Section 5.2 The Definite Integral
2010 Kiryl Tsishchanka
The Definite Integral
DEFINITION: The area A of the region S that lies under the graph of the continuous nonnegative function
f is the limit of the sum of the areas of approximating rectangles:
A = lim Rn = lim [f (x1 )∆x + f (x2 )∆x + . . . + f (xn )∆x] = lim
n→∞
n→∞
n→∞
n
X
f (xi )∆x
(1)
i=1
REMARK 1: It can be proved that the above limit always exists, since we are assuming that f is continuous.
REMARK 2: It can also be shown that we get the same value if we use left endpoints or midpoints.
Moreover, instead of using left endpoints, right endpoints or midpoints, we could take the height of the
ith rectangle to be the value of f at any number x∗i in the ith subinterval [xi−1 , xi ]. We call the numbers
x∗1 , x∗2 , . . . , x∗n the sample points. So a more general expression for the area of S is
A = lim [f (x∗1 )∆x + f (x∗2 )∆x + . . . + f (x∗n )∆x] = lim
n→∞
n→∞
n
X
f (x∗i )∆x
(2)
i=1
In this section we consider limits similar to (2) but in which f need not to be positive
or continuous and the subintervals don’t necessarily have the same length.
DEFINITION OF A DEFINITE INTEGRAL: If f is a function defined on [a, b], the definite integral
of f from a to b is a number
Zb
n
X
f (x)dx = lim
f (x∗i )∆xi
max ∆xi →0
a
|i=1
{z
Riemann sum
}
provided that this limit exists. If it does exist, we say that f is integrable on [a, b].
1
Section 5.2 The Definite Integral
2010 Kiryl Tsishchanka
THEOREM: If f is continuous on [a, b], or if f has only a finite number of jump discontinuities, then f is
Zb
integrable on [a, b]; that is, the definite integral f (x)dx exists.
a
THEOREM: If f is integrable on [a, b], then
Z
b
f (x)dx = lim
n→∞
a
where
∆x =
n
X
f (xi )∆x
i=1
b−a
n
and xi = a + i∆x
REMARK: A definite integral can be interpreted as a net area, that is, a difference of areas:
Z b
f (x)dx = A1 − A2
a
where A1 is the area of the region above the x-axis and below the graph of f , and A2 is the area of the
region below the x-axis and above the graph of f.
EXAMPLE: Evaluate
Z
0
4
(x3 − 2x)dx.
2
Section 5.2 The Definite Integral
EXAMPLE: Evaluate
2010 Kiryl Tsishchanka
Z
0
4
(x3 − 2x)dx.
Solution: We partition [0, 4] into n subintervals. We have
∆x =
Thus
x1 =
b−a
4−0
4
=
=
n
n
n
4
8
4i
4n
, x2 = , . . . , xi = , . . . , xn =
=4
n
n
n
n
Therefore
Z4
3
(x − 2x)dx = lim
n→∞
0
n
X
i=1
n
X
4i 4
f
f (xi )∆x = lim
n→∞
n n
i=1
" #
" #
n
n
3
3
X
4i
4i
4i 4
4i
4X
= lim
−2
= lim
−2
n→∞
n→∞ n
n
n
n
n
n
i=1
i=1
" n
#
n n
4 X 64i3 X 8i
4 X 64i3 8i
= lim
−
−
= lim
3
3
n→∞ n
n→∞ n
n
n
n
n
i=1
i=1
i=1
"
#
n
n
8 n(n + 1)
4 64 n2 (n + 1)2
4 64 X 3 8 X
= lim
i = lim
i −
−
n→∞ n n3
n→∞ n n3
n
4
n
2
i=1
i=1
= lim
n→∞
" 2
#
1
1
64(n + 1)2 16(n + 1)
= lim 64 1 +
−
− 16 1 +
n→∞
n2
n
n
n
= 64 − 16 = 48
3
Section 5.2 The Definite Integral
EXAMPLE: Evaluate
2010 Kiryl Tsishchanka
Z1 √
1 − x2 dx by interpreting it in terms of areas.
0
Solution: We have
Z1 √
π
1
1 − x2 dx = π · 12 =
4
4
0
EXAMPLE: Evaluate
Z3
(x − 1)dx by interpreting it in terms of areas.
0
Solution: We have
Z3
1
1
(x − 1)dx = A1 − A2 = (2 · 2) − (1 · 1) = 1.5
2
2
0
EXAMPLE: Evaluate
Z
5
Z
4
Z
7
Z
4
Z
4
Z
2
3dx.
1
EXAMPLE: Evaluate
xdx.
0
EXAMPLE: Evaluate
xdx.
2
EXAMPLE: Evaluate
(2x + 1)dx.
1
EXAMPLE: Evaluate
(2x + 1)dx.
−1
EXAMPLE: Evaluate
−2
√
4 − x2 dx.
4
Section 5.2 The Definite Integral
EXAMPLE: Evaluate
2010 Kiryl Tsishchanka
Z
5
3dx.
1
Solution: We have
EXAMPLE: Evaluate
Z
Z
5
3dx = (5 − 1) · 3 = 12
1
4
xdx.
0
Solution: We have
Z
4
xdx =
0
EXAMPLE: Evaluate
Z
4·4
=8
2
7
xdx.
2
Solution: We have
EXAMPLE: Evaluate
Z
Z
7
xdx =
2
9
45
(2 + 7)
· (7 − 2) = · 5 =
2
2
2
4
(2x + 1)dx.
1
Solution: We have
Z
4
(2x + 1)dx =
1
EXAMPLE: Evaluate
Z
12
(3 + 9)
· (4 − 1) =
· 3 = 18
2
2
4
(2x + 1)dx.
−1
Solution 1: We have
Z
4
2
√
(2x + 1)dx =
−1
(−1 + 9)
8
· (4 − (−1)) = · 5 = 20
2
2
Solution 2: We have
Z 4
9
1
·1
·9
1 81
−1 + 81
80
2
2
+
=− +
=
=
= 20
(2x + 1)dx = −
2
2
4
4
4
4
−1
EXAMPLE: Evaluate
Z
−2
Solution: We have
4 − x2 dx.
Z
2
−2
√
1
4 − x2 dx = π · 22 = 2π
2
5
Section 5.2 The Definite Integral
2010 Kiryl Tsishchanka
The Midpoint Rule
MIDPOINT RULE:
Zb
f (x)dx ≈
a
n
X
f (xi )∆x = ∆x[f (x1 ) + . . . + f (xn )]
i=1
where
∆x =
and
b−a
n
1
xi = (xi−1 + xi ) = midpoint of [xi−1 , xi ]
2
EXAMPLE: Use the Midpoint Rule with n = 5 to approximate
Z2
1
dx.
x
1
Solution: The endpoints of the five subintervals are
1,
1.2,
1.4,
1.6,
1.8,
and 2.0
so the midpoints are
1.1,
1.3, 1.5, 1.7, and 1.9
2−1
1
The width of the subintervals is ∆x =
= , so the Midpoint Rule gives
5
5
2
Z
1
dx ≈ ∆x[f (1.1) + f (1.3) + f (1.5) + f (1.7) + f (1.9)]
x
1
1 1
1
1
1
1
≈ 0.691908
=
+
+
+
+
5 1.1 1.3 1.5 1.7 1.9
Note that
L5 ≈ 0.745635,
R5 ≈ 0.645635,
Z2
1
6
1
dx = ln 2 ≈ 0.693147
x
Section 5.2 The Definite Integral
2010 Kiryl Tsishchanka
Properties of the Definite Integral
1.
Z
b
Z
a
Z
b
Z
c
f (x)dx = −
a
2.
Z
a
f (x)dx
b
f (x)dx = 0
a
3.
cdx = c(b − a)
a
4.
f (x)dx +
a
5.
Z
a
Z
c
b
f (x)dx =
Z
b
[c1 f (x) ± c2 g(x)]dx = c1
b
f (x)dx
a
Z
b
a
6. If f (x) ≥ 0 for a ≤ x ≤ b, then
f (x)dx ± c2
Z
Z
b
g(x)dx
a
b
a
7. If f (x) ≥ g(x) for a ≤ x ≤ b, then
f (x)dx ≥ 0.
Z
b
a
f (x)dx ≥
Z
b
g(x)dx.
a
8. If m ≤ f (x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤
Zb
a
EXAMPLE: Use Property 8 to estimate
Z4
√
xdx.
1
7
f (x)dx ≤ M(b − a).
Section 5.2 The Definite Integral
2010 Kiryl Tsishchanka
EXAMPLE: Use Property 8 to estimate
Z4
√
xdx.
1
Solution: Since f (x) =
√
x is an increasing function on [1, 4], its absolute minimum value on [1, 4] is
√
m = f (1) = 1 = 1
and its absolute maximum value on [1, 4] is
M = f (4) =
√
4=2
Thus, by Property 8,
1(4 − 1) ≤
Z4
√
Z4
√
xdx ≤ 2(4 − 1)
1
or
3≤
xdx ≤ 6
1
EXAMPLE: Use Property 8 to estimate
Z1
2
e−x dx.
0
2
Solution: Since f (x) = e−x is a decreasing function on [0, 1], its absolute maximum value on [0, 1] is
2
M = f (0) = e−0 = 1
and its absolute minimum value on [0, 1] is
2
m = f (1) = e−1 = e−1
Thus, by Property 8,
e (1 − 0) ≤
−1
Z1
e−x dx ≤ 1(1 − 0)
Z1
e−x dx ≤ 1
2
0
or
e−1 ≤
2
0
Since e−1 ≈ 0.3679, we can write
0.367 ≤
Z1
2
e−x dx ≤ 1
0
8