Computing integrals with Riemann sums is like computing derivatives with limits. The calculus of integrals turns out to come from antidifferentiation. This startling fact is the Second Fundamental Theorem of Calculus!
(pre-class handout)
1. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Section 5.3
Notes
Evaluating Definite Integrals
V63.0121.006/016, Calculus I
New York University
April 20, 2010
Announcements
April 16: Quiz 4 on §§4.1–4.4
April 29: Movie Day!!
April 30: Quiz 5 on §§5.1–5.4
Monday, May 10, 12:00noon (not 10:00am as previously announced)
Final Exam
Image credit: docman
Announcements
Notes
April 16: Quiz 4 on
§§4.1–4.4
April 29: Movie Day!!
April 30: Quiz 5 on
§§5.1–5.4
Monday, May 10, 12:00noon
(not 10:00am as previously
announced) Final Exam
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 2 / 48
Homework: The Good
Notes
Most got problems 1 and 3 right.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 3 / 48
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2. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Homework: The Bad (steel pipe)
Notes
Problem
A steel pipe is being carried down a hallway 9 ft wide. At the end of the
hall there is a right-aangled turn into a narrower hallway 6 ft wide. What is
the length of the longest pipe that can be carried horizontally around the
corner?
θ
θ 6 6
6 sec
9
θ θ
9 csc
9
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 4 / 48
Solution
Notes
Solution
The longest pipe that barely fits is the smallest pipe that almost doesn’t
fit. We want to find the minimum value of
f (θ) = a sec θ + b csc θ
on the interval 0 < θ < π/2. (a = 9 and b = 6 in our problem.)
f (θ) = a sec θ tan θ − b csc θ cot θ
sin θ cos θ a sin3 θ − b cos3 θ
=a −b 2 =
cos2 θ sin θ sin2 θ cos2 θ
So the critical point is when
b
a sin3 θ = b cos3 θ =⇒ tan3 θ =
a
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 5 / 48
Finding the minimum
Notes
If f (θ) = a sec θ tan θ − b csc θ cot θ, then
f (θ) = a sec θ tan2 θ + a sec3 θ + b csc θ cot2 θ + b csc3 θ
which is positive on 0 < θ < π/2.
So the minimum value is
f (θmin ) = a sec θmin + b csc θmin
1/3
b b
where tan3 θmin = =⇒ tan θmin = .
a a
Using
1 + tan2 θ = sec2 θ 1 + cot2 θ = csc2 θ
We get the minimum value is
2/3
b a 2/3
min = a 1+ +b 1+
a b
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 6 / 48
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3. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Simplifying
Notes
2/3
b a 2/3
min = a 1+ +b 1+
a b
b 2/3 a2/3 a2/3 b 2/3
=b + +a + 2/3
b 2/3 b 2/3 a2/3 a
b a
= b 2/3 + a2/3 + 1/3 a2/3 + b 2/3
b 1/3 a
= b 2/3 b 2/3 + a2/3 + a2/3 a2/3 + b 2/3
= (b 2/3 + a2/3 ) b 2/3 + a2/3
2/3 2/3 3/2
= (a +b )
If a = 9 and b = 6, then min ≈ 21.070.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 7 / 48
Homework: The Bad (Diving Board)
Notes
Problem
If a diver of mass m stands at the end of a diving board with length L and
linear density ρ, then the board takes on the shape of a curve y = f (x),
where
EIy = mg (L − x) + 2 ρg (L − x)2
1
E and I are positive constants that depend of the material of the board
and g < 0 is the acceleration due to gravity.
(a) Find an expression for the shape of the curve.
(b) Use f (L) to estimate the distance below the horizontal at the end of
the board.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 8 / 48
Notes
Solution
We have
EIy (x) = mg (L − x) + 1 ρg (L − x)2
2
Antidifferentiating once gives
EIy (x) = − 1 mg (L − x)2 − 1 ρg (L − x)3 + C
2 6
Once more:
EIy (x) = 1 mg (L − x)3 +
6
1
24 ρg (L − x)4 + Cx + D
where C and D are constants.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 9 / 48
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4. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Don’t stop there!
Notes
Plugging y (0) = 0 into
EIy (x) = 1 mg (L − x)3 +
6
1
24 ρg (L − x)4 + Cx + D
gives
1 1
0 = 1 mgL3 +
6
1
24 ρgL
4
+ D =⇒ D = − mgL3 − ρgL4
6 24
Plugging y (0) = 0 into
EIy (x) = − 1 mg (L − x)2 − 1 ρg (L − x)3 + C
2 6
gives
1 1
0 = − 1 mgL2 − 6 ρgL3 + C =⇒ C = mgL2 + ρgL3
2
1
2 6
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 10 / 48
Solution completed
Notes
So
EIy (x) = 1 mg (L − x)3 +
6
1
24 ρg (L − x)4
1 1 1 1
+ mgL2 + L3 x − mgL3 − ρgL4
2 6 6 24
which means
1 1 1 1
EIy (L) = mgL2 + L3 L − mgL3 − ρgL4
2 6 6 24
1 3 1 4 1 1
= mg L + L − mgL3 − ρgL4
2 6 6 24
1 1
= mgL3 + ρgL4
3 8
gL3 m ρL
=⇒ y (L) = +
EI 3 8
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 11 / 48
Homework: The Ugly
Notes
Some students have gotten their hands on a solution manual and are
copying answers word for word.
This is very easy to catch: the graders are following the same solution
manual.
This is not very productive: the best you will do is ace 10% of your
course grade.
This is a violation of academic integrity. I do not take it lightly.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 12 / 48
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5. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Objectives
Notes
Use the Evaluation Theorem
to evaluate definite integrals.
Write antiderivatives as
indefinite integrals.
Interpret definite integrals as
“net change” of a function
over an interval.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 13 / 48
Outline
Notes
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
The Theorem of the Day
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 14 / 48
The definite integral as a limit
Notes
Definition
If f is a function defined on [a, b], the definite integral of f from a to b
is the number
b n
f (x) dx = lim f (ci ) ∆x
a n→∞
i=1
b−a
where ∆x = , and for each i, xi = a + i∆x, and ci is a point in
n
[xi−1 , xi ].
Theorem
If f is continuous on [a, b] or if f has only finitely many jump
discontinuities, then f is integrable on [a, b]; that is, the definite integral
b
f (x) dx exists and is the same for any choice of ci .
a
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 15 / 48
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6. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Notation/Terminology
Notes
b n
f (x) dx = lim f (ci ) ∆x
a n→∞
i=1
— integral sign (swoopy S)
f (x) — integrand
a and b — limits of integration (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an infinitesimal? a variable?)
The process of computing an integral is called integration
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 16 / 48
Properties of the integral
Notes
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a, b] and c a constant. Then
b
1. c dx = c(b − a)
a
b b b
2. [f (x) + g (x)] dx = f (x) dx + g (x) dx.
a a a
b b
3. cf (x) dx = c f (x) dx.
a a
b b b
4. [f (x) − g (x)] dx = f (x) dx − g (x) dx.
a a a
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 17 / 48
More Properties of the Integral
Notes
Conventions:
a b
f (x) dx = − f (x) dx
b a
a
f (x) dx = 0
a
This allows us to have
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 18 / 48
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7. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Definite Integrals We Know So Far
Notes
If the integral computes an
area and we know the area,
we can use that. For
y
instance,
1
π
1 − x 2 dx =
0 4
By brute force we computed x
1 1
1 1
x 2 dx = x 3 dx =
0 3 0 4
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 19 / 48
Example
1
4 Notes
Estimate dx using the midpoint rule and four divisions.
0 1 + x2
Solution
Dividing up [0, 1] into 4 pieces gives
1 2 3 4
x0 = 0, x1 = , x2 = , x3 = , x4 =
4 4 4 4
So the midpoint rule gives
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
150, 166, 784
= ≈ 3.1468
47, 720, 465
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 20 / 48
Comparison Properties of the Integral
Notes
Theorem
Let f and g be integrable functions on [a, b].
b
6. If f (x) ≥ 0 for all x in [a, b], then f (x) dx ≥ 0
a
b b
7. If f (x) ≥ g (x) for all x in [a, b], then f (x) dx ≥ g (x) dx
a a
8. If m ≤ f (x) ≤ M for all x in [a, b], then
b
m(b − a) ≤ f (x) dx ≤ M(b − a)
a
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 21 / 48
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8. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Example
2 Notes
1
Estimate dx using Property 8.
1 x
Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
y
we have
2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or
1 2
1 x
≤ dx ≤ 1
2 1 x
(Not a very good estimate)
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 22 / 48
Outline
Notes
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
The Theorem of the Day
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 23 / 48
Socratic dialogue
Notes
The definite integral of
velocity measures
displacement (net distance)
The derivative of
displacement is velocity
So we can compute
displacement with the
definite integral or an
antiderivative of velocity
But any function can be a
velocity function, so . . .
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 24 / 48
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9. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Theorem of the Day
Notes
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F for another function F , then
b
f (x) dx = F (b) − F (a).
a
Note
In Section 5.3, this theorem is called “The Evaluation Theorem”. Nobody
else in the world calls it that.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 25 / 48
Proving the Second FTC
Notes
Proof.
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual. For
n
each i, F is continuous on [xi−1 , xi ] and differentiable on (xi−1 , xi ). So
there is a point ci in (xi−1 , xi ) with
F (xi ) − F (xi−1 )
= F (ci ) = f (ci )
xi − xi−1
Or
f (ci )∆x = F (xi ) − F (xi−1 )
See if you can spot the invocation of the Mean Value Theorem!
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 26 / 48
Proving the Second FTC
Notes
We have for each i
f (ci )∆x = F (xi ) − F (xi−1 )
Form the Riemann Sum:
n n
Sn = f (ci )∆x = (F (xi ) − F (xi−1 ))
i=1 i=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)
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 27 / 48
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10. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Proving the Second FTC
Notes
We have shown for each n,
Sn = F (b) − F (a)
so in the limit
b
f (x) dx = lim Sn = lim (F (b) − F (a)) = F (b) − F (a)
a n→∞ n→∞
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 28 / 48
Verifying earlier computations
Notes
Example
Find the area between y = x 3 the
x-axis, x = 0 and x = 1.
Solution
1 1
x4 1
A= x 3 dx = =
0 4 0 4
Here we use the notation F (x)|b or [F (x)]b to mean F (b) − F (a).
a a
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 29 / 48
Verifying Archimedes
Notes
Example
Find the area enclosed by the parabola y = x 2 and y = 1.
1
−1 1
Solution
1 1
x3 1 1 4
A=2− x 2 dx = 2 − =2− − − =
−1 3 −1 3 3 3
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 30 / 48
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11. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Computing exactly what we earlier estimated
Notes
Example
1
4
Evaluate the integral dx.
0 1 + x2
Solution
1 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
π
=4 −0 =π
4
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 31 / 48
Example
1
4 Notes
Estimate dx using the midpoint rule and four divisions.
0 1 + x2
Solution
Dividing up [0, 1] into 4 pieces gives
1 2 3 4
x0 = 0, x1 = , x2 = , x3 = , x4 =
4 4 4 4
So the midpoint rule gives
1 4 4 4 4
M4 = + + +
4 1 + (1/8)2 1 + (3/8)2 1 + (5/8)2 1 + (7/8)2
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
150, 166, 784
= ≈ 3.1468
47, 720, 465
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 32 / 48
Computing exactly what we earlier estimated
Notes
Example
1
4
Evaluate the integral dx.
0 1 + x2
Solution
1 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
= 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)
π
=4 −0 =π
4
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 33 / 48
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12. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Computing exactly what we earlier estimated
Notes
Example
2
1
Evaluate dx.
1 x
Solution
2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 34 / 48
Example
2 Notes
1
Estimate dx using Property 8.
1 x
Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
y
we have
2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or
1 2
1 x
≤ dx ≤ 1
2 1 x
(Not a very good estimate)
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 35 / 48
Computing exactly what we earlier estimated
Notes
Example
2
1
Evaluate dx.
1 x
Solution
2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 36 / 48
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13. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Outline
Notes
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
The Theorem of the Day
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 37 / 48
The Integral as Total Change
Notes
Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a
or, the integral of a derivative along an interval is the total change over
that interval. This has many ramifications:
Theorem
If v (t) represents the velocity of a particle moving rectilinearly, then
t1
v (t) dt = s(t1 ) − s(t0 ).
t0
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
The Integral as Total Change
Notes
Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a
or, the integral of a derivative along an interval is the total change over
that interval. This has many ramifications:
Theorem
If MC (x) represents the marginal cost of making x units of a product, then
x
C (x) = C (0) + MC (q) dq.
0
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
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14. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
The Integral as Total Change
Notes
Another way to state this theorem is:
b
F (x) dx = F (b) − F (a),
a
or, the integral of a derivative along an interval is the total change over
that interval. This has many ramifications:
Theorem
If ρ(x) represents the density of a thin rod at a distance of x from its end,
then the mass of the rod up to x is
x
m(x) = ρ(s) ds.
0
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 38 / 48
Outline
Notes
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
The Theorem of the Day
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 39 / 48
A new notation for antiderivatives
Notes
To emphasize the relationship between antidifferentiation and integration,
we use the indefinite integral notation
f (x) dx
for any function whose derivative is f (x). Thus
x 2 dx = 1 x 3 + C .
3
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 40 / 48
14
15. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
My first table of integrals
Notes
[f (x) + g (x)] dx = f (x) dx + g (x) dx
x n+1
x n dx = + C (n = −1) cf (x) dx = c f (x) dx
n+1
1
e x dx = e x + C dx = ln |x| + C
x
ax
sin x dx = − cos x + C ax dx = +C
ln a
cos x dx = sin x + C csc2 x dx = − cot x + C
sec2 x dx = tan x + C csc x cot x dx = − csc x + C
1
sec x tan x dx = sec x + C √ dx = arcsin x + C
1 − x2
1
dx = arctan x + C
1 + x2
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 41 / 48
Outline
Notes
Last time: The Definite Integral
The definite integral as a limit
Properties of the integral
Comparison Properties of the Integral
Evaluating Definite Integrals
The Theorem of the Day
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 42 / 48
Example
Notes
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the
vertical lines x = 0 and x = 3.
Solution
3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
and (2, 3], and negative on (1, 2). If we want the area of the region, we
have to do
1 2 3
A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
0 1 2
1 3 1 2 3
= 3x − 2 x 2 + 2x
3
0
− 1 3
3 x − 3 x 2 + 2x
2 1
+ 1 3
3x − 2 x 2 + 2x
3
2
5 1 5 11
= − − + = .
6 6 6 6
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 43 / 48
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16. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
Graph
Notes
y
x
1 2 3
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 44 / 48
Example
Notes
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the
vertical lines x = 0 and x = 3.
Solution
3
Consider (x − 1)(x − 2) dx. Notice the integrand is positive on [0, 1)
0
and (2, 3], and negative on (1, 2). If we want the area of the region, we
have to do
1 2 3
A= (x − 1)(x − 2) dx − (x − 1)(x − 2) dx + (x − 1)(x − 2) dx
0 1 2
1 3 1 2 3
= 3x − 2 x 2 + 2x
3
0
− 1 3
3x − 3 x 2 + 2x
2 1
+ 1 3
3x − 2 x 2 + 2x
3
2
5 1 5 11
= − − + = .
6 6 6 6
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 45 / 48
Interpretation of “negative area” in motion
Notes
There is an analog in rectlinear motion:
t1
v (t) dt is net distance traveled.
t0
t1
|v (t)| dt is total distance traveled.
t0
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 46 / 48
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17. V63.0121.006/016, Calculus I Section 5.3 : Evaluating Definite Integrals April 20, 2010
What about the constant?
Notes
It seems we forgot about the +C when we say for instance
1 1
x4 1 1
x 3 dx = = −0=
0 4 0 4 4
But notice
1
x4 1 1 1
+C = +C − (0 + C ) = +C −C =
4 0 4 4 4
no matter what C is.
So in antidifferentiation for definite integrals, the constant is
immaterial.
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 47 / 48
Summary
Notes
Second FTC:
b b
f (x) dx = F (x)
a a
where F is an antiderivative
of f .
Computes any “net change”
over an interval
Proving the FTC requires
the MVT
V63.0121.006/016, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals April 20, 2010 48 / 48
Notes
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