MATH , Calculus 2, Fall 2018


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1 MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly one of the most importnt theorems in ll of mthemtics. In essence, it sttes tht differentition nd integrtion re inverse processes. There re two prts to the FTC, the second of which is the most difficult to understnd. The Fundmentl Theorem of Clculus (FTC Suppose tht f is continuous function over closed intervl x b. Prt : Prt 2: If A(x = f(xdx = F (b F (, where F is ny ntiderivtive of f. x f(t dt is the re function giving the re under f from to x, then A(x is differentible nd its derivtive is just f(x. In other words, A (x = f(x or ( d x f(t dt = f(x. ( dx Some Importnt Notes Concerning the FTC: (i The first prt of the FTC will be used over nd over gin throughout the course. It sttes tht to find the signed re under f from to b, we need only find n ntiderivtive of f, evlute it t the endpoints nd b, nd then subtrct. There is something fundmentlly surprising going on here: somehow, the re only depends on the vlues of the ntiderivtive t the endpoints, not in between. How is this possible? It is deep fct nd one tht hs useful generliztion into higher dimensions vi n importnt theorem known s Stokes Theorem. (ii Since F is n ntiderivtive of F (by definition, nother wy to write prt of the FTC is F (t dt = F (b F (. (2 Notice the integrl sign nd derivtive sign cnceling out here. Eqution (2 bsiclly sys tht integrting rte of chnge gives the net (or totl chnge. We hve lredy seen this in terms of Physics with the reltionship between position s(t nd velocity v(t: s (t dt = v(t dt = s(b s(. In words, the integrl of velocity over t b gives the totl chnge in position (sometimes clled the net displcement. An exmple from Economics would be something like C (x dx = C(b C(, which sttes tht the integrl of the mrginl cost gives the totl chnge in cost when incresing the number of items produced from to b units.
2 (iii In prt 2 of the FTC, the vrible is x nd it is the upper limit of integrtion. The lower limit is n rbitrry constnt. As x vries, the re under the curve vries, which is why A(x is relly function. The mzing spect of prt 2 is tht this function hs derivtive equl to the function we re finding the re under, nmely f. (iv A simple wy to remember prt 2, nd in prticulr eqution (, is tht the derivtive nd integrl sign cncel out. In other words, differentition nd integrtion re inverse processes doing one opertion nd then the other gets you bck to where you strted. But be creful, the vrible x must be limit of integrtion! Check out one of my fvorite finl exm questions below: Exercise : Clculte The nswer is NOT sin(e x2. ( d 7 sin(e t2 dt dx (v It is prticulrly importnt to understnd the difference between f(x dx nd f(x dx. The limits of integrtion, or lck thereof, re criticl. The f(x dx is clled n indefinite integrl nd represents the generl ntiderivtive of f(x. Here you cn think of the integrl sign s telling you to find the generl ntiderivtive. For exmple, we hve x 2 dx = 3 x3 + c nd cos x dx = sin x + c. The quntity f(x dx is clled the definite integrl, nd represents the signed re under f from to b. The definite integrl is number! Exercises: (problems # 3 focus on prt of the FTC, while the remining problems use prt 2. Use prt of the FTC to compute 2 2x + dx. Check tht your nswer grees with the one you obtined for Exmple.3 on the worksheet for Section 5.2 (The Definite Integrl. 2. Use prt of the FTC to compute Interpret your nswers grphiclly. π sin θ dθ nd 2π sin θ dθ. 2
3 3. Evlute ech integrl using prt of the FTC: ( 5e 3t + t dt (b 3 2 x 5 x 2 dx (c π/2 π/3 sin(3θ + 4 cos(2θ dθ (d 4 x 2 4x + 3 dx Hint: Drw grph of the integrnd nd then brek the integrl up into three pieces without ny bsolute vlue signs. 3
4 x 3 4. If F (x = cos2 (3t + 5 dt, find F (x Find d ( x e t2 dt, nd then find d ( 8 e t2 dt. dx 8 dx x ( 6. Find d x sin(t 2 dt. Hint: Use the chin rule where the inside function is x. dx 4 7. If G(x = 5 x 3 t5 + 5 dt, find G (x. t5 + 5 dt, find H (x. Hint: Brek the integrl into two integrls. Then differ 8. If H(x = entite. e 2x x 3 4
5 9. Suppose tht G(x = x f(t dt, where the grph of f is shown below. ( Find G(, G(2, G(3, G(4 nd G(5. (b Find G (2, G (3 nd G (5. (c Where is G(x concve up? Does G (3 exist? Explin. (d Sketch the grph of G over the intervl x 5. 5
x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
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