Sections 5.2: The Definite Integral


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1 Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x) is continuous function on the intervl [, b] nd let n be some positive integer. Do the following: (i) Subdivide the intervl [, b] into n equl sized pieces, ech of length x = (b )/n. Let the endpoints of these intervls be = x < x < x 2 < < x n = b. (ii) From the ith subintervl [x i, x i ], choose reference point, cll it x i (do this for ech subintervl). We define the definite integrl of f(x) from to b to be = lim n n f(x i) x provided this limit exists. If this limit exists, we sy tht f(x) is integrble on [, b]. Before we strt looking t exmples of how to clculte the definite integrl, we need some terminology nd some bsic fcts. i= Terminology. Suppose tht b is definite integrl. (i) We cll f(x) the integrnd. (ii) We cll nd b the limits of integrtion  the lower limit nd b the upper limit. (iii) The procedure of finding the limit is clled integrtion. (iv) The sum is clled Riemnn sum. Fct. Suppose tht is definite integrl. (i) If f(x) on [, b], then mesures the re bounded between f(x) nd the xxis between nd b.
2 2 (ii) In generl, mesures the weighted re bounded between f(x) nd the xxis between nd b  by weighted we men tht ll re below the xxis is counted negtively nd ll re bove the xxis is counted positively. (iii) It f(x) is continuous on [, b] except for t finitely mny discontinuities, then f(x) is integrble on [, b]. This lst observtion mens tht nerly ll functions we shll consider will be integrble. We re now redy to consider some explicit exmples. Note tht the book emphsizes the use of formuls to clculte integrls  we shll void this insted emphsizing on how to clculte them by hnd, geometriclly nd using the clcultor. Exmple.2. Approximte the following integrls using the specified rules. (i) (ii) (iii) (x 2 + 2x)dx with n = 3 using the left hnd sum. We hve x = /3, so Left Hnd Sum = ( f () + f ( ) + f 3 = ( ) 3 = sin (x) dx 3 x with n = 2 using the midpoint sum. ( )) We hve x = /2, so if f(x) = sin (x)/x then ( ( ) ( )) 3 5 Midpoint Sum = f + f = ( ) 2 = ln (x) x dx with n = 4 using the right hnd sum.
3 We hve x = (3 )/4 = /2, so if f(x) = ln (x)/x then ( ( ) ( ) ) 3 5 Right Hnd Sum = f + f(2) + f + f(3) = ( ) 2 =.675. Exmple.3. Suppose tht the vlues of f(x) re given in the tble below. x f(x) Approximte the following integrls using the specified rules. (i) 4 with n = 4 using the left hnd sum. We hve x =, so Left Hnd Sum = (f () + f() + f(2) + f(3)) 3 (ii) (iii) = ( ) =. 8 with n = 4 using the midpoint sum. We hve x = 2, so Midpoint Sum = (f() + f(3) + f(5) + f(7)) 2 = ( ) 2 = 4. 8 with n = 4 using the right hnd sum. We hve x = 2, so Right Hnd Sum = (f(2) + f(4) + f(6) + f(8)) 2 = (4 3 9) 2 = 8. Exmple.4. Evlute the following integrls exctly.
4 4 (i) x dx. First we look t the grph of f(x): K2 K 2 x Since the definite integrl is equl to the weighted re bounded between the xxis nd the grph of f(x) = x, we simply need to evlute the re of the two tringles in the grph. i.e. (ii) 2 x dx =. 4 x 2 dx. As before, we look t the grph of f(x): K2 K 2 x K.5 K. K.5 K2. Since the definite integrl is equl to the weighted re bounded between the xxis nd the grph of f(x) = 4 x 2, we simply need to evlute the re of the qurter circle in the third qudrnt nd negte this vlue (since it is below the xis). i.e. (iii) 2 4 x 2 dx = π. π 2 π 2 sin (x)dx. As before, we look t the grph of f(x):
5 5..5 K3 K2 K 2 3 x K.5 K. Note tht between x = π/2 nd x = π/2, there is n equl mount of re bounded between the xxis nd f(x) = sin (x) bove nd below the xxis. Since the definite integrl is equl to the weighted re bounded between the xxis it follows tht the integrl must be zero. i.e. π 2 π 2 sin (x)dx =. 2. Properties of the Definite Integrl As with derivtives, there re mny useful properties of definite integrls llowing us to clculte them more esily. Result 2.. Suppose tht f(x) nd g(x) re continuous functions on n intervl [, b] nd suppose tht c is some constnt. Then we hve the following: (i) (ii) (iii) (iv) (v) (vi) = (f(x) + g(x))dx = cdx = c(b ). c = c (f(x) g(x))dx = c + c b. +. =. g(x)dx. g(x)dx.
6 Exmple 2.4. Show tht 6 (vii) = b. In ddition to the bsic lgebric properties of integrls, there re number of methods we cn use to help pproximte vlues of integrls. Specificlly, we hve the following. Result 2.2. (Comprisons for Integrls) (i) If f(x) for x b, then. (ii) If f(x) g(x) for x b, then g(x)dx. (iii) If m f(x) M for x b, then m(b ) M(b ). We finish by illustrting with some exmples of how to use these properties nd comprisons. Exmple 2.3. If = 2 nd = find. We know tht + = so + = 2 or =. sin (x)dx 2 by using the fct tht sin (x) x on the intervl [, ].
7 7 Using comprison, we know sin (x)dx xdx = x2 2 = 2. Exmple 2.5. Fill in the following tble for F(x) given tht F(x) = x where the grph of f(x) is given below. x F(x) 2π 4 π We clculte ech prt individully by determining pproprite res between the xxis nd the function. We hve F( 5) = F( 3) = 5 3 F( ) = = = F(2) = 5 3 = F() = F() = 2 = ( π ) = 2π 4. = ( π ) = π 2. = ( 4 4 ) =. =. = 2. = =.
8 8 F(5) = F(3) = F(4) = 5 4 = =. = = 2. = = 5. Exmple 2.6. Estimte the definite integrl e x dx. In this cse, we need to use Riemnn sums since the re we re trying to determine is not n elementry geometric shpe. Therefore using the clcultor, using left hnd sums with subdivisions, we get e x dx.7
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