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1 Techer Notes Activity 17 Fundmentl Theorem of Clculus Objectives Explore the connections between n ccumultion function, one defined by definite integrl, nd the integrnd Discover tht the derivtive of the ccumultor is the integrnd Mterils TI84 Plus / TI83 Plus Teching Time 5 minutes Abstrct This ctivity follows the Accumultion Function ctivity nd explores grphic nd numeric mnifesttions of the Fundmentl Theorem of Clculus. The version covered here sttes tht if f is function continuous on the closed intervl [, b], nd x is ny number in [, b], then the function is n ntiderivtive for f. Another wy of expressing this result is Students evlute n ccumultion function using list vribles in the grphing hndheld. They note the loction of reltive mxim in the tble of vlues for the ccumultion function nd see tht ech corresponds to sign chnge from positive to negtive in the vlues of the integrnd. Similrly, they see tht minimum on the ccumultor occurs where the integrnd chnges sign from negtive to positive. With the vlues of the ccumultor in list vrible, it is possible to verify the Fundmentl Theorem of Clculus by forming difference quotient. This is suggested s followup ctivity. Evidence of Lerning Students will: x Fx ( ) = ft ( ) dt d x ft ( ) dt dx = be ble to predict the loction of locl extrem on the grph of functions of the x form gx ( ) = ft ( ) dt by inspecting the grph of y = f(x). fx ( ) recognize tht chnging the lower limit of integrtion in the function x gx ( ) = ft ( ) dt results in verticl trnsltion of the grph of y = g(x).
2 168 Clculus Activities be ble to differentite functions defined by definite integrls, such s d. dx x = ft ( ) dt Mngement Tips nd Hints Prerequisites Students should: be ble to grph functions nd nvigte the CALC Menu to find zeroes of function. be ble to produce sctter plot. The ctivity introduces them to the seq commnd. know the nottion for defining definite integrl; n integrl s vlue is positive when the integrnd is positive over the intervl of integrtion, nd n integrl s vlue is negtive when the integrnd is negtive. The ccumultion function ctivity or some other experience with functions defined by definite integrl is necessry preprtion for students. Common Student Errors/Misconceptions Students my hve difficulty with the ide of dummy vrible (the vrible of integrtion) nd distinguishing it from the independent vrible of the ccumultion function (usully n upper limit of integrtion). Students cn lose trck of where the vlues in the list L2 ctully come from. Ech results from the evlution of definite integrl. Extension #1 After students hve seen tht the derivtive of Y2(X) is Y1(X), it is sensible to confirm this numericlly. Follow the steps outlined below. 1. Hve hlf of the students copy L2 into L4, nd hve the other hlf copy L3 into L4. Shown re the screenshots tht cme from using L2.
3 Activity 17: Fundmentl Theorem of Clculus Position the cursor on the first vlue in L4, nd press to delete it. Do the sme for the lst vlue in L2. Ech list will then contin totl of 47 vlues, with the vlues of L2 (or L3) nd L4 offset by one. 3. Define L5 s (L4 L2)/(2*DX) s the difference quotient for the ccumultion function, Y2 or Y3. Replce L2 with L3 for hlf of the students. This will pproximte its derivtive. 4. Delete the first element of L1. Define PLOT3 s sctter plot of L5 versus L1. Select only Y1 nd PLOT3 for grphing. Drw the grph. If you get n error messge bout DIM MISMATCH, check tht the lengths of L1 nd L5 re the sme. The sctter plot nd the grph of Y1 coincide. Extension #2 Ask students to exmine the loctions of points of inflection on the grphs of PLOT1 nd PLOT2. They should see tht these points occur t extrem on the grph of Y1. Points of inflection on the grph of function lwys occur t extrem on the grph of the derivtive. Moreover, point of inflection is plce where outputs re loclly chnging the fstest. Becuse the function is ccumulting re, it should chnge the fstest where the integrnd is frthest from the xxis.
4 17 Clculus Activities Activity Solutions 1. There is no re to clculte. 2. The integrnd, Y1(X), is positive over the intervl [,.1638] x x 2 cos.5dt > 2cos dt 3 3 becuse the integrnd is positive over the intervl [.1638,.21277]. More positive re hs been ccumulted x x 2 cos.5dt < 2cos dt 3 3 becuse the integrnd, Y1(X), is negtive on the intervl [2.213, ]. Negtive re is ccumulting, so the ccumultion function is decresing. 7. The nswers to Questions 4 nd 5 re pproximtely the sme becuse the mximum vlue of the ccumultion function, Y2(X), will occur when ccumultion of the positive re stops nd ccumultion of the negtive re strts. This occurs where the integrnd chnges sign from positive to negtive (the nswer to Question 5). The nswer to Question 4 is the pproximte loction where the vlues of Y2(X) rech mximum The nswers to Questions 8 nd 9 re similr becuse the ccumultion function reches minimum when negtive re stops ccumulting nd positive re strts ccumulting. As in Question 7, this occurs where the integrnd chnges sign from negtive to positive Y2(X) reches nother locl mximum t tht point. Looking t the tble, the mximum occurs ner Answers will vry. Focusing on x =, x 2 Y 2 ( ) = cos.5dt 3 is zero becuse the upper nd lower limits of integrtion re equl. However, x 2 Y 3 ( ) = cos.5dt < 3 1 becuse the integrnd is positive, but the integrtion occurs from right to left. This plces the grph of L3 below the grph of L2.
5 Activity 17: Fundmentl Theorem of Clculus PLOT1 nd PLOT2 hve the sme shpe becuse ech increses where the integrnd is positive, nd ech decreses where the integrnd is negtive. In fct, t ech vlue of x, the two plots re chnging t exctly the sme rte becuse ech ccumultes re under the grph of Y1. The Fundmentl Theorem sys tht the two ccumultion functions, Y2 nd Y3, hve the sme derivtive, Y1. d ( Y dx 2 ( X) ) = Y 1 ( X)
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