Chapter 8.2: The Integral


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1 Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in the differentil prt of the house which contined the mthemtics for descriing how things chnge or grow t specific point or moment in time, i.e. slope. In integrl clculus, we re interested in finding things like res nd volumes of irregulr ojects. These two seemingly completely different ides, slopes nd res, re connected y the sme ide of the limit, which you ll eplore more net yer in clculus course. Ich in Hppy. It is the nture of mthemtics to use things for which we hve keen nd thorough understnding to help us eplin nd develop tools for things we don t yet fully understnd. Just s we used lines to help us tlk out how curves chnge in differentil clculus, similrly with integrl clculus, we will use the res of simple oject like rectngle to help us find res of irregulr regions. This method is of prtitioning region into rectngles nd dding up ll the res is clled the method of Riemnn Sums, fter Germn Mthemticin Georg Friedrich Bernhrd Riemnn. Here s the Ide: Emple 1: Identify the region ounded y the function f, the is, nd the verticl line 4. Prtition this region into 4 equl widths. By using oth the left nd right endpoints of ech intervl, drw 4 rectngles of equl width. In ech cse, find the sum of the res of the 4 rectngles to pproimte the re of the region. Compre this with the ctul re. How cn we mke the pproimtion of rectngulr res get closer to the ctul re?? Pge 1 of 6
2 Emple : Find the re of the region ounded y f, the is, nd the verticl lines nd, using four suintervls of equl width using oth Left nd Right endpoint Riemnn Sums. Using this Jv Applet, determine the limit of the upper nd lower sums s nd n. We cn void this finite, numeric pproimtion process y using the Fundmentl Theorem of Clculus. Not only will this theorem llow us to compute the re y single clcultion, ut it will give us the ect re s well, not n pproimtion. Here s the essence of the Fundmentl Theorem of Clculus. The Fundmentl Theorem of Clculus (FTOC) Suppose we hve function ounded y the is from to. y f. We re interested in finding the re under the curve of y f, 1. Find function F such tht F f. F is clled n ntiderivtive of. Then the re of the region is simply F F. Amzing, mthemgicl result, eh? f. Pge of 6
3 Emple 3: Find the re of the region from Emples 1 nd using the FTOC. The process in Emple 3 hs precise mthemticl nottion. We cn rewrite the instructions from Emple 3 in the following wy: 4 d nd d We red this first one s the integrl of from zero to four with respect to. The integrl symol, not surprisingly, resemles the letter S for... Sum!... the sum of infinitely mny res of infinitely mny rectngles. It looks like this mthemticlly: lim f f d The right side of this eqution is known s definite integrl, which gives definite nswer which, in the emples we re working, represents the re of the region ounded y the grph of f, the is from to. There is n intermedite step involved when using this new, concise nottion. f d F F F Pge 3 of 6 When r, or Evlution r
4 Emple 4: Evlute sin d, then illustrte wht this represents geometriclly. Emple 5: Evlute ln 4 e d, then illustrte wht this represents geometriclly. Emple 6: Evlute 1 d for, then illustrte wht this represents geometriclly. So fr we hve limited our study of re to regions ABOVE the is. These hve een positive numers (s re CANNOT e negtive, right?). The process in the ove emples using the integrl symol is clled, in generl, integrtion. When integrting from left to right, regions BELOW the is re counted negtively. Pge 4 of 6
5 Emple 7: Evlute () geometriclly. sin d, () 3 / sin d, nd (c) 5 / sin d, then illustrte wht these represent To understnd why nd how the FTOC works, you ll hve to wit until net yer (or you cn sk Khn or Google it.) Here, though, is n importnt piece to understnding why it works. Question: If f g, does it necessrily follow tht Emple 8: List severl functions whose derivtive is 3. f g? **In generl, if F f, then f d F C, where C is some constnt. Tht is, if two functions hve the sme derivtive, then those two functions differ only y constnt C. Note 1: Note : f d is known s n indefinite integrl, since its nswer is n indefinite, vrile nswer. F C is known s the generl ntiderivtive, s opposed to n ntiderivtive. Pge 5 of 6
6 Emple 9: 4 Evlute () 3 d nd () 3 d Trying to find ntiderivtives intuitively cn sometimes e n rduous tsk. Luckily, we cn use our knowledge of finding derivtives using the power rule (multiply, then sutrct) in reverse order using inverse opertions (dd, then divide) to find ntiderivtives. The Power Rule for ntidifferentition (or integrtion) n1 n d C n 1 As it ws for differentition, similr rule pply for integrtion rewriting prior to using the rule is often the key. The integrl of the sum is the sum of the integrls Emple 1: 7 Evlute () 3 d () d (c) 3 1 5sin 3 d Emple 11: Evlute () 3 3 d () 3 4 cos e d Pge 6 of 6
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