Evaluating definite integrals

..
Sec on 5.3
Evalua ng Deﬁnite Integrals
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
April 27, 2011

Announcements
Today: 5.3
Thursday/Friday: Quiz on
4.1–4.4
Monday 5/2: 5.4
Wednesday 5/4: 5.5
Monday 5/9: Review and
Movie Day!
Thursday 5/12: Final
Exam, 2:00–3:50pm

Objectives
Use the Evalua on
Theorem to evaluate
definite integrals.
Write an deriva ves as
indefinite integrals.
Interpret definite
integrals as “net change”
of a func on over an
interval.

Outline
Last me: The Definite Integral
The definite integral as a limit
Proper es of the integral
Evalua ng Definite Integrals
Examples
The Integral as Net Change
Indefinite Integrals
My first table of integrals
Compu ng Area with integrals

Defini on
If f is a func on defined on [a, b], the definite integral of f from a to
b is the number
∫ b
a
f(x) dx = lim
n→∞
n∑
i=1
f(ci) ∆x
where ∆x =
b − a
n
, and for each i, xi = a + i∆x, and ci is a point in
[xi−1, xi].

Theorem
If f is con nuous on [a, b] or if f has only ﬁnitely many jump
discon nui es, then f is integrable on [a, b]; that is, the deﬁnite
integral
∫ b
a
f(x) dx exists and is the same for any choice of ci.

Notation/Terminology
∫ b
a
f(x) dx
∫
— integral sign (swoopy S)
f(x) — integrand
a and b — limits of integra on (a is the lower limit and b the
upper limit)
dx — ??? (a parenthesis? an inﬁnitesimal? a variable?)
The process of compu ng an integral is called integra on

Example
Es mate
∫ 1
0
4
1 + x2
dx using M4.

Example
Es mate
∫ 1
0
4
1 + x2
dx using M4.
Solu on
We have x0 = 0, x1 =
1
4
, x2 =
1
2
, x3 =
3
4
, x4 = 1.
So c1 =
1
8
, c2 =
3
8
, c3 =
5
8
, c4 =
7
8
.

Example
Es mate
∫ 1
0
4
1 + x2
dx using M4.
Solu on
M4 =
1
4
(
4
1 + (1/8)2
+
4
1 + (3/8)2
+
4
1 + (5/8)2
+
4
1 + (7/8)2
)

Example
Es mate
∫ 1
0
4
1 + x2
dx using M4.
Solu on
M4 =
1
4
(
4
1 + (1/8)2
+
4
1 + (3/8)2
+
4
1 + (5/8)2
+
4
1 + (7/8)2
)
=
1
4
(
4
65/64
+
4
73/64
+
4
89/64
+
4
113/64
)

Example
Es mate
∫ 1
0
4
1 + x2
dx using M4.
Solu on
M4 =
1
4
(
4
1 + (1/8)2
+
4
1 + (3/8)2
+
4
1 + (5/8)2
+
4
1 + (7/8)2
)
=
1
4
(
4
65/64
+
4
73/64
+
4
89/64
+
4
113/64
)
=
64
65
+
64
73
+
64
89
+
64
113
≈ 3.1468

Properties of the integral
Theorem (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
a
c dx = c(b − a)
2.
∫ b
a
[f(x) + g(x)] dx =
∫ b
a
f(x) dx +
∫ b
a
g(x) dx.
3.
∫ b
a
cf(x) dx = c
∫ b
a
f(x) dx.
4.
∫ b
a
[f(x) − g(x)] dx =
∫ b
a
f(x) dx −
∫ b
a
g(x) dx.

More Properties of the Integral
Conven ons: ∫ a
b
f(x) dx = −
∫ b
a
f(x) dx
∫ a
a
f(x) dx = 0
This allows us to have
Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
f(x) dx for all a, b, and c.

Illustrating Property 5
Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
..
x
.
y
..
a
..
b
..
c

Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
..
x
.
y
..
a
..
b
..
c
.
∫ b
a
f(x) dx

Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
..
x
.
y
..
a
..
b
..
c
.
∫ b
a
f(x) dx
.
∫ c
b
f(x) dx

Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
..
x
.
y
..
a
..
b
..
c
.
∫ b
a
f(x) dx
.
∫ c
b
f(x) dx
.
∫ c
a
f(x) dx

Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
..
x
.
y
..
a
..
b
..
c
.
∫ c
b
f(x) dx =
−
∫ b
c
f(x) dx

Theorem
5.
∫ c
a
f(x) dx =
∫ b
a
f(x) dx +
∫ c
b
..
x
.
y
..
a
..
b
..
c
.
∫ c
b
f(x) dx =
−
∫ b
c
f(x) dx
.
∫ c
a
f(x) dx

Deﬁnite Integrals We Know So Far
If the integral computes an area
and we know the area, we can
use that. For instance,
∫ 1
0
√
1 − x2 dx =
π
4
By brute force we computed
∫ 1
0
x2
dx =
1
3
∫ 1
0
x3
dx =
1
4
..
x
.
y

Comparison Properties of the Integral
Theorem
Let f and g be integrable func ons on [a, b].

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

Theorem
∫ b
a
f(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b
a
f(x) dx ≥
∫ b
a
g(x) dx

Theorem
∫ b
a
f(x) dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a, b], then
∫ b
a
f(x) dx ≥
∫ b
a
g(x) dx
8. If m ≤ f(x) ≤ M for all x in [a, b], then
m(b − a) ≤
∫ b
a
f(x) dx ≤ M(b − a)

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 and choice
of 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
a
f(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 all
x in [a, b]. So by the previous
property
∫ b
a
h(x) dx ≥ 0 .. x.
f(x)
.
g(x)
.
h(x)
This means that
∫ b
a
f(x) dx −
∫ b
a
g(x) dx =
∫ b
a
(f(x) − g(x)) dx =
∫ b
a
h(x) dx ≥ 0

Bounding the integral
Proof.
If m ≤ f(x) ≤ M on for all x in [a, b], then by
the previous property
∫ b
a
m dx ≤
∫ b
a
f(x) dx ≤
∫ b
a
M dx
By Property 8, the integral of a constant
func on is the product of the constant and
the width of the interval. So:
m(b − a) ≤
∫ b
a
f(x) dx ≤ M(b − a)
.. x.
y
.
M
.
f(x)
.
m
..
a
..
b

Example
Es mate
∫ 2
1
1
x
dx using the comparison proper es.

Example
Es mate
∫ 2
1
1
x
dx using the comparison proper es.
Solu on
Since
1
2
≤
1
x
≤
1
1
for all x in [1, 2], we have
1
2
· 1 ≤
∫ 2
1
1
x
dx ≤ 1 · 1

Ques on
Es mate
∫ 2
1
1
x
dx with L2 and R2. Are your es mates overes mates?
Underes mates? Impossible to tell?

Ques on
Es mate
∫ 2
1
1
x
dx with L2 and R2. Are your es mates overes mates?
Underes mates? Impossible to tell?
Answer
Since the integrand is decreasing,
Rn <
∫ 2
1
1
x
dx < Ln
for all n. So
7
12
<
∫ 2
1
1
x
dx <
5
6
.

Socratic proof
The deﬁnite integral of velocity
measures displacement (net
distance)
The deriva ve of displacement
is velocity
So we can compute
displacement with the deﬁnite
integral or the an deriva ve of
velocity
But any func on can be a
velocity func on, so . . .

Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′
for another func on F,
then ∫ b
a
f(x) dx = F(b) − F(a).

Theorem of the Day
Theorem (The Second Fundamental Theorem of Calculus)
Suppose f is integrable on [a, b] and f = F′
for another func on F,
then ∫ b
a
f(x) dx = F(b) − F(a).
Note
In Sec on 5.3, this theorem is called “The Evalua on Theorem”.
Nobody else in the world calls it that.

Proving the Second FTC
Proof.
Divide up [a, b] into n pieces of equal width ∆x =
b − a
n
as
usual.

Proof.
b − a
n
as
usual.
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)

Proof.
b − a
n
as
usual.
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)

Proof.
Form the Riemann Sum:

Proof.
Sn =
n∑
i=1
f(ci)∆x =
n∑
i=1
(F(xi) − F(xi−1))

Proof.
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))

Proof.
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)

Proof.
We have shown for each n,
Sn = F(b) − F(a)
Which does not depend on n.

Proof.
We have shown for each n,
Sn = F(b) − F(a)
Which does not depend on n.
So in the limit
∫ b
a
f(x) dx = lim
n→∞
Sn = lim
n→∞
(F(b) − F(a)) = F(b) − F(a)

Computing area with the 2nd FTC
Example
Find the area between y = x3
and the x-axis, between x = 0 and
x = 1.
.

Example
x = 1.
Solu on
A =
∫ 1
0
x3
dx =
x4
4
1
0
=
1
4 .

Example
x = 1.
Solu on
A =
∫ 1
0
x3
dx =
x4
4
1
0
=
1
4 .
Here we use the nota on F(x)|b
a or [F(x)]b
a to mean F(b) − F(a).

Example
Find the area enclosed by the parabola y = x2
and the line y = 1.

Example
and the line y = 1.
...
−1
..
1
..
1

Example
and the line y = 1.
Solu on
A = 2 −
∫ 1
−1
x2
dx = 2 −
[
x3
3
]1
−1
= 2 −
[
1
3
−
(
−
1
3
)]
=
4
3
...
−1
..
1
..
1

Computing an integral we
estimated before
Example
Evaluate the integral
∫ 1
0
4
1 + x2
dx.

estimated before
Example
∫ 1
0
4
1 + x2
dx.
Solu on
∫ 1
0
4
1 + x2
dx = 4
∫ 1
0
1
1 + x2
dx

estimated before
Example
∫ 1
0
4
1 + x2
dx.
Solu on
∫ 1
0
4
1 + x2
dx = 4
∫ 1
0
1
1 + x2
dx = 4 arctan(x)|1
0

estimated before
Example
∫ 1
0
4
1 + x2
dx.
Solu on
∫ 1
0
4
1 + x2
dx = 4
∫ 1
0
1
1 + x2
dx = 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0)

estimated before
Example
∫ 1
0
4
1 + x2
dx.
Solu on
∫ 1
0
4
1 + x2
dx = 4
∫ 1
0
1
1 + x2
dx = 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0) = 4
(π
4
− 0
)

estimated before
Example
∫ 1
0
4
1 + x2
dx.
Solu on
∫ 1
0
4
1 + x2
dx = 4
∫ 1
0
1
1 + x2
dx = 4 arctan(x)|1
0
= 4 (arctan 1 − arctan 0) = 4
(π
4
− 0
)
= π

estimated before
Example
Evaluate
∫ 2
1
1
x
dx.

estimated before
Example
Evaluate
∫ 2
1
1
x
dx.
Solu on
∫ 2
1
1
x
dx

estimated before
Example
Evaluate
∫ 2
1
1
x
dx.
Solu on
∫ 2
1
1
x
dx = ln x|2
1

estimated before
Example
Evaluate
∫ 2
1
1
x
dx.
Solu on
∫ 2
1
1
x
dx = ln x|2
1 = ln 2 − ln 1

estimated before
Example
Evaluate
∫ 2
1
1
x
dx.
Solu on
∫ 2
1
1
x
dx = ln x|2
1 = ln 2 − ln 1 = ln 2

Another way to state this theorem is:
∫ b
a
F′
(x) dx = F(b) − F(a),
or the integral of a deriva ve along an interval is the net change
over that interval. This has many interpreta ons.

Corollary
If v(t) represents the velocity of a par cle moving rec linearly, then
∫ t1
t0
v(t) dt = s(t1) − s(t0).

Corollary
If MC(x) represents the marginal cost of making x units of a product,
then
C(x) = C(0) +
∫ x
0
MC(q) dq.

Corollary
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
m(x) =
∫ x
0
ρ(s) ds.

A new notation for antiderivatives
To emphasize the rela onship between an diﬀeren a on and
integra on, we use the indeﬁnite integral nota on
∫
f(x) dx
for any func on whose deriva ve is f(x).

A new notation for antiderivatives
To emphasize the rela onship between an diﬀeren a on and
integra 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.

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 + x2
dx = arctan x + C
∫
cf(x) dx = c
∫
f(x) dx
∫
1
x
dx = 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

Computing Area with integrals
Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex
.

Example
Find the area of the region bounded by the lines x = 1, x = 4, the
x-axis, and the curve y = ex
.
Solu on
The answer is ∫ 4
1
ex
dx = ex
|4
1 = e4
− e.

Example
Find the area of the region bounded by the curve y = arcsin x, the
x-axis, and the line x = 1.

Example
Solu on
The answer is
∫ 1
0
arcsin x dx, but
we do not know an an deriva ve
for arcsin.
..
x
.
y
..
1
..
π/2

Example
Solu on
Instead compute the area as
π
2
−
∫ π/2
0
sin y dy
..
x
.
y
..
1
..
π/2

Example
Solu on
π
2
−
∫ π/2
0
sin y dy =
π
2
−[− cos x]
π/2
0
..
x
.
y
..
1
..
π/2

Example
Solu on
π
2
−
∫ π/2
0
sin y dy =
π
2
−[− cos x]
π/2
0 =
π
2
−1
..
x
.
y
..
1
..
π/2

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.

Example
Solu on
No ce the func on
y = (x − 1)(x − 2) is posi ve on [0, 1)
and (2, 3], and nega ve on (1, 2).
.. x.
y
..
1
..
2
..
3

Example
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

Example
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

Example
Solu on
A =
[1
3x3
− 3
2x2
+ 2x
]1
0
−
[1
3x3
− 3
2x2
+ 2x
]2
1
+
[1
3x3
− 3
2x2
+ 2x
]3
2
=
11
6
.. x.
y
..
1
..
2
..
3

Interpretation of “negative area”
in motion
There is an analog in rectlinear mo on:
∫ t1
t0
v(t) dt is net distance traveled.
∫ t1
t0
|v(t)| dt is total distance traveled.

What about the constant?
It seems we forgot about the +C when we say for instance
∫ 1
0
x3
dx =
x4
4
1
0
=
1
4
− 0 =
1
4
But no ce
[
x4
4
+ C
]1
0
=
(
1
4
+ C
)
− (0 + C) =
1
4
+ C − C =
1
4
no ma er what C is.
So in an diﬀeren a on for deﬁnite integrals, the constant is
immaterial.

Summary
The second Fundamental Theorem of Calculus:
∫ b
a
f(x) dx = F(b) − F(a)
where F′
= f.
Deﬁnite integrals represent net change of a func on over an
interval.
We write an deriva ves as indeﬁnite integrals
∫
f(x) dx

Evaluating definite integrals

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