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..
Sec on 5.3Evalua ng Definite Integrals
V63.0121.011: Calculus IProfessor Ma hew Leingang
New York University
April 27, 2011
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AnnouncementsI Today: 5.3I Thursday/Friday: Quiz on4.1–4.4
I Monday 5/2: 5.4I Wednesday 5/4: 5.5I Monday 5/9: Review andMovie Day!
I Thursday 5/12: FinalExam, 2:00–3:50pm
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ObjectivesI Use the Evalua onTheorem to evaluatedefinite integrals.
I Write an deriva ves asindefinite integrals.
I Interpret definiteintegrals as “net change”of a func on over aninterval.
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OutlineLast me: The Definite Integral
The definite integral as a limitProper es of the integral
Evalua ng Definite IntegralsExamples
The Integral as Net Change
Indefinite IntegralsMy first table of integrals
Compu ng Area with integrals
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The definite integral as a limitDefini 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
n→∞
n∑i=1
f(ci)∆x
where∆x =b− an
, and for each i, xi = a+ i∆x, and ci is a point in[xi−1, xi].
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The definite integral as a limit
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 and is the same for any choice of ci.
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Notation/Terminology∫ b
af(x) dx
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
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Example
Es mate∫ 1
0
41+ x2
dx usingM4.
Solu on
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Example
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.
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Example
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
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Example
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
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Example
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
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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.
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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.
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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
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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
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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
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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
.
∫ c
af(x) dx
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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
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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
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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
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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
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Definite Integrals We Know So FarI If the integral computes an areaand we know the area, we canuse that. For instance,∫ 1
0
√1− x2 dx =
π
4
I By brute force we computed∫ 1
0x2 dx =
13
∫ 1
0x3 dx =
14
..x
.
y
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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)
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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)
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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)
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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)
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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
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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
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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
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Example
Es mate∫ 2
1
1xdx using the comparison proper es.
Solu onSince
12≤ 1
x≤ 1
1for all x in [1, 2], we have
12· 1 ≤
∫ 2
1
1xdx ≤ 1 · 1
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Example
Es mate∫ 2
1
1xdx using the comparison proper es.
Solu onSince
12≤ 1
x≤ 1
1for all x in [1, 2], we have
12· 1 ≤
∫ 2
1
1xdx ≤ 1 · 1
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Ques on
Es mate∫ 2
1
1xdx with L2 and R2. Are your es mates overes mates?
Underes mates? Impossible to tell?
AnswerSince the integrand is decreasing,
Rn <
∫ 2
1
1xdx < Ln
for all n. So712
<
∫ 2
1
1xdx <
56.
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Ques on
Es mate∫ 2
1
1xdx with L2 and R2. Are your es mates overes mates?
Underes mates? Impossible to tell?
AnswerSince the integrand is decreasing,
Rn <
∫ 2
1
1xdx < Ln
for all n. So712
<
∫ 2
1
1xdx <
56.
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OutlineLast me: The Definite Integral
The definite integral as a limitProper es of the integral
Evalua ng Definite IntegralsExamples
The Integral as Net Change
Indefinite IntegralsMy first table of integrals
Compu ng Area with integrals
![Page 36: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/36.jpg)
Socratic proofI The definite integral of velocitymeasures displacement (netdistance)
I The deriva ve of displacementis velocity
I So we can computedisplacement with the definiteintegral or the an deriva ve ofvelocity
I But any func on can be avelocity func on, so . . .
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Theorem of the DayTheorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another func on F,then ∫ b
af(x) dx = F(b)− F(a).
NoteIn Sec on 5.3, this theorem is called “The Evalua on Theorem”.Nobody else in the world calls it that.
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Theorem of the DayTheorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′ for another func on F,then ∫ b
af(x) dx = F(b)− F(a).
NoteIn Sec on 5.3, this theorem is called “The Evalua on Theorem”.Nobody else in the world calls it that.
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Proving the Second FTCProof.
I Divide up [a, b] into n pieces of equal width∆x =b− an
asusual.
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Proving the Second FTCProof.
I Divide up [a, b] into n pieces of equal width∆x =b− an
asusual.
I For each i, F is con nuous on [xi−1, xi] and differen able on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
= F(xi)− F(xi−1)
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Proving the Second FTCProof.
I Divide up [a, b] into n pieces of equal width∆x =b− an
asusual.
I For each i, F is con nuous on [xi−1, xi] and differen able on(xi−1, xi). So there is a point ci in (xi−1, xi) with
F(xi)− F(xi−1)
xi − xi−1= F′(ci) = f(ci)
=⇒ f(ci)∆x = F(xi)− F(xi−1)
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Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
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Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
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Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 45: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/45.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 46: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/46.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 47: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/47.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 48: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/48.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 49: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/49.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 50: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/50.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 51: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/51.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 52: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/52.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 53: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/53.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 54: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/54.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 55: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/55.jpg)
Proving the Second FTCProof.
I Form the Riemann Sum:
Sn =n∑
i=1
f(ci)∆x =n∑
i=1
(F(xi)− F(xi−1))
= (F(x1)− F(x0)) + (F(x2)− F(x1)) + (F(x3)− F(x2)) + · · ·· · · + (F(xn−1)− F(xn−2)) + (F(xn)− F(xn−1))
= F(xn)− F(x0) = F(b)− F(a)
![Page 56: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/56.jpg)
Proving the Second FTCProof.
I We have shown for each n,
Sn = F(b)− F(a)
Which does not depend on n.
I So in the limit∫ b
af(x) dx = lim
n→∞Sn = lim
n→∞(F(b)− F(a)) = F(b)− F(a)
![Page 57: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/57.jpg)
Proving the Second FTCProof.
I We have shown for each n,
Sn = F(b)− F(a)
Which does not depend on n.I So in the limit∫ b
af(x) dx = lim
n→∞Sn = lim
n→∞(F(b)− F(a)) = F(b)− F(a)
![Page 58: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/58.jpg)
Computing area with the 2nd FTCExample
Find the area between y = x3 and the x-axis, between x = 0 andx = 1.
Solu on
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14
.
Here we use the nota on F(x)|ba or [F(x)]ba to mean F(b)− F(a).
![Page 59: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/59.jpg)
Computing area with the 2nd FTCExample
Find the area between y = x3 and the x-axis, between x = 0 andx = 1.
Solu on
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14 .
Here we use the nota on F(x)|ba or [F(x)]ba to mean F(b)− F(a).
![Page 60: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/60.jpg)
Computing area with the 2nd FTCExample
Find the area between y = x3 and the x-axis, between x = 0 andx = 1.
Solu on
A =
∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14 .
Here we use the nota on F(x)|ba or [F(x)]ba to mean F(b)− F(a).
![Page 61: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/61.jpg)
Computing area with the 2nd FTCExample
Find the area enclosed by the parabola y = x2 and the line y = 1.
Solu on
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43 ...
−1..
1..
1
![Page 62: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/62.jpg)
Computing area with the 2nd FTCExample
Find the area enclosed by the parabola y = x2 and the line y = 1.
Solu on
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43
...−1
..1
..
1
![Page 63: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/63.jpg)
Computing area with the 2nd FTCExample
Find the area enclosed by the parabola y = x2 and the line y = 1.
Solu on
A = 2−∫ 1
−1x2 dx = 2−
[x3
3
]1−1
= 2−[13−
(−13
)]=
43 ...
−1..
1..
1
![Page 64: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/64.jpg)
Computing an integral weestimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solu on∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10
= 4 (arctan 1− arctan 0) = 4(π4− 0
)
= π
![Page 65: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/65.jpg)
Example
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
![Page 66: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/66.jpg)
Computing an integral weestimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solu on∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx
= 4 arctan(x)|10
= 4 (arctan 1− arctan 0) = 4(π4− 0
)
= π
![Page 67: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/67.jpg)
Computing an integral weestimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solu on∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx = 4 arctan(x)|10
= 4 (arctan 1− arctan 0) = 4(π4− 0
)
= π
![Page 68: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/68.jpg)
Computing an integral weestimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solu on∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx = 4 arctan(x)|10
= 4 (arctan 1− arctan 0)
= 4(π4− 0
)
= π
![Page 69: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/69.jpg)
Computing an integral weestimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solu on∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx = 4 arctan(x)|10
= 4 (arctan 1− arctan 0) = 4(π4− 0
)
= π
![Page 70: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/70.jpg)
Computing an integral weestimated before
Example
Evaluate the integral∫ 1
0
41+ x2
dx.
Solu on∫ 1
0
41+ x2
dx = 4∫ 1
0
11+ x2
dx = 4 arctan(x)|10
= 4 (arctan 1− arctan 0) = 4(π4− 0
)= π
![Page 71: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/71.jpg)
Computing an integral weestimated before
Example
Evaluate∫ 2
1
1xdx.
Solu on ∫ 2
1
1xdx
= ln x|21 = ln 2− ln 1 = ln 2
![Page 72: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/72.jpg)
Example
Es mate∫ 2
1
1xdx using the comparison proper es.
Solu onSince
12≤ 1
x≤ 1
1for all x in [1, 2], we have
12· 1 ≤
∫ 2
1
1xdx ≤ 1 · 1
![Page 73: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/73.jpg)
Computing an integral weestimated before
Example
Evaluate∫ 2
1
1xdx.
Solu on ∫ 2
1
1xdx
= ln x|21 = ln 2− ln 1 = ln 2
![Page 74: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/74.jpg)
Computing an integral weestimated before
Example
Evaluate∫ 2
1
1xdx.
Solu on ∫ 2
1
1xdx = ln x|21
= ln 2− ln 1 = ln 2
![Page 75: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/75.jpg)
Computing an integral weestimated before
Example
Evaluate∫ 2
1
1xdx.
Solu on ∫ 2
1
1xdx = ln x|21 = ln 2− ln 1
= ln 2
![Page 76: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/76.jpg)
Computing an integral weestimated before
Example
Evaluate∫ 2
1
1xdx.
Solu on ∫ 2
1
1xdx = ln x|21 = ln 2− ln 1 = ln 2
![Page 77: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/77.jpg)
OutlineLast me: The Definite Integral
The definite integral as a limitProper es of the integral
Evalua ng Definite IntegralsExamples
The Integral as Net Change
Indefinite IntegralsMy first table of integrals
Compu ng Area with integrals
![Page 78: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/78.jpg)
The Integral as Net Change
Another way to state this theorem is:∫ b
aF′(x) dx = F(b)− F(a),
or the integral of a deriva ve along an interval is the net changeover that interval. This has many interpreta ons.
![Page 79: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/79.jpg)
The Integral as Net Change
![Page 80: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/80.jpg)
The Integral as Net Change
Corollary
If v(t) represents the velocity of a par cle moving rec linearly, then∫ t1
t0v(t) dt = s(t1)− s(t0).
![Page 81: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/81.jpg)
The Integral as Net Change
Corollary
If MC(x) represents the marginal cost of making x units of a product,then
C(x) = C(0) +∫ x
0MC(q) dq.
![Page 82: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/82.jpg)
The Integral as Net Change
Corollary
If ρ(x) represents the density of a thin rod at a distance of x from itsend, then the mass of the rod up to x is
m(x) =∫ x
0ρ(s) ds.
![Page 83: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/83.jpg)
OutlineLast me: The Definite Integral
The definite integral as a limitProper es of the integral
Evalua ng Definite IntegralsExamples
The Integral as Net Change
Indefinite IntegralsMy first table of integrals
Compu ng Area with integrals
![Page 84: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/84.jpg)
A new notation for antiderivativesTo emphasize the rela onship between an differen a on andintegra on, we use the indefinite integral nota on∫
f(x) dx
for any func on whose deriva ve is f(x).
Thus∫x2 dx = 1
3x3 + C.
![Page 85: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/85.jpg)
A new notation for antiderivativesTo emphasize the rela onship between an differen a on andintegra on, we use the indefinite integral nota on∫
f(x) dx
for any func on whose deriva ve is f(x). Thus∫x2 dx = 1
3x3 + C.
![Page 86: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/86.jpg)
My first table of integrals..∫
[f(x) + g(x)] dx =∫
f(x) dx+∫
g(x) dx∫xn dx =
xn+1
n+ 1+ C (n ̸= −1)∫
ex dx = ex + C∫sin x dx = − cos x+ C∫cos x dx = sin x+ C∫sec2 x dx = tan x+ C∫
sec x tan x dx = sec x+ C∫1
1+ x2dx = arctan x+ C
∫cf(x) dx = c
∫f(x) dx∫
1xdx = ln |x|+ C∫
ax dx =ax
ln a+ C∫
csc2 x dx = − cot x+ C∫csc x cot x dx = − csc x+ C∫
1√1− x2
dx = arcsin x+ C
![Page 87: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/87.jpg)
OutlineLast me: The Definite Integral
The definite integral as a limitProper es of the integral
Evalua ng Definite IntegralsExamples
The Integral as Net Change
Indefinite IntegralsMy first table of integrals
Compu ng Area with integrals
![Page 88: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/88.jpg)
Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.
Solu onThe answer is ∫ 4
1ex dx = ex|41 = e4 − e.
![Page 89: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/89.jpg)
Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, thex-axis, and the curve y = ex.
Solu onThe answer is ∫ 4
1ex dx = ex|41 = e4 − e.
![Page 90: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/90.jpg)
Computing Area with integralsExample
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solu on
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
![Page 91: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/91.jpg)
Computing Area with integralsExample
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solu on
I The answer is∫ 1
0arcsin x dx, but
we do not know an an deriva vefor arcsin.
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
![Page 92: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/92.jpg)
Computing Area with integralsExample
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solu on
I Instead compute the area as
π
2−∫ π/2
0sin y dy
=π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
![Page 93: Lesson 25: Evaluating Definite Integrals (slides](https://reader033.vdocument.in/reader033/viewer/2022042614/5597ee631a28ab9c378b479f/html5/thumbnails/93.jpg)
Computing Area with integralsExample
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solu on
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20
=π
2−1
..x
.
y
..1
..
π/2
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Computing Area with integralsExample
Find the area of the region bounded by the curve y = arcsin x, thex-axis, and the line x = 1.
Solu on
I Instead compute the area as
π
2−∫ π/2
0sin y dy =
π
2−[− cos x]π/20 =
π
2−1
..x
.
y
..1
..
π/2
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Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis,and the ver cal lines x = 0 and x = 3.
Solu on
.. x.
y
..1
..2
..3
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Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis,and the ver cal lines x = 0 and x = 3.
Solu onNo ce the func ony = (x− 1)(x− 2) is posi ve on [0, 1)and (2, 3], and nega ve on (1, 2).
.. x.
y
..1
..2
..3
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Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis,and the ver cal lines x = 0 and x = 3.
Solu on
A =
∫ 1
0(x2 − 3x+ 2) dx
−∫ 2
1(x2 − 3x+ 2) dx
+
∫ 3
2(x2 − 3x+ 2) dx
.. x.
y
..1
..2
..3
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Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis,and the ver cal lines x = 0 and x = 3.
Solu on
A =
∫ 1
0(x− 1)(x− 2) dx
−∫ 2
1(x− 1)(x− 2) dx
+
∫ 3
2(x− 1)(x− 2) dx
.. x.
y
..1
..2
..3
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Example
Find the area between the graph of y = (x− 1)(x− 2), the x-axis,and the ver cal lines x = 0 and x = 3.
Solu on
A =[13x
3 − 32x
2 + 2x]10
−[13x
3 − 32x
2 + 2x]21
+[13x
3 − 32x
2 + 2x]32
=116
.. x.
y
..1
..2
..3
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Interpretation of “negative area”in motion
There is an analog in rectlinear mo on:
I
∫ t1
t0v(t) dt is net distance traveled.
I
∫ t1
t0|v(t)| dt is total distance traveled.
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What about the constant?I It seems we forgot about the+C when we say for instance∫ 1
0x3 dx =
x4
4
∣∣∣∣10=
14− 0 =
14
I But no ce[x4
4+ C
]10=
(14+ C
)− (0+ C) =
14+ C− C =
14
no ma er what C is.I So in an differen a on for definite integrals, the constant isimmaterial.
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SummaryI The second Fundamental Theorem of Calculus:∫ b
af(x) dx = F(b)− F(a)
where F′ = f.I Definite integrals represent net change of a func on over aninterval.
I We write an deriva ves as indefinite integrals∫
f(x) dx