lesson 24: areas and distances, the definite integral (handout)
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Sec on 5.1–5.2Areas and Distances, The Definite
Integral
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
April 25, 2011
.
Announcements
I Quiz 5 on Sec ons4.1–4.4 April 28/29
I Final Exam Thursday May12, 2:00–3:50pm
I cumula veI loca on TBDI old exams on common
website
.
Objectives from Section 5.1I Compute the area of a region byapproxima ng it with rectanglesand le ng the size of therectangles tend to zero.
I Compute the total distancetraveled by a par cle byapproxima ng it as distance =(rate)( me) and le ng the meintervals over which oneapproximates tend to zero.
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Notes
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Notes
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Notes
. 1.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Objectives from Section 5.2
I Compute the definite integralusing a limit of Riemann sums
I Es mate the definite integralusing a Riemann sum (e.g.,Midpoint Rule)
I Reason with the definite integralusing its elementary proper es.
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OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
Easy Areas: RectangleDefini onThe area of a rectangle with dimensions ℓ and w is the productA = ℓw.
..ℓ
.
w
It may seem strange that this is a defini on and not a theorem butwe have to start somewhere.
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Notes
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Notes
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Notes
. 2.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Easy Areas: ParallelogramBy cu ng and pas ng, a parallelogram can be made into a rectangle.
..b
.b
.
h
SoFactThe area of a parallelogram of base width b and height h is
A = bh
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Easy Areas: TriangleBy copying and pas ng, a triangle can be made into a parallelogram.
..b
.
h
SoFactThe area of a triangle of base width b and height h is
A =12bh
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Easy Areas: Other PolygonsAny polygon can be triangulated, so its area can be found bysumming the areas of the triangles:
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Notes
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Notes
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Notes
. 3.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Hard Areas: Curved Regions
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???
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Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (a er Euclid)
I GeometerI Weapons engineer
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Archimedes and the Parabola
..
1
.
18
.
18
.
164
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164
.164
.164
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18+ 4 · 1
64+ · · · = 1+
14+
116
+ · · ·+ 14n
+ · · ·
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Notes
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Notes
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Notes
. 4.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Summing a geometric seriesFactFor any number r and any posi ve integer n,
(1− r)(1+ r+ r2 + · · ·+ rn) = 1− rn+1.
Proof.
(1− r)(1+ r+ r2 + · · ·+ rn)= (1+ r+ r2 + · · ·+ rn)− r(1+ r+ r2 + · · ·+ rn)= (1+ r+ r2 + · · ·+ rn)− (r+ r2 + r3 · · ·+ rn + rn+1)
= 1− rn+1
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Summing a geometric seriesFactFor any number r and any posi ve integer n,
(1− r)(1+ r+ r2 + · · ·+ rn) = 1− rn+1.
Corollary
1+ r+ · · ·+ rn =1− rn+1
1− r
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Summing the series
We need to know the value of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
Using the corollary,
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=
43as n → ∞.
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Notes
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Notes
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Notes
. 5.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Cavalieri
I Italian,1598–1647
I Revisitedthe areaproblemwith adifferentperspec ve
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Cavalieri’s method
..
y = x2
..0..
1
..12
..
Divide up the interval into pieces andmeasure the area of the inscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
16125
=30125
Ln =?
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The Square Pyramidial Numbers
FactLet n be a posi ve integer. Then
1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)
6
This formula was known to the Arabs and discussed by Fibonacci inhis book Liber Abaci.
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Notes
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Notes
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Notes
. 6.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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What is Ln?Divide the interval [0, 1] into n pieces. Then each has width
1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · ·+ (n− 1)2
n3=
1+ 22 + 32 + · · ·+ (n− 1)2
n3
SoLn =
n(n− 1)(2n− 1)6n3
→ 13
as n → ∞.
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Cavalieri’s method for different functionsTry the same trick with f(x) = x3. We have
Ln =1n· f(1n
)+
1n· f(2n
)+ · · ·+ 1
n· f(n− 1n
)=
1n· 1n3
+1n· 2
3
n3+ · · ·+ 1
n· (n− 1)3
n3
=1+ 23 + 33 + · · ·+ (n− 1)3
n4
=n2(n− 1)2
4n4→ 1
4as n → ∞.
.
Nicomachus’s Theorem
Fact (Nicomachus 1st c. CE, Aryabhata 5th c., Al-Karaji 11th c.)
1+ 23 + 33 + · · ·+ (n− 1)3 = [1+ 2+ · · ·+ (n− 1)]2
=[12n(n− 1)
]2
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Notes
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Notes
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Notes
. 7.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Cavalieri’s method with different heights
.
Rn =1n· 1
3
n3+
1n· 2
3
n3+ · · ·+ 1
n· n
3
n3
=13 + 23 + 33 + · · ·+ n3
n4
=1n4
[12n(n+ 1)
]2=
n2(n+ 1)2
4n4→ 1
4as n → ∞.
So even though the rectangles overlap, we s ll get the same answer.
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OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
Cavalieri’s method in generalProblem
.. x..x0..x1
..xi
..xn−1
..xn
.. . .
.. . .
Let f be a posi ve func on definedon the interval [a, b]. Find thearea between x = a, x = b, y = 0,and y = f(x).
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Notes
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Notes
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Notes
. 8.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
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Cavalieri’s method in generalFor each posi ve 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 stepbetween a and b.
.. 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
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Forming Riemann Sums
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
Thus we approximate area under a curve by a sum of areas ofrectangles.
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Forming Riemann sumsWe have many choices of representa ve points to approximate thearea in each subinterval.
…even random points!
.. x.......
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Notes
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Notes
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Notes
. 9.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Theorem of the DayTheoremIf f is a con nuous func on on[a, b] or has finitely many jumpdiscon nui es, then
limn→∞
Sn = limn→∞
{n∑
i=1
f(ci)∆x
}
exists and is the same value noma er what choice of ci we make.
.... x.
M15 = 7.49968
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AnalogiesThe Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over be er andbe er approxima ons
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over be er andbe er approxima ons
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OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
Notes
.
Notes
.
Notes
. 10.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Distances
Just like area = length× width, we have
distance = rate× me.
So here is another use for Riemann sums.
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Application: Dead Reckoning
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Computing position by Dead ReckoningExampleA sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s posi on and velocity are recorded, butshortly therea er a storm blows in and posi on is impossible tomeasure. The velocity con nues to be recorded at thirty-minuteintervals.
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Notes
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Notes
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Notes
. 11.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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Computing position by Dead ReckoningExample
Time 12:00 12:30 1:00 1:30 2:00Speed (knots) 4 8 12 6 4Direc on E E E E WTime 2:30 3:00 3:30 4:00Speed 3 3 5 9Direc on W E E E
Es mate the ship’s posi on at 4:00pm.
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SolutionSolu onWe es mate that the speed of 4 knots (nau cal miles per hour) ismaintained from 12:00 un l 12:30. So over this me interval theship travels (
4 nmihr
)(12hr)
= 2 nmi
We can con nue for each addi onal half hour and get
distance = 4× 1/2 + 8× 1/2 + 12× 1/2
+ 6× 1/2 − 4× 1/2 − 3× 1/2 + 3× 1/2 + 5× 1/2 = 15.5
So the ship is 15.5 nmi east of its original posi on.
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Analysis
I This method of measuring posi on by recording velocity wasnecessary un l global-posi oning satellite technology becamewidespread
I If we had velocity es mates at finer intervals, we’d get be eres mates.
I If we had velocity at every instant, a limit would tell us ourexact posi on rela ve to the last me we measured it.
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Notes
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Notes
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Notes
. 12.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
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Other uses of Riemann sums
Anything with a product!I Area, volumeI Anything with a density: Popula on, massI Anything with a “speed:” distance, throughput, powerI Consumer surplusI Expected value of a random variable
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OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
The definite integral as a limit
Defini onIf f is a func on defined on [a, b], the definite integral of f from a tob is the number ∫ b
af(x) dx = lim
∆x→0
n∑i=1
f(ci)∆x
.
Notes
.
Notes
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Notes
. 13.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
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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 integra on (a is the lower limit and b theupper limit)
I dx— ??? (a parenthesis? an infinitesimal? a variable?)I The process of compu ng an integral is called integra on orquadrature
.
The limit can be simplified
TheoremIf f is con nuous on [a, b] or if f has only finitely many jumpdiscon nui es, then f is integrable on [a, b]; that is, the definite
integral∫ b
af(x) dx exists.
So we can find the integral by compu ng the limit of any sequenceof Riemann sums that we like,
.
Example
Find∫ 3
0x dx
Solu on
For any n we have∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the func on on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)= lim
n→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
.
Notes
.
Notes
.
Notes
. 14.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Example
Find∫ 3
0x2 dx
Solu on
For any n and i we have∆x =3nand xi =
3in. So∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)= lim
x→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
.
Example
Find∫ 3
0x3 dx
Solu on
For any n we have∆x =3nand xi =
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
.
OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
Notes
.
Notes
.
Notes
. 15.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Estimating the Definite IntegralExample
Es mate∫ 1
0
41+ x2
dx usingM4.
Solu on
We have x0 = 0, x1 =14, x2 =
12, x3 =
34, x4 = 1.
So c1 =18, c2 =
38, c3 =
58, c4 =
78.
.
Estimating the Definite IntegralExample
Es mate∫ 1
0
41+ x2
dx usingM4.
Solu on
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 IntegralExample
Es mate∫ 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
.
Notes
.
Notes
.
Notes
. 16.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Estimating the Definite IntegralExample
Es mate∫ 1
0
41+ x2
dx using L4 and R4
Answer
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
.
OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
Properties of the integralTheorem (Addi ve Proper es of the Integral)
Let f and g be integrable func ons 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.
.
Notes
.
Notes
.
Notes
. 17.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
ProofsProofs.
I When integra ng a constant func on 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 integralof a sum is the sum of the integrals.
I Di o for constant mul plesI Di o for differences
.
Example
Find∫ 3
0
(x3 − 4.5x2 + 5.5x+ 1
)dx
Solu on
∫ 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 func on we were es ma ng the integral of before)
.
More Properties of the IntegralConven ons: ∫ a
bf(x) dx = −
∫ b
af(x) dx∫ a
af(x) dx = 0
This allows us to haveTheorem
5.∫ c
af(x) dx =
∫ b
af(x) dx+
∫ c
bf(x) dx for all a, b, and c.
.
Notes
.
Notes
.
Notes
. 18.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Illustrating Property 5Theorem
5.∫ c
af(x) dx =
∫ b
af(x) dx+
∫ c
bf(x) dx for all a, b, and c.
..x
.
y
..a
..b
..c
.
∫ b
af(x) dx
.
∫ c
bf(x) dx
.
Illustrating Property 5Theorem
5.∫ c
af(x) dx =
∫ b
af(x) dx+
∫ c
bf(x) dx for all a, b, and c.
..x
.
y
..a..
b..
c.
∫ c
bf(x) dx =
−∫ b
cf(x) dx
.
∫ c
af(x) dx
.
Using the PropertiesExampleSuppose f and g are func onswith
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.
.
Notes
.
Notes
.
Notes
. 19.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Solu on
We 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
.
OutlineArea through the Centuries
EuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applica ons
The definite integral as a limitEs ma ng the Definite IntegralProper es of the integralComparison Proper es of the Integral
.
Comparison Properties of the IntegralTheoremLet f and g be integrable func ons 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)
.
Notes
.
Notes
.
Notes
. 20.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Integral of a nonnegative function is nonnegativeProof.If f(x) ≥ 0 for all x in [a, b], then forany number of divisions n and choiceof sample points {ci}:
Sn =n∑
i=1
f(ci)︸︷︷︸≥0
∆x ≥n∑
i=1
0 ·∆x = 0
.. x.......Since Sn ≥ 0 for all n, the limit of {Sn} is nonnega ve, too:∫ b
af(x) dx = lim
n→∞Sn︸︷︷︸≥0
≥ 0
.
The 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 for allx in [a, b]. So by the previousproperty ∫ b
ah(x) dx ≥ 0 .. x.
f(x)
.
g(x)
.
h(x)
This means that∫ b
af(x) dx−
∫ b
ag(x) dx =
∫ b
a(f(x)− g(x)) dx =
∫ b
ah(x) dx ≥ 0
.
Bounding the integralProof.Ifm ≤ f(x) ≤ M on for all x in [a, b], then bythe previous property∫ b
amdx ≤
∫ b
af(x) dx ≤
∫ b
aMdx
By Property 8, the integral of a constantfunc on is the product of the constant andthe width of the interval. So:
m(b− a) ≤∫ b
af(x) dx ≤ M(b− a)
.. x.
y
.
M
.
f(x)
.
m
..a
..b
.
Notes
.
Notes
.
Notes
. 21.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
.
.
Example
Es mate∫ 2
1
1xdx using the comparison proper es.
Solu on
Since12≤ x ≤ 1
1for all x in [1, 2], we have
12· 1 ≤
∫ 2
1
1xdx ≤ 1 · 1
.
Summary
I We can compute the area of a curved region with a limit ofRiemann sums
I We can compute the distance traveled from the velocity with alimit of Riemann sums
I Many other important uses of this process.
.
Summary
I The definite integral is a limit of Riemann SumsI The definite integral can be es mated with Riemann SumsI The definite integral can be distributed across sums andconstant mul ples of func ons
I The definite integral can be bounded using bounds for thefunc on
.
Notes
.
Notes
.
Notes
. 22.
. Sec on 5.1–5.2: Areas, Distances, Integral. V63.0121.001: Calculus I . April 25, 2011
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