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 non-liner, 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:
Integrls - Liner Functions If f(x) is liner, we cn still use geometry Also for f(x) = 2 x 2
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
Exmple - Right Endpoints
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 i-th sub-intervl
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 end-point of the i-th subintervl In the limiting cse, we cn choose ny point in ech sub-intervl! The method of Riemnn sums is completely generl Cn be used with generl f(x) (not just liner)
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
Riemnn Sum - Exmple
Sigm Nottion The Riemnn sum pproximtion leds to expressions of long sums To simplify the process we introduce the sigm (Σ) nottion
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
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
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?
Useful Sums Using different pproch (mthemticl induction) we get similr expressions for the sum of squres nd cubes 1. 2. 3. n k=1 n k=1 k =1 + 2 + 3 + 4 + 5 +... + n = n(n+1) 2 k 2 =1 + 4 + 9 + 16 + 25 +... + n 2 = n(n+1)(2n+1) 6 n k 3 =1 + 8 + 27 + 64 + 125 +... + n 3 = k=1 ( ) n(n+1) 2 2
Integrls - Definite Integrl The re bounded by f(x), the x-xis, 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 jump-discontinuities in x b The bove gives useful reltion for breking up integrls!
Summtion Exmples Ex: If I 1 = I 3 = ) 9 4 9 2 4 0 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
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 =
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: 1. 2. 3. 4. 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
Integrl Exmples Ex: If I 1 = I 3 = ) 9 4 2 0 4 0 r(q)dq = 10, I 2 = r(x)dx = 2 nd I 4 = r(q)dq 4 2 9 0 r(m)dm = 1, p(y)dy = 3 clculte the following: b) 9 0 [9r(z) 4p(z)] dz
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 2 2 4 r 2 dr f(t)dt b f(t)dt
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 x-xis, x = nd x = b
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
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!
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
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!
FTC II - Exmple Ex: Use FTC II to evlute the following integrls: ) 3 0 (t 2 + 1)dt
FTC II - Exmple Ex: Use FTC II to evlute the following integrls: b) π 4 π sin(r)dr
FTC II - Exmple Ex: Use FTC II to evlute the following integrls: 8 dt c) 1 + t 2
Integrls We do integrls by recognizing the integrnd s the derivtive of known function Ex: (3t 15) dt Ex: 9 1 1 q 2dq Ex: ds s
Integrls We do integrls by recognizing the integrnd s the derivtive of known function
Integrls - Substitution Wht bout other integrls? Ex: tn θdθ
Integrls - Substitution Wht bout other integrls? t Ex: 1 t 2 dt
Integrls - Substitution Wht bout other integrls? 2t Ex: 9 dt
Integrls - Substitution Wht bout other integrls? Ex: x 2x 9 dx
Integrls - Substitution Wht bout other integrls? 1 Ex: e r + e rdr