riemann sums, definite integral - utah state universityrheal/math1210/lecture_notes/... · 2015....
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RiemannSums,DefiniteIntegral
Howshouldweapproximatewithareasofrectangles?1. Weneedtopartitiontheinterval[ a ,b ]intosmallsubintervals.
2. Wemustthenusethefunctionftodeterminetheheightofeachrectangleanddecidewhethertocounttheareapositivelyornegatively.
DefinitionApartitionof[ a ,b ]isasetofpoints{ x0 , x1 , x2 , x3 ... xk−1 , xk , ..., xn−1 , xn } suchthata= x0 < x1 < x2 < x3 < ... < xk−1 < xk < ...< xn−1 < xn = b .
Examplesofpartitions:Ourgoalistoapproximateareaswitheverincreasingdegreeofaccuracy;sowewillwantourpartitionstodefinealargenumberofsubintervalswithsmallwidth.IfPisapartitionof[ a ,b ]thendeterminehowgoodthepartitionisbyconsideringthelengthofthelargestsubinterval.Somemorenotation:LetP = { x0 , x1 , x2 , x3 ... xk−1 , xk , ..., xn−1 , xn } beapartitionof[ a ,b ].Soa= x0 < x1 < x2 < x3 < ... < xk−1 < xk < ...< xn−1 < xn = b .Thesubintervalsare[ x0 , x1 ],[ x1 , x2 ],[ x2 , x3 ], ...,[ xk−1 , xk ], ...,[ xn−1 , xn ].LetΔ x1 = x1 − x0 , Δ x2 = x2 − x1 , ... , Δ xk = xk − xk−1 , ... , Δ xn = xn − xn−1
Thelengthofthelargestsubintervalisequaltomax{Δ x1 ,Δ x2 , ..., Δ xk , ... , Δ xn }
andisdenotedby P ,calledthenormofthepartitionP.
Ifwewantourapproximationtobeaccurate,thenwewantP tobesmall(closetozero).Fortheheightsoftherectangleswewillchooseapointck from each[ xk−1 , xk ] and evaluate f ( ck ) .WenowobtainaRiemannsum:
f ( ck )Δkk=1
n∑ .Whathappenswhen P → 0 ?
DefinitionSupposefisacontinuousfunctionon[ a ,b ].Thedefiniteintegraloffover[ a ,b ]is
limP →0
f ( ck )Δkk=1
n∑ andisdenotedby f (x) dx
a
b∫ .
Weread“theintegralofffromatobwithrespecttox”.
Formally,
limP →0
f ( ck )Δkk=1
n∑ = L means
for each ε > 0 , there exists δ > 0 such that
f ( ck )Δkk=1
n
∑ − L < ε whenever P < δ
Aslongasthenormofthepartitionissmallenough(normlessthandelta),itdoesn’tmatterwhatpointyouchoosefromeachsubinterval.TheRiemannsumwillbewithinepsilonofL.
Supposefisacontinuousfunctionon[ a ,b ].Theaveragevalueoffon[ a ,b ]canbecomputedintermsofadefiniteintegral.
Averagevalueoffon[ a ,b ]isequalto 1b − a
f (x)dxa
b∫
Howdowecompute f (x) dxa
b∫ ?
Wearegoingtoshowthatiffiscontinuouson[ a ,b ],then
f (x) dxa
b∫ = F(x)
b
a= F(b)− F(a) whereFisany
antiderivativeoff.
Let’s find the area below the graph of y= f (x) = 1− x2 between x = 0 and x =1 .
(1− x2 ) dx01∫ = x − x
3
3⎛
⎝⎜
⎞
⎠⎟1
0= 23
FundamentalTheoremofCalculus
I.Supposefisacontinuousfunctionon[ a ,b ].LetF (x)= f (t)dt
a
x∫ for a≤ x ≤b .ThenF '(x) = f (x) foreachx.
AlsonotethatF (b)= f (x)dxa
b∫ .
Proof:
F '(x)= limh→0
F (x+h)− F (x)h = lim
h→0
f (t)dta
x+h∫ − f (t)dt
a
x∫
h
limh→0
f (t)dtx
x+h∫
h = limh→0
f (th)(h)h = lim
h→0f (th) where x ≤ th ≤ x+h .
As h→ 0 , th→ x , so limh→0
f (th)= f (x) and weget F '(x)= f (x).
II.Supposefisacontinuousfunctionon[ a ,b ].LetG(x) beanyantiderivativeof f (x) .Then
f (x)dxa
b∫ = G(x)
b
a= G(b)−G(a) .
Proof.
IfF (x)= f (t)dta
x∫ for a≤ x ≤b thenGandFdifferbyatmost
aconstantandwehave
G(x)= F(x)+ c for a≤ x ≤b .Now,
G(a)= F(a)+ c = 0+ c = c , and G(b) = F(b)+ c = F(b)+G(a)
WenowhaveG(b)−G(a) = F (b) = f (x)dxa
b∫ .
Evaluatethefollowingdefiniteintegrals:
a) sinxdx0π∫
b) ( 3x2 + 4cos2x )dx0
π /4∫
c) t et2dt03∫
d)01∫ 61+4x2 dx
e)01/2∫ 7
1− x2dx