DefiniteIntegral

THE DEFINITE INTEGRAL A PRECISE DEFINITION OF THE DEFINITE INTEGRAL Recall from the previous section that the limit

n

lim ∑

(1)

f(xi *)∆ x

n→∞ i=1

consistently shows up whenever we want to compute the area under a curve or the distance covered by an object.

This particular type of limit has several practical applications, such as work, volume, the length of a curve, just to mention a few. Because of its versatility, this limit has been given a special name: The Definite Integral. First, we’ll be dealing with a definite Integral, whose definition is given below (taken directly from my textbook):

If f is a continuous function defined for a ≤ x ≤ b, we divide the interval [a, b] into n subintervals of equal width ∆x

= (b – a)/n.

We let x0, x1, x2......xn be the endpoints of these subintervals (where x0 = a, xn = b) and we choose sample points x1*, x2*.....xn* in these subintervals. Therefore the definite integral of f from a to b is

∫ The symbol

b

f(x) d(x)

a

n

lim ∑

=

n→∞ i=1

f(x*i ) ∆x

(2)

∫ was introduced by Gottfried Wilhelm Leibniz and it's called the Integral Sign. This choice of symbol

is much appreciated especially when we see Integration as a process of summation (the integral sign is basically a “stretched” S and is chosen specifically because an Integral is a limit of sums). The process of computing an Integral is called Integration. A definite integral is generally represented by the symbol

∫

b a

f(x) d(x)

where

∫

is the Integral sign,

b

is the upper limit of Integration

a

is the lower limit of Integration

f(x)

is the Integrand

d(x)

HAS NO MEANING BY ITSELF.

Note that the definite integral is actually a finite number, not a function (unlike an indefinite integral which can represent an entire family of functions). This means it does not depend on x.

The definition of a definite integral could take any of these three forms (which basically depends on which sample points we choose to use; left endpoints, right endpoints or midpoints): If we take sample points to be right hand endpoints, then the definite integral would be defined as:

∫

b a

f(x) d(x)

=

n

lim ∑

n→∞ i=1

f(xi ) ∆x

(3)

On the other hand, if the sample points happen to be left hand endpoints, the definition becomes

∫

b a

f(x) d(x)

n

lim ∑

=

n→∞ i=1

(4)

f(xi-1) ∆x

Then again, the sample points could be midpoints, or ANY other number between the subinterval [xi–1, xi]:

∫

b a

f(x) d(x)

n

lim ∑

=

n→∞ i=1

(5)

f(xi *) ∆x

The sum which occurs in the definition of a definite integral, the same sum that constantly pops up when we want to calculate areas or distances, i.e.:

n

lim ∑

n→∞ i=1

f(xi *) ∆x

is called a Riemann Sum, after the German Mathematician Bernhard Riemann (1826-1866). From what we know about areas, if f is a positive function, then the Riemann Sum is basically the sum of areas of approximating rectangles.

The graph above is that of f(x)=√5–x

n

∑ i=1

drawn on the interval [0,5].Observe the

f(x*i) ∆x

10 approximating rectangles used in estimating the required area.

Thus, based on the definition of a Riemann sum, if f(x) ≥ 0, then the sum

10

∑ f(x ) ∆x *

i=1

i

represents the sum of the areas of the approximating rectangles.

Also, If f(x) ≥ 0, the integral

∫

b a

f(x) d(x)

≡

∫

5 0

√5 − x

d(x)

represents the area under the curve f(x) = √ 5–x on [0, 5]. (In this case, a = 0 and b = 5).

Based on the illustration above, we say that the definite integral is simply the area of the region P under the function y = f(x) on the specific interval [a, b]. So far, we have dealt with positive functions; i.e. Functions that lie above the xaxis. We know for sure that functions can also have negative values; either the entire function lies below the x-axis, or simply part of the function lies below the x axis. So, how do we find the area of any given region under such functions? Consider the illustration below: The graph below is that of f(x) = 2x2 – 5x plotted on the interval [0, 4]. Here, 8 approximating rectangles are used.

Observe the layout: the first five triangles are below the x-axis, while the remaining three are above the x-axis.

Therefore, we say that f has taken on both positive and negative values.

In cases like this we also use the Riemann Sum (which in this case is the sum of the areas of the rectangles that lie above the x axis and the negatives of the areas of the rectangles that lie below the x axis. In other words, the area of the region is the area of the region above the x axis minus the area of the region below the x axis). So, if f is not entirely a positive function, (like the one above) then its definite integral can be defined a its net area; i.e., a difference of areas:

This means

âˆŤ

b a

f(x) d(x)

=

A1

- A2

where A1 represents the area above the x-axis and A2 represents the area under the x-axis.

In summary, the integral can be interpreted in the following ways: As the area under the curve of a continuous function y = f(x) from a to b.

As the limit of the sums of areas of approximating rectangles (based on left-hand, or right-hand midpoints, or ANY number within the given sub-interval) As a NET AREA (if and only if f takes on both positive and negative values).

As the limit of the sums of areas of approximating rectangles (assuming the sub-intervals are not equally spaced).

These interpretations of an integral are adequately represented by equations (1) through (5) above. Let’s go back to the equation that defines a definite integral:

∫

b a

n

lim ∑

=

f(x) d(x)

n→∞ i=1

f(x*i ) ∆x

Having established that an integral is a limit of sums, Leibniz did the the this: he replaced

Σ

with

∫

xi*

with

x

dx

with

∆x

He specifically chose these symbols so that it would serve as a constant reminder that an integral is indeed a limit of sums. In other words, the symbols were chosen as a reminder of the limiting process.

So now, we’ll be examining three examples, each which will illustrate how limits can be expressed as definite integrals on the given intervals.

EXAMPLE 1 n

lim ∑ xi sin xi ∆x n→∞ i=1

[0, π]

Solution If you examine the above limit, you’ll notice that the interval is [0, p]. This implies that [0, π] = [a, b] Which means that a = 0 (which represents the lower limit), and b = π (which represents the upper limit). If we decide to go by Leibniz’s notations, then, we replace

∑

by

∫,

xi

by

x,

∆x

by

dx.

and

Therefore, the limit can be expressed as the definite integral

∫

b a

∫

=

f(x) dx

b a

x sin x dx

Here’s one way of explaining the result we just obtained above: If we compare the given limit with the limit in the definition of a definite integral, we then have:

n

n

lim ∑ xi sin xi ∆x n→∞ i=1

lim ∑ f(xi) ∆x n→∞ i=1

=

This implies that the sample points used in the limit above are right endpoints. This means

xi*

=

xi

(We already know the intervals: a = 0, and b = π). So, using the definition of an integral, we have

n

n

lim ∑ xi sin xi ∆x n→∞ i=1

lim ∑ f(xi) ∆x n→∞ i=1

=

Which corresponds to

∫

b a

f(x) dx

∫

=

b a

x sin x dx

EXAMPLE 2 n

lim ∑ n→∞ i=1

xi

∆x

1 + xi

[1, 5]

Solution (Here, note that the sample points used are right-end points). By making the appropriate symbol replacements, and let a = 1, and b = 5, then we have

n

lim ∑ n→∞ i=1

xi 1 + xi

∆x

=

∫

5 1

x 1+ x

dx

EXAMPLE 3 n

lim ∑ n→∞ i=1

[2(x*i)2

– 5(x*i)] ∆x

[0, 1]

Solution As usual, if we make the proper symbol replacements, and let a = 0, and b = 1, therefore,

n

lim ∑ n→∞ i=1

[2(x*i)2

– 5(x*i)] ∆x

=

∫

1 0

(2x2 – 5x) dx

At this point, we now understand that an integral is a limit of sums.

Therefore, in order to fully understand the concept of the definite integral, we need to know how to work with large sums. In the previous sections, we found that sums are written more compactly using the sigma notation. It is recommended that you understand how to use the sigma notation to write large sums before you can really understand the idea behind evaluating definite integrals. CLICK HERE for a brief discussion on sigma notation, including additional topics (taken from stewartcalculus.com, written by James Stewart). I will also like you to take a look at some problems; just to brush up your math skills. CLICK HERE (also taken from stewartcalculus.com).

Visit this site for more resources.

Let’s treat some examples and see how definite integrals are evaluated.

Example 1 (a)

Evaluate the Riemann sum for f(x) = 2x2 – 5x taking the sample points to be right hand endpoints and a = 0, b = 1, and n = 5.

(b)

Evaluate

∫

1 0

(2x2 – 5x) dx

SOLUTION (a)

With n = 5, the sub-interval width (∆x) is ∆x

=

(b – a)/n

∆x

=

(1 – 0)/5

= 1/5

Hence, the width of each subinterval is ∆x = 1/5. And so, the right hand endpoints would be

x1

=

a + ∆x

= 1/5

=

0.2

x2

=

a + 2∆x

= 2/5

=

0.4

x3

=

a + 3∆x

= 3/5

=

0.6

x4

=

a + 4∆x

= 4/5

=

0.8

x5

=

a + 5∆x

= 5/5

=

1.0

Therefore, the Riemann sum is

R5

=

R5

=

R5

=

R5

=

R5

=

n

∑

i=1

f(xi )∆x

f(0.2)∆x + f(0.4)∆x + f(0.6)∆x + f(0.8)∆x + f(1)∆x ∆x[f(0.2) +

f(0.4) +

f(0.6) +

1/5[-0.92 – 1.68 – 2.28 – 2.72 - 3] 1/5[-10.6]

=

-2.12

f(0.8) +

f(1.0)]

The above result should not be surprising because f is a negative function (you will find that, on the interval [0, 1], the graph lies below the x-axis): its graph is drawn below:

So, the Riemann sum computed above represents the sum of the areas of the five approximating rectangles lying below the x-axis.

(b)

We computed the Riemann sum for n = 5 (i.e., using 5 approximating rectangles), and we obtained -2.12. This

time, we do not give n any value. So, with n sub-intervals, the width of each subinterval (∆x) is

∆x

= (b – a)/n

∆x

= (1 – 0)/n

= 1/n

Hence, the sub-interval widths are

x1

=

a + ∆x

= 1/n

x2

=

a + 2∆x

= 2/n

x3

=

a + 3∆x

= 3/n

and so on. In general,

xi

=

a + i∆x

=

0 + (1/n)i

Thus,

xi

=

i/n

Since, we are using right endpoints, we can then say that

∫

1 0

(2x2 – 5x) dx

n

∑ i=1

and

n

lim ∑ f(x )∆x i n→∞ i=1

=

n

=

f(xi )∆x

lim ∑ f(i/n) 1/n n→∞ i=1

this gives

n

∑ [2(i/n)2

lim 1/n n→∞

– 5(i/n)]

i=1

In the square brackets, we bring out i to give

lim 1/n n→∞

n

∑ [(2/n2)i2

– (5/n)i]

i=1

Expanding the expression gives

lim 1/n n→∞

n

∑ [(2/n2)i2

i=1

n

=

lim ∑ n→∞ i=1

=

lim n→∞

– (5/n)i]

[(2/n3)i2

=

n

∑ i [(2/n ) i=1

2

n

∑ [(1/n)(2/n2)i2

– (1/n)(5/n)i]

i=1

– (5/n2)i]

n

3

lim n→∞

– (5/n2)

∑ i=1

]

i

We know that

n

∑ i=1

i2

=

i

=

n(n+1)(2n+1) 2

and

n

∑ i=1

n(n+1) 2

So that we have

=

(2/n3)[(n(n+1)(2n + 1)]/6

lim n→∞

(2n)((n+1)(2n + 1))/6n3

=

lim n→∞

=

lim n→∞

=

[[ [[ [[ [[ [[

lim n→∞

lim n→∞

[(5/n )[(n(n + 1))/2]]] 2

[((5n)(n + 1))/2n ]]

–

–

[(5(n + 1))/2n]]

]

]

–

]

((n+1)(2n + 1))/3n2

(2n2 + 3n + 2)/3n2

]

–

2

[(5n + 5)/2n]] ]

(2n2/3n2) + (3n/3n2) + (2/3n2)

–

[(5n/2n) + (5/2n)]]

=

[[

lim n→∞

]

(2/3) + (1/n) + 2/3(1/n2)

–

[(5/2) + 5/2(1/n)]]

Remember that limits at infinity equals 0, that is,

lim (1/n) = 0 n→∞ This results in

lim [2/3] n→∞

lim [1/n] n→∞

+

+

lim [2/3][1/n2] n→∞

=

[2/3] + [0] + [(2/3)(0)] – [5/2] – [(2/3)(0)]

=

2/3 + 0 + 0 – 5/2 – 0

=

2/3 –

5/2

=

lim [5/2] n→∞

–

–

lim [2/3][1/n2] n→∞

– 11/6

So, the actual value of the area under the graph is -11/6. In other words,

∫

1 0

(2x2 – 5x) dx

=

-11/6

Example 2 Express the integral as a limit of sums. Then evaluate, using a CAS (Computer Algebra System) to find both the sum and the limit.

∫

π 0

(sin 5x) dx

SOLUTION Here, we have f(x) = sin 5x, and a = 0 and b = π. Therefore, the sub-interval width (∆x) is

∆x ∆x

= =

(b – a)/n (π – 0)/n

=

π/n

Hence, endpoints of the sub-intervals are

x0

=

a

=

0

x1

=

a + ∆x

=

π/n

x2

=

a + 2∆x

=

2π/n

x3

=

a + 3∆x

x4

=

a + 4∆x

= =

3π/n 4π/n

and so on. In general,

xi

=

a +∆xi

Thus, from equation 3, we have

∫

π 0

n

(sin 5x) dx

=

∑

lim f(xi )∆x n→∞ i=1

=

πi/n

and

n

n

lim ∑ f(xi )∆x n→∞ i=1

=

lim ∑ f(πi/n) π/n n→∞ i=1

∫

Using a CAS, we find that

π 0

(sin 5x) dx

=

=

lim π/n n→∞

n

∑

[sin 5(πi/n)]

i=1

0.4

Example 3 Express the integral as a limit of sums. Then evaluate, using a CAS (Computer Algebra System) to find both the sum and the limit.

∫

10 2

x6 dx

SOLUTION For the given integral,

f(x) = x6 a = 2 and b = 10. Hence, the sub-interval width (∆x) is

∆x ∆x

= =

(b – a)/n (10 – 2)/n

=

=

2

Therefore, the endpoints of the sub-intervals are

x0 x1

= =

a + ∆x

=

(2n + 8)/n

x2

=

a + 2∆x

=

(2n + 16)/n

x3

=

a + 3∆x

=

(2n + 24)/n

x4

=

a + 4∆x

=

(2n + 32)/n

xi

=

a + ∆xi

xi

=

2 + (8i/n)

a

and so on. In general,

This means

From equation 3, we have

∫

10 2

n

x dx 6

=

lim ∑ f(xi )∆x n→∞ i=1

Putting xi = 2 + (8i/n) and ∆x = 8/n in the expression above gives

∫

10 2

n

x6 dx

=

∑ f(2 + (8i/n))

lim n→∞ i=1

6

× 8/n

Then we have

∫

10 2

n

x dx 6

=

lim 8/n ∑ f(2 + (8i/n))6 n→∞ i=1

8/n

If we ask a CAS to evaluate and simplify the sum, we obtain

∫

10 2

x6 dx

=

1,428,553.14286

So far, we have been able to express integrals as limits of sums. The following examples will illustrate how we can evaluate integrals by interpreting them in terms of areas.

Example 4 Evaluate the integral by interpreting it in terms of area.

∫

3 1

(1 + 2x) dx

SOLUTION One way (and probably the best way) of solving this problem is by first drawing a graph of the integrand. The figure below shows the graph of y = 1 + 2x drawn and shaded on the interval [1, 3].

From the figure, we compute the integral as the area under the straight line y = f(x) = 1 + 2x from 1 to 3. The graph shows that the area under the region is a trapezium (coordinates shown):

3, 7 Therefore, the area of the trapezium is

1, 3

7

3 1, 0

A

=

½(sum of parallel sides) × height

A

=

½(3 + 7) × 2

A

=

10

Thus, the area under the line y = 1 + 2x from 1 to 3 is 10. In other words,

2

3, 0

∫

3 1

(1 + 2x) dx

=

10

Example 5 Evaluate the integral by interpreting it in terms of area.

∫

2 -2

√(4 - x2) dx

SOLUTION Note that F(x) = √(4 - x2)

≥

0. Therefore, we can interpret this integral as the area under the curve y = F(x)

from -2 to 2. The integrand can be re-written as

y2 = 4 – x2

Which means x2 + y2 = 4 This implies that the graph of f would be semicircle with radius 2:

The graph has been to view the graph of F on the interval [-2, 2]. Therefore, the integral can be expressed as the area of the semicircle. So, we have

∫

2 -2

√(4 - x2) dx

= =

½

πR2

½

π(2)2

≈

6.2832

Thus, the area under the curve x2 + y2 = 4 from -2 to 2 is approximately 6.2832.

Example 6 Evaluate the integral by interpreting it in terms of area.

∫

0 -3

(1 + √(9 - x2)) dx

SOLUTION From the given integral above, the integrand is

f(x) = 1 + √(9 – x2). Its graph is drawn below (on the interval

[-3,0]):

If we “extract” the required area from the graph, we have

4, 0

A - 3, 1

B

- 3, 0 Which then gives

A 3 1

B 3

1, 0 0, 0

Thus, the integral can be interpreted as the area of the quarter-circle plus the area of the rectangle. So, then we have:

A1 =

Area of semicircle =

¼ πr2 =

¼ π(3)2 =

9π/4

A2 =

Area of rectangle

lw

3×1

3

=

=

=

Thus, the sum of areas equals

=

A1 + A2

=

=

3 + (9π/4) ≈

¼ πr2 + lw =

(9π/4) + 3

10.0865

Therefore, the area under the graph y = (1 + √(9 - x2)) from -3 to 0 is approximately 10.0865.

Example 7 Evaluate the integral by interpreting it in terms of area.

∫

3 -1

(2 – x) dx

SOLUTION From the given integral, we see that f(x) = 2 – x. Its graph is shown below:

If you look at the graph carefully, you’ll find that we’ll have to compute two areas A and B (both of which are triangles):

-1, 3

-1, 0 This gives

2, 0

A

B

3, 0 3, -1

2, 0

1

B

1

1

A 3

We can therefore interpret the integral as the area under the region from -1 to 3, which is represented by the two rectangles. Thus, Area under region = Sum of areas of triangles A and B. Area of A

=

½ bh

=

½(3 × 3)

=

½(9)

=

4.5

Area of B

=

½ bh

=

½(1 × -1)

=

½(-1)

=

-0.5

Therefore, Total Area

=

Area of A + Area of B

=

4.5 + (-0.5)

=

4.0

Thus, the area under the line y = 2 – x from -1 to 3 is 4.

Example 8 Evaluate the integral by interpreting it in terms of area.

∫

2 -2

( 1 – |x|) dx

SOLUTION We start by graphing the function f(x) = 1 - |x| on the interval [-2, 2]. (See the next page). From the graph, we see that the integral can be interpreted as the sum of the areas of the triangles A, B and C.

We break the required area apart into three triangles:

0, 1

A -1, 0

-2, 0

1, 0

-1, 0

B

1, 0

C

-2, -1

2, 0

2, -1

We then have

1

A 2

1

-1

B

1

C

-1

So,

Area of A = ½ bh =

½ (1 × 2)

=1

Area of B = ½ bh =

½ (1 × -1)

= - 0.5

Area of C = ½ bh =

½ (1 × -1)

= - 0.5

Hence,

TOTAL AREA

=

Area of ∆A + Area of ∆B + Area of ∆C

=

1.0 + (-0.5) + (-0.5)

=

1.0 – 0.5 – 0.5

=

1.0 – 1.0

=

0

This implies that the area under the line y = 1 - |x| from -2 to 2 is zero.

When I used a graphing calculator, the answer I obtained was not exactly zero. I got 1.09674266305 × 10-14

Well, that’s practically zero anyway! The result indicates that the definite integral is the limiting value of the sum of the areas of the triangles.

Example 9 Evaluate the integral by interpreting it in terms of area.

∫

3 0

( |3x – 5|) dx

SOLUTION On the next page, you will see the graph of the integrand f(x) = |3x – 5| plotted on the interval [0, 3].

From the graph above, we can say that the integral can be expressed as the sum of the areas of triangles A and B:

3, 4

5, 0

1.66, 0

0, 0

1.66, 0

3, 0

4

5 1.66

1.34

Therefore, Area of A

=

Area of B

=

½ bh ½ bh

=

½ (5 × 1.66)

=

4.15

=

½ (4 × 1.34)

=

2.68

Hence,

TOTAL AREA

=

Area of ∆A + Area of ∆B

=

4.15 + 2.68

=

6.83

Therefore, the definite integral, expressed as a sum of the areas of two triangles, is approximately equal to 6.83.

In the next section, we’ll be looking at a more efficient way of finding estimates of definite integrals, using a method called the MIDPOINT RULE.

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