mind. We will focus on the function Exmple g(x) = x Let g(x) = x f (t) dt. Find the following: g() 2 g() 3 g(2) 4 g(3) 5 g(4) f (t) dt re so fr (University of Bhrin) Integrls 3 / 24
The Fundmentl Theorem of Clculus, Prt Question: Wht is the derivtive of g(x) = x Theorem 2 In generl, Theorem 3 d dx ( h(x) g(x) ( d x ) f (t) dt = f (x) dx ) f (t) dt f (t) dt? = f (h(x))(h(x)) f (g(x))(g(x)) (University of Bhrin) Integrls 4 / 24
Exmple 4 Find the derivtive of g(x) = x Solution: t 3 + dt. ( ) ( ) g (x) = x 3 (x) + () 3 () + = x 3 + (University of Bhrin) Integrls 5 / 24
Exercise 5 Find the derivtive of g(x) = 2 x Solution: + sec t dt. g (x) = ( + sec 2)(2) ( + sec x)(x) = + sec x (University of Bhrin) Integrls 6 / 24
Exmple 6 Find the derivtive of g(x) = tn x Solution: t + t dt. ( g (x) = tn x + ) ( tn x ( tn x) + ) () ( = tn x + ) tn x sec 2 x (University of Bhrin) Integrls 7 / 24
Exmple 7 Find the derivtive of g(x) = x 3 Solution: sec x et + 5t 2 dt. ( g (x) = e x3 + 5(x 3 ) 2) ( x 3 ) ( e sec x + 5(sec x) 2) ( sec x) ( = e x3 + 5x 6) ( 3x 2 ) ( e sec x + 5 sec 2 x ) ( sec x tn x) (University of Bhrin) Integrls 8 / 24
Exercise 8 Find the derivtive of g(x) = x 9 sin x tn9 t dt. (University of Bhrin) Integrls 9 / 24
The Fundmentl Theorem of Clculus, Prt 2 Question: How to evlute the definite integrl? Theorem 9 If f is continuous on the intervl [, b] nd F is the nti-derivtive of f, then b f (x) dx = F (x) }{{} ntiderivtive b = F (b) F () Definite integrl b f (x) dx gives number representing the re. 2 Indefinite integrl f (x) dx gives function. (University of Bhrin) Integrls / 24
Exmple Find 2 (x 3 6x) dx. Solution: 2 ( 4 (2)4 3(2) 2 [ (x 3 6x) dx = 4 x 4 3x 2 ) ( ) 4 ( )4 3( ) 2 ] 2 = 2 4 Direct evlution (University of Bhrin) Integrls / 24
Exercise Find 4 ( x 3 + ) x dx. (University of Bhrin) Integrls 2 / 24
Exmple 2 Find 9 6 x dx. Solution: 2 9 9 6 x dx = 6x 2 dx = [6 2 3 x 3 2 ( ) ( ) 4(9) 3 2 4() 2 3 = 4 ] 9 2 Direct evlution (University of Bhrin) Integrls 3 / 24
Exercise 3 Find (x + )2 dx. (University of Bhrin) Integrls 4 / 24
Exmple 4 Find 2 Solution: 3 x 5 +3x 3 x 4 2 dx. x 5 + 3x 3 2 x 4 dx = x + 3 [ ] 2 x dx = 2 x 2 + 3 ln x ( ) ( ) 2 (2)2 + 3 ln 2 2 ()2 + 3 ln = 3 2 + 3 ln 2 3 Direct evlution (University of Bhrin) Integrls 5 / 24
Exercise 5 Find Solution: 4 2 +x 2 dx. 2 + x 2 dx = [ 2 tn x ] ( 2 tn ) ( 2 tn ) = π 2 4 Direct evlution (University of Bhrin) Integrls 6 / 24
Exmple 6 If 2 (x + )2 dx = 9, then find the vlue of. Solution: 5 9 = 9 = 2 ( 26 3 (x + ) 2 dx = ) 2 ( 3 3 + 2 + 9 = 3 3 2 + 26 3 = 3 3 2 + 2 3 = 2 (x 2 + 2x + ) dx = ) [ ] 2 3 x 3 + x 2 + x 5 Finding limit of integrtion (University of Bhrin) Integrls 7 / 24
Exercise 7 If 3 (3x 2 + 2x) dx = 36, then find the vlue of. Solution: 6 36 = 2 (3x 2 + 2x) dx = 36 = (36) ( 3 + 2) 36 = 3 2 + 36 = 3 2 = 2 ( + ) = or = 3 (3x 2 + 2x) dx = [ x 3 + x 2] 3 6 Finding limit of integrtion (University of Bhrin) Integrls 8 / 24
Properties of Integrtion Recll: Definite integrls compute the re under the curve, i.e., 2 3 4 5 Are = b f (x) dx b [c f (x)] dx = c b f (x) dx. b [f (x) + g(x)] dx = b f (x) dx + b g(x) dx. f (x) dx =. b f (x) dx = f (x) dx. b b f (x) dx = c f (x) dx + b g(x) dx. c 6 If f (x) g(x) on [, b], then b f (x) dx b g(x) dx. (University of Bhrin) Integrls 9 / 24
Exmple 8 If 2 f (x) dx = 3, 2 g(x) dx = 2, then find 2 Solution: 7 2 [4f (x) + g(x)] dx. 2 2 [4f (x) + g(x)] = 4 [f (x) dx] + [g(x)] dx = 4(3) + 2 = 4 7 Properties of integrl (University of Bhrin) Integrls 2 / 24
Exercise 9 If 5 f (x) dx = 3, 3 f (x) dx =, nd 3 h(x) dx = 5 then find 5 2f (x) dx. 2 3 4 5 6 7 3 3 5 5 3 3 [f (x) + h(x)] dx. [2f (x) 5h(x)] dx. f (x) dx. f (x) dx. [h(x) f (x)] dx. 3 3 [h(x) f (x)] dx. Properties of integrl (University of Bhrin) Integrls 2 / 24
Exmple 2 Given Evlute 4 f (x) dx Solution: 8 4 f (x) dx = 4x + 2, x < 2 f (x) = 3x 2 2, 2 x < 6 6, x 6 2 2 f (x) dx + 4 2 4 f (x) dx = 4x + 2 dx + 2 3x 2 2 dx = [ 2x 2 + 2x ] 2 + [ x 3 2x ] 4 2 = 6 + 56 4 = 68 8 Properties of integrtion (University of Bhrin) Integrls 22 / 24
Exercise 2 Given Evlute 2 f (x) dx f (x) = { 3x 2, x < 2x +, x Exercise 22 Evlute 7 x dx 5 (University of Bhrin) Integrls 23 / 24