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