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 x-xis nd function over n intervl [, ] the entire region or some portion of it my e elow the x-xis. The re of the region elow the x-xis 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 x-intercepts in the given intervl [, ]. Identify ny regions tht re elow the x-xis. If there re regions ove nd elow the x-xis 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 x-xis 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 x-intercepts in the given intervl [, ] f (x) = x = x = x There is n x-intercept t x =, which cn e seen in the grph. Step : Identify ny regions tht re elow the x-xis The region over the intervl [-, ] is elow the x-xis. 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 x-xis 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 x-xis 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 x-intercept 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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