lesson 25: the definite integral

Section 5.2The Definite Integral

V63.0121.002.2010Su, Calculus I

New York University

June 17, 2010

Announcements

I

. . . . . .

. . . . . .

Announcements

I

V63.0121.002.2010Su, Calculus I (NYU) Section 5.2 The Definite Integral June 17, 2010 2 / 32

. . . . . .

Objectives

I Compute the definiteintegral using a limit ofRiemann sums

I Estimate the definiteintegral using a Riemannsum (e.g., Midpoint Rule)

I Reason with the definiteintegral using itselementary properties.


. . . . . .

Outline

Recall

The definite integral as a limit

Estimating the Definite Integral

Properties of the integral

Comparison Properties of the Integral


. . . . . .

Cavalieri's method in general

Let f be a positive function defined on the interval [a,b]. We want tofind the area between x = a, x = b, y = 0, and y = f(x).For each positive integer n, divide up the interval into n pieces. Then

∆x =b− an

. For each i between 1 and n, let xi be the ith step betweena and b. So

. .x..x0

..x1

..xi

..xn−1

..xn.. . . .. . .

x0 = a

x1 = x0 +∆x = a+b− an

x2 = x1 +∆x = a+ 2 · b− an

. . .

xi = a+ i · b− an

. . .

xn = a+ n · b− an

= b


. . . . . .

Forming Riemann sums

We have many choices of representative points to approximate thearea in each subinterval.

left endpoints…

Ln =n∑

i=1

f(xi−1)∆x

. .x. . . . . . .

In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .



right endpoints…

Rn =n∑

i=1

f(xi)∆x

. .x. . . . . . .


Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .



midpoints…

Mn =n∑

i=1

f(xi−1 + xi

2

)∆x

. .x. . . . . . .


Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .



the minimum value on theinterval…

. .x. . . . . . .


Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .



the maximum value on theinterval…

. .x. . . . . . .


Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .



…even random points!

. .x. . . . . . .


Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .



…even random points!

. .x. . . . . . .In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum

Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑

i=1

f(ci)∆x


. . . . . .

Theorem of the (previous) Day

Theorem

If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then

limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}

exists and is the same value nomatter what choice of ci we make. .

.

.

.x


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L1 = 3.0


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L2 = 5.25


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L3 = 6.0


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L4 = 6.375


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L5 = 6.59988


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L6 = 6.75


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L7 = 6.85692


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L8 = 6.9375


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L9 = 6.99985


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L10 = 7.04958


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L11 = 7.09064


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L12 = 7.125


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L13 = 7.15332


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L14 = 7.17819


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L15 = 7.19977


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L16 = 7.21875


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L17 = 7.23508


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L18 = 7.24927


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L19 = 7.26228


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L20 = 7.27443


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L21 = 7.28532


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L22 = 7.29448


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L23 = 7.30406


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L24 = 7.3125


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L25 = 7.31944


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L26 = 7.32559


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L27 = 7.33199


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L28 = 7.33798


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L29 = 7.34372


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.left endpoints

.L30 = 7.34882


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R1 = 12.0


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R2 = 9.75


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R3 = 9.0


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R4 = 8.625


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R5 = 8.39969


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R6 = 8.25


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R7 = 8.14236


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R8 = 8.0625


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R9 = 7.99974


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R10 = 7.94933


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R11 = 7.90868


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R12 = 7.875


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R13 = 7.84541


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R14 = 7.8209


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R15 = 7.7997


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R16 = 7.78125


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R17 = 7.76443


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R18 = 7.74907


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R19 = 7.73572


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R20 = 7.7243


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R21 = 7.7138


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R22 = 7.70335


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R23 = 7.69531


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R24 = 7.6875


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R25 = 7.67934


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R26 = 7.6715


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R27 = 7.66508


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R28 = 7.6592


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R29 = 7.65388


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.right endpoints

.R30 = 7.64864


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M1 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M2 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M3 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M4 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M5 = 7.4998


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M6 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M7 = 7.4996


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M8 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M9 = 7.49977


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M10 = 7.49947


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M11 = 7.49966


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M12 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M13 = 7.49937


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M14 = 7.49954


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M15 = 7.49968


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M16 = 7.49988


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M17 = 7.49974


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M18 = 7.49916


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M19 = 7.49898


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M20 = 7.4994


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M21 = 7.49951


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M22 = 7.49889


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M23 = 7.49962


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M24 = 7.5


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M25 = 7.49939


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M26 = 7.49847


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M27 = 7.4985


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M28 = 7.4986


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M29 = 7.49878


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M30 = 7.49872


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U1 = 12.0


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U2 = 10.55685


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U3 = 10.0379


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U4 = 9.41515


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U5 = 8.96004


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U6 = 8.76895


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U7 = 8.6033


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U8 = 8.45757


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U9 = 8.34564


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U10 = 8.27084


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U11 = 8.20132


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U12 = 8.13838


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U13 = 8.0916


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U14 = 8.05139


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U15 = 8.01364


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U16 = 7.98056


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U17 = 7.9539


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U18 = 7.92815


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U19 = 7.90414


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U20 = 7.88504


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U21 = 7.86737


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U22 = 7.84958


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U23 = 7.83463


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U24 = 7.82187


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U25 = 7.80824


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U26 = 7.79504


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U27 = 7.78429


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U28 = 7.77443


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U29 = 7.76495


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.maximum points

.U30 = 7.7558


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L1 = 3.0


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L2 = 4.44312


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L3 = 4.96208


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L4 = 5.58484


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L5 = 6.0395


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L6 = 6.23103


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L7 = 6.39577


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L8 = 6.54242


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L9 = 6.65381


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L10 = 6.72797


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L11 = 6.7979


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L12 = 6.8616


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L13 = 6.90704


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L14 = 6.94762


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L15 = 6.98575


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L16 = 7.01942


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L17 = 7.04536


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L18 = 7.07005


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L19 = 7.09364


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L20 = 7.1136


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L21 = 7.13155


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L22 = 7.14804


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L23 = 7.16441


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L24 = 7.17812


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L25 = 7.19025


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L26 = 7.2019


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L27 = 7.21265


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L28 = 7.22269


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L29 = 7.23251


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.minimum points

.L30 = 7.24162


. . . . . .

Outline

Recall






. . . . . .


DefinitionIf f is a function defined on [a,b], the definite integral of f from a to bis the number ∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x


. . . . . .

Notation/Terminology

∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫

— integral sign (swoopy S)

I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)

I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature


. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫


I f(x) — integrand

I a and b — limits of integration (a is the lower limit and b theupper limit)



. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫





. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫



I dx — ??? (a parenthesis? an infinitesimal? a variable?)

I The process of computing an integral is called integration orquadrature


. . . . . .


∫ b

af(x)dx = lim

∆x→0

n∑i=1

f(ci)∆x

I∫





. . . . . .

The limit can be simplified

TheoremIf f is continuous on [a,b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a,b]; that is, the definite integral∫ b

af(x) dx exists.

TheoremIf f is integrable on [a,b] then∫ b

af(x)dx = lim

n→∞

n∑i=1

f(xi)∆x,

where∆x =

b− an

and xi = a+ i∆x


. . . . . .

Example: Integral of x

Example

Find∫ 3

0x dx

Solution

For any n we have ∆x =3nand xi =

3in. So

Rn =n∑

i=1

f(xi)∆x =n∑

i=1

(3in

)(3n

)=

9n2

n∑i=1

i

=9n2

· n(n+ 1)2

−→ 92

So∫ 3

0x dx =

92= 4.5


. . . . . .

Example: Integral of x2

Example

Find∫ 3

0x2 dx

Solution


3in. So

Rn =n∑

i=1

f(xi)∆x =n∑

i=1

(3in

)2(3n

)=

27n3

n∑i=1

i2

=27n3

· n(n+ 1)(2n+ 1)6

−→ 273

= 9

So∫ 3

0x2 dx = 9


. . . . . .

Example: Integral of x3

Example

Find∫ 3

0x3 dx

Solution


3in. So

Rn =n∑

i=1

f(xi)∆x =n∑

i=1

(3in

)3(3n

)=

81n4

n∑i=1

i3

=81n4

· n2(n+ 1)2

4−→ 81

4

So∫ 3

0x3 dx =

814

= 20.25


. . . . . .

Outline

Recall






. . . . . .


Example

Estimate∫ 1

0

41+ x2

dx using M4.

Solution

We have x0 = 0, x1 =14, x2 =

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)

=14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

6465

+6473

+6489

+64113

≈ 3.1468


. . . . . .


Example

Estimate∫ 1

0

41+ x2

dx using M4.

Solution

We have x0 = 0, x1 =14, x2 =

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)

=6465

+6473

+6489

+64113

≈ 3.1468


. . . . . .


Example

Estimate∫ 1

0

41+ x2

dx using M4.

Solution

We have x0 = 0, x1 =14, x2 =

12, x3 =

34, x4 = 1.

So c1 =18, c2 =

38, c3 =

58, c4 =

78.

M4 =14

(4

1+ (1/8)2+

41+ (3/8)2

+4

1+ (5/8)2+

41+ (7/8)2

)=

14

(4

65/64+

473/64

+4

89/64+

4113/64

)=

6465

+6473

+6489

+64113

≈ 3.1468


. . . . . .

Estimating the Definite Integral (Continued)

Example

Estimate∫ 1

0

41+ x2

dx using L4 and R4

Answer

L4 =14

(4

1+ (0)2+

41+ (1/4)2

+4

1+ (1/2)2+

41+ (3/4)2

)= 1+

1617

+45+

1625

≈ 3.38118

R4 =14

(4

1+ (1/4)2+

41+ (1/2)2

+4

1+ (3/4)2+

41+ (1)2

)=

1617

+45+

1625

+12≈ 2.88118


. . . . . .

Outline

Recall






. . . . . .


Theorem (Additive Properties of the Integral)

Let f and g be integrable functions on [a,b] and c a constant. Then

1.∫ b

ac dx = c(b− a)

2.∫ b

a[f(x) + g(x)] dx =

∫ b

af(x)dx+

∫ b

ag(x)dx.

3.∫ b

acf(x)dx = c

∫ b

af(x)dx.

4.∫ b

a[f(x)− g(x)] dx =

∫ b

af(x)dx−

∫ b

ag(x)dx.


. . . . . .

Proofs

Proofs.

I When integrating a constant function c, each Riemann sumequals c(b− a).

I A Riemann sum for f+ g equals a Riemann sum for f plus aRiemann sum for g. Using the sum rule for limits, the integral of asum is the sum of the integrals.

I Ditto for constant multiplesI Ditto for differences


. . . . . .

Example

Find∫ 3

0

(x3 − 4.5x2 + 5.5x+ 1

)dx

Solution

∫ 3

0(x3−4.5x2 + 5.5x+ 1)dx

=

∫ 3

0x3 dx− 4.5

∫ 3

0x2 dx+ 5.5

∫ 3

0x dx+

∫ 3

01 dx

= 20.25− 4.5 · 9+ 5.5 · 4.5+ 3 · 1 = 7.5

(This is the function we were estimating the integral of before)


. . . . . .


Theorem


limn→∞

Sn = limn→∞

{ n∑i=1

f(ci)∆x

}


.

.

.x.midpoints

.M15 = 7.49968


. . . . . .

More Properties of the Integral

Conventions: ∫ a

bf(x)dx = −

∫ b

af(x)dx

∫ a

af(x)dx = 0

This allows us to have

5.∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c

bf(x)dx for all a, b, and c.


. . . . . .

Example

Suppose f and g are functions with

I∫ 4

0f(x)dx = 4

I∫ 5

0f(x)dx = 7

I∫ 5

0g(x)dx = 3.

Find

(a)∫ 5

0[2f(x)− g(x)] dx

(b)∫ 5

4f(x)dx.


. . . . . .

SolutionWe have(a) ∫ 5

0[2f(x)− g(x)] dx = 2

∫ 5

0f(x)dx−

∫ 5

0g(x)dx

= 2 · 7− 3 = 11

(b) ∫ 5

4f(x)dx =

∫ 5

0f(x)dx−

∫ 4

0f(x)dx

= 7− 4 = 3


. . . . . .

Outline

Recall






. . . . . .


TheoremLet f and g be integrable functions on [a,b].

6. If f(x) ≥ 0 for all x in [a,b], then∫ b

af(x)dx ≥ 0

7. If f(x) ≥ g(x) for all x in [a,b], then∫ b

af(x)dx ≥

∫ b

ag(x)dx

8. If m ≤ f(x) ≤ M for all x in [a,b], then

m(b− a) ≤∫ b

af(x)dx ≤ M(b− a)


. . . . . .


TheoremLet f and g be integrable functions on [a,b].6. If f(x) ≥ 0 for all x in [a,b], then∫ b

af(x)dx ≥ 0

7. If f(x) ≥ g(x) for all x in [a,b], then∫ b

af(x)dx ≥

∫ b

ag(x)dx

8. If m ≤ f(x) ≤ M for all x in [a,b], then

m(b− a) ≤∫ b



. . . . . .

The integral of a nonnegative function is nonnegative

Proof.If f(x) ≥ 0 for all x in [a,b], then for any number of divisions n andchoice of sample points {ci}:

Sn =n∑

i=1

f(ci)︸︷︷︸≥0

∆x ≥n∑

i=1

0 ·∆x = 0

Since Sn ≥ 0 for all n, the limit of {Sn} is nonnegative, too:∫ b

af(x)dx = lim

n→∞Sn︸︷︷︸≥0

≥ 0


. . . . . .

The definite integral is “increasing"

Proof.Let h(x) = f(x)− g(x). If f(x) ≥ g(x) for all x in [a,b], then h(x) ≥ 0 forall x in [a,b]. So by the previous property∫ b

ah(x)dx ≥ 0

This means that∫ b

af(x)dx−

∫ b

ag(x)dx =

∫ b

a(f(x)− g(x)) dx =

∫ b

ah(x)dx ≥ 0

So ∫ b

af(x)dx ≥

∫ b

ag(x)dx


. . . . . .

Bounding the integral using bounds of the function

Proof.If m ≤ f(x) ≤ M on for all x in [a,b], then by the previous property∫ b

amdx ≤

∫ b

af(x)dx ≤

∫ b

aMdx

By Property ??, the integral of a constant function is the product of theconstant and the width of the interval. So:

m(b− a) ≤∫ b



. . . . . .

Example

Estimate∫ 2

1

1xdx using the comparison properties.

SolutionSince

12≤ x ≤ 1

1for all x in [1,2], we have

12· 1 ≤

∫ 2

1

1xdx ≤ 1 · 1


. . . . . .

Summary

I The definite integral is a limit of Riemann SumsI The definite integral can be estimated with Riemann SumsI The definite integral can be distributed across sums and constantmultiples of functions

I The definite integral can be bounded using bounds for the function


lesson 25: the definite integral

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