THE DEFINITE INTEGRAL PROPERTIES OF THE DEFINITE INTEGRAL In this section, we will discuss the basic properties of the definite integral. We will then move on to illustrate each property with a number of examples.
Property 1 When we defined the definite integral
∫
b
f(x) dx
a
we did so with the understanding that a < b (since a is the lower limit and b is the upper limit). On the contrary, the definite integral is also defined even when a > b. Recall that
∆x
=
(b – a)/n
(if a < b)
But, if a > b, then ∆x will be NEGATIVE. Therefore, if a > b, then
∫
b a
n
f(x) dx
∑ – i=1
=
f(x*i ) ∆x
=
–∫
a b
f(x) dx
This leads us to Property 1 of definite integrals:
∫
b a
– ∫
=
f(x) dx
a b
f(x) dx
(if a > b)
Property 2 If a = b, then
∆x
=
(b – a)/n
∆x
=
0/n
∆x
=
0
Therefore,
∫
b a
n
f(x) dx
∑
=
i=1
f(x*i ) ∆x
and this gives
n
n
∑ i=1
f(x i ) ∆x *
∑
=
i=1
f(x*i ) 0
=
0
This leads us to property 2 of definite integrals:
∫
b a
f(x) dx
=
0
(if a = b)
Property 3 This property deals with the integral of the constant function f(x) = c. Property 3 says that the integral of a constant function is the constant times the interval length (i.e. b – a). So, assuming the c > 0 and a < b, then c(b – a) represents the area under the graph (whose shape is a rectangle):
Therefore, property 3 can be stated like this:
∫
b a
c dx
c(b – a)
=
(where c is a constant)
Property 4 Just as the limit of a sum is a sum of limits, the integral of a sum is the sum of individual integrals. Below is proof of this property:
∫ =
b a
[f(x) +
]
g(x) ∆x
i
n→∞ i=1 n
lim
∑
n→∞ =
n
lim ∑ [f(x ) +
=
∫
i=1 b a
n
f(xi )
f(x) dx
So, we have property 4:
]
g(xi ) ∆x
∑
+
i=1
+
∫
b a
g(xi )
g(x) dx
∆x
=
n
lim ∑
n→∞ i=1
n
f(xi )∆x
+
lim ∑ g(x )∆x
n→∞ i=1
i
∫
b a
[ f(x) +
]
∫
=
g(x) dx
b a
+
f(x) dx
∫
b a
g(x) dx
(For easy reference, we’ll call property 4 the “Addition property of definite integrals”)
Property 5 In pre-calculus, I believe you understand that concept of transformations of functions. Recall that, if y = f(x), then, To obtain y = cf(x), stretch the graph of y = f(x) vertically by a factor of c. To obtain y = (1/c)f(x), compress the graph of y = f(x) vertically by a factor of c. The two transformation types above play a role in defining property 5, which states:
∫
b a
c
=
cf(x) dx
∫
b a
f(x) dx
(where c is a constant; integer or fraction) From an intuitive point of view, property 5 makes sense because we know that multiplying a given function f by a positive number c stretches or compresses the graph vertically by a factor of c. Consequentially, each approximating rectangle involved is either stretched or compressed vertically by a factor of c. This has the effect of multiplying the entire area by that factor c.
Property 6 This property deals with the difference between integrals. It can be proved a way similar to the addition property, but in this case, we simply put f – g as f + (-g).
∫
b a
[f(x) –
n
]
g(x) ∆x
lim ∑ [f(x ) +
=
i
n→∞ i=1
]
n
lim ∑ [f(x )∆x +
=
[–g(xi )] ∆x
n→∞ i=1
Splitting the expression gives
n
n
lim ∑
n→∞ i=1
f(xi )∆x
lim ∑ –
+
n→∞ i=1
g(xi )∆x
At this point, we apply property 5 (i.e., we let c = -1). So that we have
n
lim ∑
n
n→∞ i=1
f(xi )∆x
–
lim ∑ g(x )∆x i
n→∞ i=1
which is equivalent to
∫
b a
f(x) dx
–
∫
b a
g(x) dx
So, we have property 6:
∫
b a
[ f(x) - g(x)]dx
=
∫
b a
f(x) dx
-
∫
b a
g(x) dx
i
]
[–g(xi )]∆x
Property 7 This is one very practical property; it allows us to easily combine the integrals of the same function over adjacent intervals. This property is stated like this:
∫
c a
f(x) dx
+
∫
b c
g(x) dx
=
∫
b a
f(x) dx
and can be graphically illustrated like so:
The graph above shows the area under a graph y = f(x). Note that the area has been split into two. One area is on the interval [a, c], while the other area lies on the interval [b, c]. Therefore, following the definition of property 7, we can say that The area under y = f(x) from a to c plus the area from c to b is the total area from a to b. So far, we have treated seven properties of the definite integral. Note that each property is true IF AND ONLY IF
a < b, a = b,
or
a > b However, the next three properties are true IF AND ONLY IF a £ b. These properties are generally called the COMPARISON PROPERTIES OF THE DEFINITE INTEGRAL.
Property 8 This property simply says that a positive function has a positive area:
∫
b a
f(x) dx
0
≥
(If f(x) ≥ 0 for a ≤ x ≤ b) This property can be proved using the Riemann concept. If a function is positive, (that is, if f(x) ≥ 0), then ALL approximating rectangles will be above the x-axis. Therefore, the limit of the sum of the areas of any given number of approximating rectangles will be a positive value. In other words, the area will have a positive value. A typical example is shown:
The graph shows a runner's motion in 3 seconds. (This example was used in a previous section). In this case, 6 approximating rectangles are used to estimate the distance covered in 3 seconds. Since the approximating rectangles are above the x-axis, then the distance covered will be a positive value. Thus, if f(x) ≥ 0, the integral
∫
b a
f(x) dx
is the positive area under the curve y = f(x) from a to b.
Property 9 Property 9 says that a bigger function has a bigger area. This property follows from properties 6 and 9. If f – g ≥ 0 , and
a ≤ x ≤ b, then,f
≥ g
Therefore, we have property 9:
∫
b a
f(x) dx
≥
∫
b a
g(x) dx
(If f(x) ≥ g(x) for a ≤ x ≤ b)
Property 10 Before we examine this property, let’s take a look at the graph on the next page. Look at the graph carefully. We find that there are two rectangles: one has height m (the orange area), and the larger one (the entire blue area plus the orange area) has a height M. Following property 3, we have
Area of rectangle of height m
=
m (b – a)
Area of rectangle of height M
=
M (b – a)
Graph of y =1/x. From the graph above, we can say intuitively that the actual area under the curve y = f(x) from a to b (orange area plus fully-shaded blue area only) is definitely greater than the area of the rectangle with height m, but much less than the area of the larger rectangle with height M. So, based on the above explanation, property 10 states
∫
≤
m(b – a)
b a
≤
f(x) dx
M(b – a)
Normally, for an increasing (positive and continuous) function, the values m and M are respectively termed the absolute minimum and absolute maximum values of the function y = f(x) on the given interval [a, b]. However, in the graph above, note that the function is decreasing on the interval [a, b]; hence in this case, m is the absolute maximum and M is the absolute minimum. Thus, property 10 can be more accurately stated like this:
m(b – a)
≤
∫
b a
f(x) dx
≤
M(b – a)
(If m ≤ f(x) ≤ M for a ≤ x ≤ b)
This property is particularly more efficient if we simply want an approximation of the value of an integral, rather than using the midpoint rule, which is considerably more tedious. To conclude this topic, we’ll apply all ten properties in evaluating definite integrals. For ease of evaluation, these two identities will come in very handy:
∫ ∫
b a
b a
x2 dx
=
(b3 – a3)/3
a
x dx
=
(b2 – a2)/2
b
Study the examples carefully
EXAMPLE 1 It is already known that
∫
1 0
=
x2 dx
1/3
Use this fact and the properties of integrals to evaluate
∫
1 0
(5 – 6x2) dx
SOLUTION Note that a = 0 and b = 1. Using the addition property of integrals, we have
∫
1
(5 – 6x2) dx
0
=
∫
1
(5) dx
0
–
∫
–
6
1 0
(6x2) dx
∫
0
Applying property 5 further simplifies the integral into
∫
1 0
∫
1 0
(5 – 6x2) dx
=
∫
(5 – 6x2) dx
=
5(1 – 0)
–
=
5
=
Therefore,
∫
1 0
(5 – 6x2) dx
=
3
1 0
(5) dx
– 2
6(1/3) 3
1
(x2) dx
EXAMPLE 2 If
∫
5
618.6
=
x4 dx
2
use the properties of integrals and the information above to evaluate
∫
5 2
(1 + 3x4) dx
SOLUTION From the integral above, we see that a = 2, and b = 5. Using the addition property of definite integrals (property 4), we have
∫
5 2
(1 + 3x4) dx
=
∫
2
=
∫
2
5
(1) dx
+
(1) dx
+
∫
5 2
(3x4) dx
Applying property 5 gives
∫
5 2
(1 + 3x4) dx
5
=
1(3)
=
3
+
+
3
∫
5 2
(3x4) dx
3(618.6)
=
1855.8
1858.8
Therefore, using properties 4 and 5, we find that
∫
5
=
(1 + 3x4) dx
2
1858.8
EXAMPLE 3 Use the properties of integrals to evaluate
∫
4 1
(2x2 – 3x + 1) dx
SOLUTION Here, we’ll simply have to apply the addition and subtraction laws (properties 4 and 6). We therefore split the integral into three individual integrals to obtain
∫
4 1
(2x2) dx
–
∫
4 1
(3x) dx
+
∫
4 1
(1) dx
Applying property 5 gives
2
∫
4 1
(x2) dx
–
3
∫
4 1
(x) dx
From equations (a)and (b) we have
2[[(4)3 – (1)3]/3] – 3[[(4)2 – (1)2]/2] + 1[4 – 1]
+
∫
4 1
(1) dx
=
2[[64 – 1]/3]
– 3[[16 – 1]/2] + 1[3]
=
2[63/3] – 3[15/2] + 1[3]
=
2[21] – 3[7.5]
+ 1[3]
=
42 – 22.5 + 3
=
22.5
Therefore, using properties 4, 5 and 6, we find that
∫
4 1
(2x2 – 3x + 1) dx
=
22.5
EXAMPLE 4 Use the properties of integrals and the fact that
∫
π/2 0
=
(cos x) dx
1
To evaluate the integral
∫
π/2 0
(2cos x – 5x) dx
SOLUTION In this case, the integrand is
f(x) = 2cos x – 5x Using the subtraction property of definite integrals, we have
∫
π/2 0
(2cos x – 5x) dx
=
(2cos x – 5x) dx
=
∫
π/2
∫
π/2
0
(2cos x) dx
–
(cos x) dx
–
∫
π/2
∫
π/2
0
(5 x) dx
From property 5,
∫
π/2 0
2
0
From equation (b),
2[1]
–
5[[(π/2)2 – (0)2]/2]
=
2[1] –
5[[(π2/2)2]/2]
=
2[1]
–
5[[π2/4]/2]
=
2[1]
–
5[π2/4 × ½]
=
2[1]
– 5[π2/8]
=
2 – 5π2/8
≈
2 – 6.1685
≈
–4.1685
Therefore,
∫
π/2 0
(2cos x – 5x) dx
=
– 4.1685
5
0
(x) dx
EXAMPLE 5 Given that
∫
9 4
=
(√x) dx
38/3
What is
∫
4
(√t) dt
9
SOLUTION Let’s not forget that a definite integral is a number, not a function. In other words,
∫
b
∫
IS THE SAME AS
(√x) dx
a
b a
(√t) dt
But, in the integrals given above, we find that the limits of the integrals have been switched. This tells us to apply property 1. So, let's assume that
∫
b
=
f(x) dx
a
∫
4
∫
a
9
=
(√x) dx
38/3
Therefore,
∫
4
(√t) dt
9
–
=
b
f(x) dx
which is equivalent to
∫
–
9 4
(√x) dx
=
– 38/3
So, if
then
∫
4
∫
9
9
=
(√x) dx
4
(√t) dt
38/3
=
-38/3
EXAMPLE 6 Evaluate
∫
1 1
(x2 cos x) dx
SOLUTION If you look at the integral above very carefully, you’ll find that a = b = 1 (that is, the lower and upper limits are equal), and thus, from property 2 of definite integrals,
∫
1 1
(x2 cos x) dx
=
0
From a geometrical perspective, this property makes perfect sense because there is no interval (that is, ∆x = 0. Thus the area under such curve will be zero).
EXAMPLE 7 Write the given sum as a single integral in the form
∫
3 1
+
f(x) dx
∫
6 3
f(x) dx
∫
b a
+
f(x) dx
∫
12 6
f(x) dx
SOLUTION The best approach to solving this problem is to graph it; from there, we’ll have a visual of what exactly we're looking for. Before we graph f, let’s assume the following:
∫ ∫ ∫
3
f(x) dx
1 6 3
f(x) dx
12 6
f(x) dx
=
Area
A
=
Area
B
=
Area
C
So, the sum of the three areas (which is what we're looking for) will be:
Area D
=
Area A + Area B + Area C
The following is a typical graph of y = f(x), showing a possible representation of Area D. For the areas A, B and C, observe that 1 is the lowest limit and 12 is the highest limit. Therefore the graph and Area D will be plotted on the interval [1, 12]:
This next graph has been partitioned into the three areas A, B and C. Observe the intervals:
Thus, the sum of the three integrals, each represented as areas A, B and C is
∫
3 1
+
f(x) dx
∫
6 3
+
f(x) dx
∫
12
f(x) dx
6
which is equal to
∫
12 1
f(x) dx
This example is an illustration of property 7. The adjacent integrals A, B and C are combined to form a single integral D, which is geometrically represented as the total area (A + B + C) under the curve y = f(x) from a = 1 and b = 12.
EXAMPLE 8 Write the given difference as a single integral in the form
∫
10 2
f(x) dx
–
∫
7 2
∫
b a
f(x) dx
f(x) dx
SOLUTION We solve this problem by graphing it. But first, let's assume that
∫ ∫
10 2 7 2
f(x) dx
=
X
f(x) dx
=
Y
Note the intervals: 2 is the lowest, 10 is the highest. Thus, intuitively, this would mean that Y is inside X. The following graph shows a typical function y = f(x) plotted and showing the area on the interval [2, 10]:
This next graph shows the graph of y = f(x) partitioned into Y and another area (which is labeled Z).
From both graphs, we see that the area we are looking for is represented by the region labeled Z. We therefore conclude that
Z = X – Y This implies that
∫
10 2
f(x) dx
–
∫
7 2
f(x) dx
=
∫
10 7
f(x) dx
EXAMPLE 9 If
∫
8 2
f(x) dx
= 1.7
∫
and
8 5
f(x) dx
= 2.5
Then find
∫
5
f(x) dx
2
SOLUTION This problem is similar to those in examples 7 and 8. So far, we solved them with the aid of graphs. We can also solve the problem with careful reasoning. First, take a look at the two integrals:
∫ ∫
8 2 8 5
f(x) dx
= 1.7
(1)
f(x) dx
= 2.5
(2)
Which one is bigger? You may automatically assume that (2) is larger because the value of the integral is larger than that of (1). But DO NOT is fooled by the values. To know which is larger, look at the upper and lower limits. For (1), the interval is [2, 8], while the interval of (2) is [5, 8]. Since the integral is interpreted as the area under a curve on the interval [a, b], then intuitively, (1) is larger than (2), because the area under (1) will be much larger than (2). We also find that (2) actually occupies a portion of (1), which happens to be the total area in this case. Thus, the integral
∫
5
f(x) dx
2
will be the result of the difference between the two integrals, that is
∫
5 2
∫
=
f(x) dx
=
8
f(x) dx
2
–
∫
8 5
f(x) dx
1.7 – 2.5 = –0.8
The value of the area is negative; implying that the region is below the x-axis.
EXAMPLE 10 Use the properties of integrals to verify the inequality without evaluating the integrals:
∫
π/4 0
(sin3 x) dx
≤
∫
π/4 0
(sin2 x) dx
SOLUTION If the inequality above were true, it would mean that
sin2 x It would also mean that f(x) =
sin2
–
sin3 x
≥
0
(from property 9)
x has a larger integral than the function g(x) = sin3 x on the interval [0,p/4].
These assumptions can be graphically illustrated and proven.
Above is the graph of h(x) = sin2x – sin3 x. Observe that the entire graph is above the x-axis. This implies that h(x) = sin2x – sin3 x. ³ 0. Check this out:
The graph above shows all three functions: f, g, and h. Here’s a more accurate visual of the area we’re looking for:
The figure above shows the graphs of f(x) = sin2 x and f(x) = sin3 x plotted on the specified interval [0, π/4]. It also shows the area bounded by each graph. So, since f(x) = sin2 x is larger than f(x) = sin3 x and h(x) = sin2x – sin3 x on the interval [0, π/4]. From property 9 of definite integrals, we can then say that
∫
π/4 0
(sin3 x) dx
∫
≤
π/4
(sin2 x) dx
0
EXAMPLE 11 Use the properties of integrals to verify the inequality without evaluating the integral:
∫
≤
2
1 -1
√(1 + x2) dx
≤
2√2
SOLUTION One way of solving this problem is by expressing the values on both sides of the integral above as definite integrals on the interval [-1, 1]. So, we have
2
=
and
2√2
=
∫ ∫
1 -1 1 -1
(1) dx
√2 dx
At this point, I believe you can see where this is heading. So, what we really want to prove is
∫
1 -1
(1) dx
≤
∫
So, we're dealing with three integrands:
1 -1
√(1 + x2) dx
≤
∫
1 -1
√2 dx
f(x)
=
1
g(x)
=
√(1 + x2)
h(x)
=
√2
At this point, one good way to prove the inequality is through a graphical illustration. The following graph shows the three functions f, g and h, all plotted on the interval [-1, 1]. See below:
You can refer to the following to the read the graph effectively:
PURPLE LINE: RED CURVE: BLUE LINE :
graph of f(x) = 1 graph of g(x) = √(1 + x2) graph of h(x) = √2
GREY SHADE ONLY GREY SHADE + ORANGE SHADE GREY SHADE + ORANGE SHADE + GREEN SHADE -
Area bounded by f Area bounded by g Area bounded by h
From the graph, you’ll find that the area under g(x) = √(1 + x2) is greater than the area under f(x) = 1, but less than the area under h(x) = √2. This is the concept of property 10 of integrals. So, this means
∫
1 -1
∫
≤
(1) dx
1 -1
√(1 + x2) dx
and
∫
1 -1
≤
√(1 + x2) dx
∫
1 -1
√2 dx
Therefore, the inequality
2 is correct.
≤
∫
1 -1
√(1 + x2) dx
≤
2√2
EXAMPLE 12 Use property 10 to estimate the value of the integral:
∫
2
(1/x) dx
1
SOLUTION The first step to solving this problem is by graphing it:
From the graph, we see that the integral above is represented by the area on the interval [1, 2], that is, the blue area plus the orange area. Note from the graph that the function is decreasing on the interval [1, 2]. So, its absolute minimum on [1, 2] is
m = f(2) = 0.5 and its absolute maximum on [1, 2] is
M = f(1) = 1 Therefore, property 10 gives
m(2 – 1)
≤
∫
=
0.5 (2 – 1)
≤
∫
=
0.5
≤
∫
2 1 2 1 2 1
(1/x) dx
≤
M(2 – 1)
(1/x) dx
≤
1(2 – 1)
(1/x) dx
≤
1
This means that the value of the integral is somewhere between 0.5 and 1. In other words, an estimate of the integral is
=
0.5
≤
∫
2 1
(1/x) dx
≤
1
EXAMPLE 13 Use property 10 to estimate the value of the integral:
∫
π/3
(cos x) dx
π/4
SOLUTION If we graph the integrand y = cos x specifically on the given interval [π/4, π/3], we get this:
This function is decreasing on the interval [p/4, p/3] . Therefore, its absolute minimum on [p/4, p/3] is m =
f(π/3)
=
cos(π/3) =
0.5
while its absolute maximum on [π/4, π/3] is
M =
f(π/4)
=
cos(π/4) =
(√ 2)/2
Thus, property 10 gives
m(π/3 – π/4)
=
0.5(π/12)
≤
∫
π/4
≤
∫
π/4
=
π/24
≤
∫
=
π/24
≤
∫
π/3 π/4 π/3 π/4
π/3
π/3
(cos x) dx
≤
M(π/3 – π/4)
(cos x) dx
≤
[(√2)/2](π/12)
(cos x) dx
≤
(cos x) dx
≤
[(√2)/2](π/12) √2(π/24)
This means that the value of the integral is somewhere between π/24 and √2(π/24). Therefore,
π/24
≤
∫
π/3 π/4
(cos x) dx
≤
√2(π/24)
OR approximately
0.1308996
≤
∫
π/3 π/4
(cos x) dx
≤
0.1851201
That is, the value of
∫
π/3 π/4
(cos x) dx
lies somewhere between the values 0.1308996 and 0.1851201.
Section summary In this section, we have highlighted the ten basic properties of the definite integral and have illustrated each property with a number of examples. Next, we treat a Theorem that adequately establishes a connection between the two branches of Calculus. We will also find that this Theorem makes the takes of evaluating integrals a whole lot easier............ But, before we proceed to the study this Theorem, let’s take a look at this exciting discovery project titled: Area functions. The primary purpose of this project is to give a glimpse of the concepts of the Theorem we want to treat next.
This Theorem, which is one of the most important in Calculus, is simply called THE FUNDAMENTAL
THEOREM OF CALCULUS.
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