1. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
Section 5.3
Notes
Evaluating Definite Integrals
V63.0121.021, Calculus I
New York University
December 7, 2010
Announcements
Today: Section 5.3
Thursday: Section 5.4
”Thursday,” December 14: Section 5.5
”Monday,” December 15: (WWH 109, 12:30–1:45pm) Review and
Movie Day!
Monday, December 20, 12:00–1:50pm: Final Exam
Announcements
Notes
Today: Section 5.3
Thursday: Section 5.4
”Thursday,” December 14:
Section 5.5
”Monday,” December 15:
(WWH 109, 12:30–1:45pm)
Review and Movie Day!
Monday, December 20,
12:00–1:50pm: Final Exam
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 2 / 41
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.
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 3 / 41
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2. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 4 / 41
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 5 / 41
Notation/Terminology
Notes
b
f (x) dx
a
— 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 6 / 41
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3. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 7 / 41
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 8 / 41
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
Theorem
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 9 / 41
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4. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
Illustrating Property 5
Notes
Theorem
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b
y
b c
f (x) dx f (x) dx
a b
a b c x
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 10 / 41
Illustrating Property 5
Notes
Theorem
c b c
5. f (x) dx = f (x) dx + f (x) dx for all a, b, and c.
a a b
y
c c
f (x) dx f (x) dx =
a b
b
− f (x) dx
c
a c x
b
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 10 / 41
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 11 / 41
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5. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
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
7. If f (x) ≥ g (x) for all x in [a, b], then
b b
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 12 / 41
Estimating an integral with inequalities
Notes
Example
2
1
Estimate dx using Property 8.
1 x
Solution
Since
1 1 1
1 ≤ x ≤ 2 =⇒ ≤ ≤
2 x 1
we have
2
1 1
· (2 − 1) ≤ dx ≤ 1 · (2 − 1)
2 1 x
or
2
1 1
≤ dx ≤ 1
2 1 x
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 13 / 41
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
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 14 / 41
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6. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
Socratic proof
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 the
antiderivative of velocity
But any function can be a
velocity function, so . . .
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 15 / 41
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 16 / 41
Proving the Second FTC
Notes
b−a
Divide up [a, b] into n pieces of equal width ∆x = as usual.
n
For 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 )
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 17 / 41
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7. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
Proof continued
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)
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 18 / 41
Proof Completed
Notes
We have shown for each n,
Sn = F (b) − F (a)
Which does not depend on n.
So in the limit
b
f (x) dx = lim Sn = lim (F (b) − F (a)) = F (b) − F (a)
a n→∞ n→∞
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 19 / 41
Computing area with the Second FTC
Notes
Example
Find the area between y = x 3 and the x-axis, between 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 20 / 41
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8. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
Computing area with the Second FTC
Notes
Example
Find the area enclosed by the parabola y = x 2 and the line 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 21 / 41
Computing an integral we estimated before
Notes
Example
1
4
Evaluate the integral dx.
0 1 + x2
Solution
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 22 / 41
Computing an integral we estimated before
Notes
Example
2
1
Evaluate dx.
1 x
Solution
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 25 / 41
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9. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 28 / 41
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 between
the sides of 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 29 / 41
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 between
the sides of 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 29 / 41
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10. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 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 between
the sides of 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 29 / 41
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
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 30 / 41
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 31 / 41
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11. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
My first table of integrals
Notes
[f (x) + g (x)] dx = f (x) dx + g (x) dx
n x n+1
x dx = + C (n = −1) cf (x) dx = c f (x) dx
n+1
x x
1
e dx = e + C dx = ln |x| + C
x
x ax
sin x dx = − cos x + C a 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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 32 / 41
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
Examples
The Integral as Total Change
Indefinite Integrals
My first table of integrals
Computing Area with integrals
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 33 / 41
Computing Area with integrals
Notes
Example
Find the area of the region bounded by the lines x = 1, x = 4, the x-axis,
and the curve y = e x .
Solution
The answer is
4
e x dx = e x |4 = e 4 − e.
1
1
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 34 / 41
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12. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
Computing Area with integrals
Notes
Example
Find the area of the region bounded by the curve y = arcsin x, the x-axis,
and the line x = 1.
Solution
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 35 / 41
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 36 / 41
Graph
Notes
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 37 / 41
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13. V63.0121.021, Calculus I Section 5.3 : Evaluating Definite Integrals December 7, 2010
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
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 39 / 41
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.
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 40 / 41
Summary
Notes
The second Fundamental Theorem of Calculus:
b
f (x) dx = F (b) − F (a)
a
where F = f .
Definite integrals represent net change of a function over an interval.
We write antiderivatives as indefinite integrals f (x) dx
V63.0121.021, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals December 7, 2010 41 / 41
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