Visual Aids For Section 5.3 The Fundamental Theorem Of .
Visual Aids for Section 5.3The Fundamental Theorem of Calculus andInterpretationsMTH 141University of Rhode IslandMTH 141 (URI)Section 5.3
Our outcomesDefinitions: Fundamental Theorem of CalculusRecall and apply the Fundamental Theorem of Calculus, forRbexample, to answer questions about the value of a f (x) dx, or toanswer questions about the amount of change of a function F (x)between x a and x b.MTH 141 (URI)Section 5.3
Review: the definite integral, so farFor continuous f over a t b,Zthe definite integralbf (t) dt is.a.the limit of a Riemann sumapproximations as n .the total signed area between thegraph and the t-axisImages: KSmrq (Wikipedia), CC BY-SA 3.0, adapted bZaf (t) dt lim n MTH 141 (URI)n 1X f (ti ) t .i 0Section 5.3
The TheoremMTH 141 (URI)Section 5.3
The TheoremThe Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenZbf (t) dt F (b) F (a).aMTH 141 (URI)Section 5.3
The TheoremThe Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).aExamples to think abouts(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
The Theorem (and two ways to think about it)The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).a A computational tool for definite integrals A statement about changes(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
The Theorem (and two ways to think about it)The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).a A computational tool for definite integrals A statement about changes(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
One way to think about the Fundamental TheoremA computational tool for definite integralsIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenZbf (t) dt F (b) F (a).aThe idea: if we can find a function F (t) whose derivative is f (t), thenRbwe can determine exactly what a f (t) dt is without needing to use anyRiemann sums!3t 2 (t 3 )01Z0MTH 141 (URI)Section 5.33t 2 dt 13 03
ExerciseZCompute1e1dx.xMTH 141 (URI)Section 5.3
ExerciseZCompute1e1dx.x(Remember,Z bf (t) dt F (b) F (a)aif F 0 (t) f (t).)MTH 141 (URI)Section 5.3
ExerciseeZCompute11dx.x(Remember,Z bf (t) dt F (b) F (a)aif F 0 (t) f (t).)Here, f (x) 1, which is the derivative of F (x) xMTH 141 (URI)Section 5.3.
ExerciseeZCompute11dx.x(Remember,Z bf (t) dt F (b) F (a)aif F 0 (t) f (t).)Here, f (x) 1, which is the derivative of F (x) ln x.xMTH 141 (URI)Section 5.3
ExerciseZCompute1e1dx.x(Remember,Z bf (t) dt F (b) F (a)aif F 0 (t) f (t).)1, which is the derivative of F (x) ln x.x1Since F 0 (x) (ln x)0 f (x), the Fundamental Theorem of Calculus tells us thatZ e x1dx ln e ln 11 xHere, f (x) MTH 141 (URI)Section 5.3
ExerciseZCompute1e1dx.x(Remember,Z bf (t) dt F (b) F (a)aif F 0 (t) f (t).)1, which is the derivative of F (x) ln x.x1Since F 0 (x) (ln x)0 f (x), the Fundamental Theorem of Calculus tells us thatZ e x1dx ln e ln 1 1 0 1.1 xHere, f (x) MTH 141 (URI)Section 5.3
Exercise(a) Find a function F (t) for which 4t 3 10t is the derivative.(b) Use the Fundamental Theorem of Calculus and your answer toZ 2part (a) to find(4t 3 10t) dt. 1MTH 141 (URI)Section 5.3
Exercise(a) Find a function F (t) for which 4t 3 10t is the derivative.(b) Use the Fundamental Theorem of Calculus and your answer toZ 2Z b(Remember,f (t) dt F (b) F (a).)part (a) to find(4t 3 10t) dt.a 1MTH 141 (URI)Section 5.3
Exercise(a) Find a function F (t) for which 4t 3 10t is the derivative.(b) Use the Fundamental Theorem of Calculus and your answer toZ 2Z b(Remember,f (t) dt F (b) F (a).)part (a) to find(4t 3 10t) dt.a 1(a) With some guessing, we arrive at F (t) t 4 5t 2 , since 0t 4 5t 2 4t 3 10t.MTH 141 (URI)Section 5.3
Exercise(a) Find a function F (t) for which 4t 3 10t is the derivative.(b) Use the Fundamental Theorem of Calculus and your answer toZ 2Z b(Remember,f (t) dt F (b) F (a).)part (a) to find(4t 3 10t) dt.a 1(a) With some guessing, we arrive at F (t) t 4 5t 2 , since 0t 4 5t 2 4t 3 10t.(b) By the Fundamental Theorem of Calculus, we know thatZ 2(4t 3 10t)dt F (2) F ( 1) 1 24 5 · 22 ( 1)4 5 · ( 1)2.MTH 141 (URI)Section 5.3
Exercise(a) Find a function F (t) for which 4t 3 10t is the derivative.(b) Use the Fundamental Theorem of Calculus and your answer toZ 2Z b(Remember,f (t) dt F (b) F (a).)part (a) to find(4t 3 10t) dt.a 1(a) With some guessing, we arrive at F (t) t 4 5t 2 , since 0t 4 5t 2 4t 3 10t.(b) By the Fundamental Theorem of Calculus, we know thatZ 2(4t 3 10t)dt F (2) F ( 1) 1 24 5 · 22 ( 1)4 5 · ( 1)2 (16 20) (1 5) 30.MTH 141 (URI)Section 5.3
The Theorem (and two ways to think about it)The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).a A computational tool for definite integrals A statement about changes(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
The Theorem (and two ways to think about it)The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).a A computational tool for definite integrals A statement about changes(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
A statement about changeWhy is the Fundamental Theorem true?MTH 141 (URI)Section 5.3
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t bMTH 141 (URI)Section 5.3
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t b n Xchange in F overi 1MTH 141 (URI)the i th subdivisionof [a,b]Section 5.3
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t b n Xchange in F overi 1the i th subdivisionof [a,b] n Xrate of changei 1of F in thei th subdivisionlength ofsubdivision Approximation improves as t 0, i.e., as n MTH 141 (URI)Section 5.3
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t b n Xchange in F overi 1the i th subdivisionof [a,b] n Xrate of changei 1of F in thei th subdivisionlength ofsubdivision Approximation improves as t 0, i.e., as n F (b) F (a) nXi 1MTH 141 (URI)Section 5.3F 0 (ti ) t
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t b n Xchange in F overi 1the i th subdivisionof [a,b] n Xrate of changei 1of F in thei th subdivisionlength ofsubdivision Approximation improves as t 0, i.e., as n F (b) F (a) limn MTH 141 (URI)Section 5.3nXi 1F 0 (ti ) t
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t b n Xchange in F overi 1the i th subdivisionof [a,b] n Xrate of changei 1of F in thei th subdivisionlength ofsubdivision Approximation improves as t 0, i.e., as n F (b) F (a) limn ZF (b) F (a) bf (t) dtaMTH 141 (URI)Section 5.3nXF 0 (ti ) ti 1if F 0 (t) f (t).
A statement about changeWhy is the Fundamental Theorem true?Change in Ffrom t a to t b n Xchange in F overi 1the i th subdivisionof [a,b] n Xrate of changei 1of F in thei th subdivisionlength ofsubdivision Approximation improves as t 0, i.e., as n F (b) F (a) limn ZF (b) F (a) bf (t) dtnXF 0 (ti ) ti 1if F 0 (t) f (t).aTotal changein quantityMTH 141 (URI)Z ab rate of changeof quantitySection 5.3dt
ExamplesTotal changein quantityZ ab rate of changeof quantitydtRecall that acceleration is the derivative of velocity. If a(t) is theacceleration an object undergoes at time t (where t is measured inseconds and a(t) is measured in meters/second2 ), what doesZ 10a(t) dt represent?0MTH 141 (URI)Section 5.3
ExamplesTotal changein quantityZ ab rate of changeof quantitydtRecall that acceleration is the derivative of velocity. If a(t) is theacceleration an object undergoes at time t (where t is measured inseconds and a(t) is measured in meters/second2 ), what doesZ 10a(t) dt represent?0ZSince a(t) is the rate of change of velocity,change in velocity between t 0 to t 10.MTH 141 (URI)Section 5.310a(t) dt equals the total0
ExamplesTotal changein quantityZ ab rate of changeof quantitydtA bacterial population starts at time t 0 with 200 cells and grows at arate of g(t) cells per hour, where t is measured in hours. Write anexpression (including a definite integral) that shows how large thepopulation will be five hours later.MTH 141 (URI)Section 5.3
ExamplesTotal changein quantityZ ab rate of changeof quantitydtA bacterial population starts at time t 0 with 200 cells and grows at arate of g(t) cells per hour, where t is measured in hours. Write anexpression (including a definite integral) that shows how large thepopulation will be five hours later.Z 5Since g(t) is the rate of change of population,g(t) dt equals the0total change in population between t 0 and t 5.MTH 141 (URI)Section 5.3
ExamplesTotal changein quantityZ ab rate of changeof quantitydtA bacterial population starts at time t 0 with 200 cells and grows at arate of g(t) cells per hour, where t is measured in hours. Write anexpression (including a definite integral) that shows how large thepopulation will be five hours later.Z 5Since g(t) is the rate of change of population,g(t) dt equals the0total change in population between t 0 and t 5. Since at t 0 thepopulation was 200, at t 5 the population is found byZ 5change ininitialPopulation population population 200 g(t) dt.0MTH 141 (URI)Section 5.3
A word about unitsTotal changein quantityMTH 141 (URI)Z ab rate of changeof quantitySection 5.3dt
A word about unitsTotal changein quantityChange invelocityLeft side: meters/second;MTH 141 (URI)bZ a rate of changeof quantitydt10Za(t) dt 0Right side: a(t) has units meters/second2Section 5.3
A word about unitsTotal changein quantityChange invelocity aLeft side: cells;MTH 141 (URI) rate of changeof quantitydt10Za(t) dt 0Right side: a(t) has units meters/second2Left side: meters/second;Change inpopulationbZZ5g(t) dt 0Right side: g(t) has units cells/hourSection 5.3
A word about unitsTotal changein quantityChange invelocity aLeft side: cells; rate of changeof quantitya(t) dt 0Right side: a(t) has units meters/second25Zg(t) dt 0Right side: g(t) has units cells/hourbZFact: The definite integralf (x) dx will have the units of f (x)amultiplied by the units of x.MTH 141 (URI)dt10ZLeft side: meters/second;Change inpopulationbZSection 5.3
The Theorem (and two ways to think about it)The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).a A computational tool for definite integrals A statement about changes(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
The Theorem (and two ways to think about it)The Fundamental Theorem of CalculusIf f is continuous on the interval [a, b] and f (t) F 0 (t), thenbZf (t) dt F (b) F (a).a A computational tool for definite integrals A statement about changes(t) positionv (t) velocity3t 2 (t 3 )0Z12333t dt 1 00MTH 141 (URI)Zbv (t) dt s(b) s(a)a change in positionSection 5.3
ExerciseThe rate, S(t), at which a company sells its widgets is approximated byS(t) 1.32e0.001t , where S(t) is measured in millions of widgets per year and t ismeasured in years since the start of January 2000. We want to know how much thecompany sold between 2000 and the start of 2020.(a) Express the amount as a definite integral.(b) Can you evaluate the definite integral from part (a)?MTH 141 (URI)Section 5.3
ExerciseThe rate, S(t), at which a company sells its widgets is approximated byS(t) 1.32e0.001t , where S(t) is measured in millions of widgets per year and t ismeasured in years since the start of January 2000. We want to know how much thecompany sold between 2000 and the start of 2020.(a) Express the amount as a definite integral.(b) Can you evaluate the definite integral from part (a)?(a) We know thatTotal changein quantitybZ rate of changeof quantitydtaHere let’s use “number of units sold” as the “quantity”. The number of units soldbetween 2000 and 2020 is thenZ 20Z 20S(t) dt 1.32e0.001t dt.0MTH 141 (URI)0Section 5.3
ExerciseThe rate, S(t), at which a company sells its widgets is approximated byS(t) 1.32e0.001t , where S(t) is measured in millions of widgets per year and t ismeasured in years since the start of January 2000. We want to know how much thecompany sold between 2000 and the start of 2020.(a) Express the amount as a definite integral.(b) Can you evaluate the definite integral from part (a)?(b) Comparing this definite integral with the Fundamental Theorem—Z201.32e0.001t dt0Zbf (t) dt F (b) F (a)if F 0 (t) f (t),a—we see that we need a function F (t) whose derivative is 1.32e0.001t .MTH 141 (URI)Section 5.3
ExerciseThe rate, S(t), at which a company sells its widgets is approximated byS(t) 1.32e0.001t , where S(t) is measured in millions of widgets per year and t ismeasured in years since the start of January 2000. We want to know how much thecompany sold between 2000 and the start of 2020.(a) Express the amount as a definite integral.(b) Can you evaluate the definite integral from part (a)?(b) Comparing this definite integral with the Fundamental Theorem—Z201.32e0.001t dt0Zbf (t) dt F (b) F (a)if F 0 (t) f (t),a—we see that we need a function F (t) whose derivative is 1.32e0.001t . Making some1.32 0.001tguesses (and checking!), we settle on F (t) 0.001e.MTH 141 (URI)Section 5.3
ExerciseThe rate, S(t), at which a company sells its widgets is approximated byS(t) 1.32e0.001t , where S(t) is measured in millions of widgets per year and t ismeasured in years since the start of January 2000. We want to know how much thecompany sold between 2000 and the start of 2020.(a) Express the amount as a definite integral.(b) Can you evaluate the definite integral from part (a)?(b) Comparing this definite integral with the Fundamental Theorem—Z201.32e0.001t dt0Zbf (t) dt F (b) F (a)if F 0 (t) f (t),a—we see that we need a function F (t) whose derivative is 1.32e0.001t . Making some1.32 0.001tguesses (and checking!), we settle on F (t) 0.001e. ThenZ201.32e0.001t dt F (20) F (0) 01.32 0.001·201.32 0e e0.0010.001 1346.67 1320 26.67.MTH 141 (URI)Section 5.3
ExerciseThe rate, S(t), at which a company sells its widgets is approximated byS(t) 1.32e0.001t , where S(t) is measured in millions of widgets per year and t ismeasured in years since the start of January 2000. We want to know how much thecompany sold between 2000 and the start of 2020.(a) Express the amount as a definite integral.(b) Can you evaluate the definite integral from part (a)?(b) Comparing this definite integral with the Fundamental Theorem—Z201.32e0.001t dt0Zbf (t) dt F (b) F (a)if F 0 (t) f (t),a—we see that we need a function F (t) whose derivative is 1.32e0.001t . Making some1.32 0.001tguesses (and checking!), we settle on F (t) 0.001e. ThenZ201.32e0.001t dt F (20) F (0) 01.32 0.001·201.32 0e e0.0010.001 1346.67 1320 26.67.This means that the company sold about 26.67 millions of widgets from 2000 to 2020.MTH 141 (URI)Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?MTH 141 (URI)Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,Z 2g(2) g(0) g 0 (t) dt0MTH 141 (URI)Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,ZZ 20g(2) g(0) g (t) dt 00MTH 141 (URI)Section 5.32cos(t 3 ) dt.
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,ZZ 20g(2) g(0) g (t) dt 00MTH 141 (URI)Section 5.32cos(t 3 ) dt.
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,ZZ 20g(2) g(0) g (t) dt 2cos(t 3 ) dt.00Since from the graph it looks like the graph’s area above the t-axis isgreater than the area below the axis, g(2) g(0) is positive.MTH 141 (URI)Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,ZZ 20g(2) g(0) g (t) dt 2cos(t 3 ) dt.00Since from the graph it looks like the graph’s area above the t-axis isgreater than the area below the axis, g(2) g(0) is positive.meaningthat g(2) is bigger than g(0).MTH 141 (URI)Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,Z 2Z0g(2) g(0) g (t) dt 0MTH 141 (URI)0Section 5.32cos(t 3 ) dt.
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,Z 2Z0g(2) g(0) g (t) dt 0Now g(2) g(0) MTH 141 (URI)R20cos(t 3 )0dtSection 5.32cos(t 3 ) dt.
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,Z 2Z0g(2) g(0) g (t) dt 0Now g(2) g(0) MTH 141 (URI)R20cos(t 3 )0dt .?Section 5.32cos(t 3 ) dt.
Z2How do we computecos(t 3 ) dt?0We could use Riemann sums.MTH 141 (URI)Section 5.3
ZHow do we compute2cos(t 3 ) dt?0We could use Riemann sums.or we could use a tool likeMathematica or WolframAlpha to help us.A screenshot from wolframalpha.com:MTH 141 (URI)Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph shown at the right,and g(0) 3. Based on the graphat the right, is which is larger, g(0)or g(2)?Can you estimate g(2) to twodecimal places?By the Fundamental Theorem of Calculus,Z 2Z0g(2) g(0) g (t) dt 0Now g(2) g(0) MTH 141 (URI)R20cos(t 3 )2cos(t 3 ) dt.0dt 3 0.855475.,Section 5.3
A difficult functionSuppose that g 0 (t) cos(t 3 ), whichhas the graph
Visual Aids for Section 5.3 The Fundamental Theorem of Calculus and Interpretations MTH 141 University of Rhode Island MTH 141 (URI) Section 5.3. Our outcomes Definitions: Fundamental Theorem of Calculus Recall and apply the Fundamental Theorem of Calculus
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