Evaluating Definite Integrals. There are a few properties that you should remember in order to assist you in evaluating definite integrals.


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1 Evluting Definite Integrls There re few properties tht you should rememer in order to ssist you in evluting definite integrls. f x dx= ; where k is ny rel constnt k f x dx= k f x dx ± = ± f x g x dx f x dx g x dx c f ( x) dx= f ( x) dx+ f ( x) dx; where c is ny rel numer in the intervl [, ] c = f x dx f x dx Exmple 1: Evlute the definite integrl ( + ) x x dx Solution: Step 1: Apply the sum/difference integrl property ( ) ( x x+ dx= x dx x) dx+ ( ) dx Step : Apply the constnt property ( x x+ ) dx= ( x ) dx ( x) dx+ dx Gerld Mnhn SLAC, Sn Antonio College, 8 1
2 Exmple 1 (Continued): Step : Find the ntiderivtives of ech term x x 1 = + ( x x+ ) dx= + ( x) ( x ) ( x ) ( x) Step : Evlute ( x x + ) dx = ( x ) ( x ) + ( x) 1 1 = + 1 = ( 7) ( 9) + 7 = ( ) ( ) ( ) = 15 7 = = As with the indefinite integrls, integrtion y sustitution cn e perform on definite integrls. The min thing to rememer when using sustitution with definite integrls is tht the upper nd lower limits of the integrl must e chnged to mtch the new vrile u. Exmple : Evlute the definite integrl x x d x Solution: Step 1: Identify the inside function nd set it equl to u. In this exmple, the inside function will e the rdicnd of 5x +. So, u = 5x +. Gerld Mnhn SLAC, Sn Antonio College, 8
3 Exmple (Continued): Step : Find the differentil of u. 5x u = + du = 1x dx du = 1x dx Step : Check if the differentil of u is present in the integrnd du = 1x dx ut the integrnd only contins x dx so you must divide oth sides y 5 in order to perform the sustitution. du = 1x dx du 1x = dx du = x dx 5 Step : Adjust the limits of the integrl to mtch with the new vrile of u. If x = 1 then If x = then u = 5x + u = 5(1) + u = 9 u = 5x + u = 5() + u = Step 5: Perform the sustitution nd chnge the integrl limits = u x 5x + dx= 5x + xdx 1 5 du Gerld Mnhn SLAC, Sn Antonio College, 8
4 Exmple (Continued): Step 6: Evlute the integrl u du = ( u) du u = = u 5 = 9 15 = 15 = = = In some prolems where you re sked to find the re of region etween the xxis nd function over n intervl [, ] the entire region or some portion of it my e elow the xxis. The re of the region elow the xxis using the Fundmentl Theorem would e negtive. Therefore, you must e creful when deling with prolems of this nture since re is suppose to e nonnegtive. Gerld Mnhn SLAC, Sn Antonio College, 8
5 The prolem with negtive res cn e voided y: 1. tking the solute vlue of the integrl, or = f x dx f x dx. pplying the rules of definite integrls to reverse the upper nd lower limits = f x dx f x dx Below re some guidelines to follow when sked to find the re of ounded region: 1. Sketch grph of the prolem. Locte ny xintercepts in the given intervl [, ]. Identify ny regions tht re elow the xxis. If there re regions ove nd elow the xxis split up the intervl nd use seprte integrls to evlute ech suregion. 5. Add the seprte res together to find the totl re. Exmple : Find the re etween the xxis nd the function f (x) = x over the intervl [, ]. Solution: Step 1: Sketch grph of the prolem Gerld Mnhn SLAC, Sn Antonio College, 8 5
6 Exmple (Continued): Step : Locte ny xintercepts in the given intervl [, ] f (x) = x = x = x There is n xintercept t x =, which cn e seen in the grph. Step : Identify ny regions tht re elow the xxis The region over the intervl [, ] is elow the xxis. Step : Split up the intervl nd use seprte integrls to evlute ech suregion Are = x dx ( ) ( ) ( ) = x dx + x dx Now since the intervl [, ] is elow the xxis we must djust the integrl for this intervl. Therefore we will tke the solute vlue of this integrl to ssure tht the vlue will e positive. + = + x dx x dx x dx x dx x x = + = + = + = + = 8 The re etween the xxis nd f(x) over the intervl [, ] would e 8. Gerld Mnhn SLAC, Sn Antonio College, 8 6
7 Let s look t wht would hppen if we did not seprte the intervl t the xintercept of zero. Are = x dx x = = = ( ) = ( ) Looking t the grph you cn ovious tell tht the re cnnot e equl to zero, ut since the intervls were not seprted the re for the intervl [, ] offset the re for the intervl [, ]. The verge vlue of function over the intervl [, ] is determined y multiplying the definite integrl y the frction of 1 over ( ). 1 Averge vlue = f x dx Exmple : Find the verge vlue of f (x) = x + on the intervl [, ]. Solution: 1 Averge vlue = f ( x) dx 1 = ( x + ) dx 1 x = + x 1 = + + Gerld Mnhn SLAC, Sn Antonio College, 8 7
8 Exmple (Continued): Averge vlue = = = + 7 = Gerld Mnhn SLAC, Sn Antonio College, 8 8
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