1. Section 5.3
Evaluating Definite Integrals
V63.0121.002.2010Su, Calculus I
New York University
June 21, 2010
Announcements
Final Exam is Thursday in class
2. Announcements
Sections 5.3–5.4 today
Section 5.5 tomorrow
Review and Movie Day
Wednesday
Final exam Thursday
roughly half-and-half
MC/FR
FR is all post-midterm
MC might have some
pre-midterm stuff on it
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 2 / 44
3. Resurrection Policy
If your final score beats your midterm score, we will add 10% to its weight,
and subtract 10% from the midterm weight.
Image credit: Scott Beale / Laughing Squid
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 3 / 44
4. Objectives
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 4 / 44
5. Outline
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 5 / 44
6. The definite integral as a limit
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 6 / 44
7. Notation/Terminology
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 7 / 44
8. Example
1
4
Estimate dx using the midpoint rule and four divisions.
0 1 + x2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
9. Example
1
4
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 = 2
+ 2
+ 2
+
4 1 + (1/8) 1 + (3/8) 1 + (5/8) 1 + (7/8)2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
10. Example
1
4
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 = 2
+ 2
+ 2
+
4 1 + (1/8) 1 + (3/8) 1 + (5/8) 1 + (7/8)2
1 4 4 4 4
= + + +
4 65/64 73/64 89/64 113/64
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 8 / 44
12. Properties of the integral
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
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13. More Properties of the Integral
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
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14. Definite Integrals We Know So Far
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 2
By brute force we computed x
1 1
1 1
x 2 dx = x 3 dx =
0 3 0 4
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 11 / 44
15. Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
6. If f (x) ≥ 0 for all x in [a, b], then
b
f (x) dx ≥ 0
a
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 12 / 44
16. 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 }:
n n
Sn = f (ci ) ∆x ≥ 0 · ∆x = 0
i=1 ≥0 i=1
Since Sn ≥ 0 for all n, the limit of {Sn } is nonnegative, too:
b
f (x) dx = lim Sn ≥ 0
a n→∞
≥0
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 13 / 44
17. Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
6. If f (x) ≥ 0 for all x in [a, b], then
b
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 14 / 44
18. The definite 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
h(x) dx ≥ 0
a
This means that
b b b b
f (x) dx − g (x) dx = (f (x) − g (x)) dx = h(x) dx ≥ 0
a a a a
So
b b
f (x) dx ≥ g (x) dx
a a
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 15 / 44
19. Comparison Properties of the Integral
Theorem
Let f and g be integrable functions on [a, b].
6. If f (x) ≥ 0 for all x in [a, b], then
b
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 16 / 44
20. Bounding the integral using bounds of the function
Proof.
If m ≤ f (x) ≤ M on for all x in [a, b], then by the previous property
b b b
m dx ≤ f (x) dx ≤ M dx
a a a
By Property 1, the integral of a constant function is the product of the
constant and the width of the interval. So:
b
m(b − a) ≤ f (x) dx ≤ M(b − a)
a
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 17 / 44
21. Estimating an integral with inequalities
Example
2
1
Estimate dx using Property 8.
1 x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 18 / 44
22. Estimating an integral with inequalities
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 18 / 44
23. Outline
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 19 / 44
24. Socratic proof
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 . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 20 / 44
25. Theorem of the Day
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 21 / 44
26. Theorem of the Day
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 21 / 44
27. Proving the Second FTC
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 )
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 22 / 44
28. Proof continued
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 23 / 44
29. Proof continued
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)
See if you can spot the invocation of the Mean Value Theorem!
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 23 / 44
30. Proof Completed
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→∞
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 24 / 44
31. Computing area with the Second FTC
Example
Find the area between y = x 3 and the x-axis, between x = 0 and x = 1.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
32. Computing area with the Second FTC
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
33. Computing area with the Second FTC
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 25 / 44
34. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x 2 and y = 1.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
35. Computing area with the Second FTC
Example
Find the area enclosed by the parabola y = x 2 and y = 1.
1
−1 1
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
36. Computing area with the Second FTC
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 26 / 44
37. Computing an integral we estimated before
Example
1
4
Evaluate the integral dx.
0 1 + x2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 27 / 44
39. Computing an integral we estimated before
Example
1
4
Evaluate the integral dx.
0 1 + x2
Solution
1 1
4 1
dx = 4 dx
0 1 + x2 0 1 + x2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
40. Computing an integral we estimated before
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
41. Computing an integral we estimated before
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)
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
42. Computing an integral we estimated before
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
4
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
43. Computing an integral we estimated before
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 29 / 44
44. Computing an integral we estimated before
Example
2
1
Evaluate dx.
1 x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 30 / 44
45. Estimating an integral with inequalities
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 31 / 44
46. Computing an integral we estimated before
Example
2
1
Evaluate dx.
1 x
Solution
2
1
dx
1 x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
47. Computing an integral we estimated before
Example
2
1
Evaluate dx.
1 x
Solution
2
1
dx = ln x|2
1
1 x
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
48. Computing an integral we estimated before
Example
2
1
Evaluate dx.
1 x
Solution
2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
49. Computing an integral we estimated before
Example
2
1
Evaluate dx.
1 x
Solution
2
1
dx = ln x|2
1
1 x
= ln 2 − ln 1
= ln 2
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 32 / 44
50. Outline
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 33 / 44
51. The Integral as Total Change
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:
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
52. The Integral as Total Change
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
53. The Integral as Total Change
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
54. The Integral as Total Change
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 34 / 44
55. Outline
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 35 / 44
56. A new notation for antiderivatives
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).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 36 / 44
57. A new notation for antiderivatives
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 36 / 44
58. My first table of integrals
[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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 37 / 44
59. Outline
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 38 / 44
60. Example
Find the area between the graph of y = (x − 1)(x − 2), the x-axis, and the
vertical lines x = 0 and x = 3.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 39 / 44
61. Example
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.
0
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62. Graph
y
x
1 2 3
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63. Example
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).
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 41 / 44
64. Example
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 41 / 44
65. Interpretation of “negative area” in motion
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
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66. What about the constant?
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.
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 43 / 44
67. Summary
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
V63.0121.002.2010Su, Calculus I (NYU) Section 5.3 Evaluating Definite Integrals June 21, 2010 44 / 44