Integrals  Motivation


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1 Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but bove the x xis) If f(x) is constnt, we cn resort to some geometry Clerly the nswer depends on the two bounding points! Nottion:
2 Integrls  Liner Functions If f(x) is liner, we cn still use geometry Also for f(x) = 2 x 2
3 Exmple Ex: Approximte the re under f(x) = 1 x between = 0 nd b = 1 using the pproximtion ) with 5 rectngles, b) with 10 rectngles
4 Exmple  Right Endpoints
5 Integrls  Approximtion By incresing the number of rectngles, we get better pproximtion! Define I n  the pproximted re (using n rectngles) Width of ech rectngle becomes x = b n Height of ech rectngle is f(c i ) for some c i in the ith subintervl
6 Integrls  Riemnn Sum We cn imgine tking the number of pproximting rectngles to be extremely lrge The resulting quntity is clled the Riemnn sum NOTE: We rbitrrily chose c i s the left endpoint of the ith subintervl In the limiting cse, we cn choose ny point in ech subintervl! The method of Riemnn sums is completely generl Cn be used with generl f(x) (not just liner)
7 Riemnn Sum  Exmple Ex: Approximte the re under f(x) = 1 x 2 between = 0 nd b = 1 by using the Riemnn sum pproximtion in the lrge n limit
8 Riemnn Sum  Exmple
9 Sigm Nottion The Riemnn sum pproximtion leds to expressions of long sums To simplify the process we introduce the sigm (Σ) nottion
10 Sigm Nottion  Terms nd Indices Sigm nottion llows for gret del of vriety nd flexibility Ex: Write down the sum of the first 10 odd integers in 3 different wys
11 Rules for Finite Sums Using the usul rules of lgebr, we get the following rules for sums: 1. n 1 = n 2. k=1 n (c k ) = c n k 3. k=1 n ( k + b k ) = k=1 n k + n b k 4. k=1 n ( k b k ) = k=1 n k  k=1 n b k k=1 k=1 k=1
12 Useful Sums Severl types of sums deserve specil mention Ex: Sum the positive integers up to ) 5 b) 7 c) 10 Cn we find generl formul?
13 Useful Sums Using different pproch (mthemticl induction) we get similr expressions for the sum of squres nd cubes n k=1 n k=1 k = n = n(n+1) 2 k 2 = n 2 = n(n+1)(2n+1) 6 n k 3 = n 3 = k=1 ( ) n(n+1) 2 2
14 Integrls  Definite Integrl The re bounded by f(x), the xxis, nd the lines x = nd x = b If this limit exists, we sy tht f(x) is integrble on x b We will find tht f(x) is integrble on x b whenever:  f(x) is continuous on x b  f(x) hs finite number of jumpdiscontinuities in x b The bove gives useful reltion for breking up integrls!
15 Summtion Exmples Ex: If I 1 = I 3 = ) r(q)dq = 10, I 2 = 4 2 r(m)dm = 1, r(x)dx = 2 clculte the following: r(q)dq b) 9 0 r(z)dz
16 Integrls  Definite Integrl If f(x) > 0, we interpret the integrl s the re under the curve Wht hppens if f(x) < 0? Wht if > b in our integrl? Using the previous results we cn check wht hppens when b =
17 Integrls  Properties These results llow us to write the following reltions for integrls: For ny f(x) nd g(x) tht re continuous on x b nd constnt c we hve tht: b b b b c dx = c (b ) [f(x) + g(x)] dx = cf(x) dx = c b [f(x) g(x)] dx = 5. If m f(x) M, then m (b ) b b f(x) dx b b f(x) dx + f(x) dx  b f(x) dx M (b ) g(x) dx g(x) dx
18 Integrl Exmples Ex: If I 1 = I 3 = ) r(q)dq = 10, I 2 = r(x)dx = 2 nd I 4 = r(q)dq r(m)dm = 1, p(y)dy = 3 clculte the following: b) 9 0 [9r(z) 4p(z)] dz
19 Averge Vlues The verge vlue of n integrl is defined s v(f) = 1 b b Interprettion: height of rectngle with the sme re s Ex: Clculte the verge vlue of r 2 dr f(t)dt b f(t)dt
20 Indefinite Integrls Recll tht f () is just number (slope of the tngent to f(x) t x = ) In similr wy, the definite integrl of f(x) between x = nd x = b is just number Net re bounded by f(x), the xxis, x = nd x = b
21 Indefinite Integrls To get the derivtive function, f (x), we left the point unspecified! We get the integrl function by leving the limits unspecified The resulting construct is known s the indefinite integrl
22 Fundmentl Theorem of Clculus I If F (x) = x f(t) dt, then we sy tht F (x) is the ntiderivtive of f(x) The Fundmentl Theorem of Clculus (FTC) reltes these two objects from integrl nd differentil clculus 1) Suppose f(x) is continuous function on x b. If F (x) = then F (x) = d dx x x f(t)dt, f(t) dt = f(x) This gives us reltionship between the indefinite integrl nd the derivtive funciton! Impliction: integrls re the inverses of derivtives!
23 FTC I  Exmple Ex: Use FTC I to clculte df ) f(x) = x (t 3 + 1)dt dx in ech cse: b) f(x) = 3 4x sin(r 2 )dr c) f(x) = sin(x) 0 dt 1 t 2
24 Fundmentl Theorem of Clculus II If F (x) = x f(t) dt, then we sy tht F (x) is the ntiderivtive of f(x) The Fundmentl Theorem of Clculus (FTC) reltes these two objects from integrl nd differentil clculus 2) Suppose f(x) is continuous function on x b nd F (x) is n ntiderivtive of f(x), then b f(t)dt = b df dt dt = F (b) F () This gives us reltionship between the definite integrl nd the derivtive t point! Impliction: integrls re the inverses of derivtives!
25 FTC II  Exmple Ex: Use FTC II to evlute the following integrls: ) 3 0 (t 2 + 1)dt
26 FTC II  Exmple Ex: Use FTC II to evlute the following integrls: b) π 4 π sin(r)dr
27 FTC II  Exmple Ex: Use FTC II to evlute the following integrls: 8 dt c) 1 + t 2
28 Integrls We do integrls by recognizing the integrnd s the derivtive of known function Ex: (3t 15) dt Ex: q 2dq Ex: ds s
29 Integrls We do integrls by recognizing the integrnd s the derivtive of known function
30 Integrls  Substitution Wht bout other integrls? Ex: tn θdθ
31 Integrls  Substitution Wht bout other integrls? t Ex: 1 t 2 dt
32 Integrls  Substitution Wht bout other integrls? 2t Ex: 9 dt
33 Integrls  Substitution Wht bout other integrls? Ex: x 2x 9 dx
34 Integrls  Substitution Wht bout other integrls? 1 Ex: e r + e rdr
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