]> Fundamentals

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 2: Definite Integrals; page 1

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Previous topic: Section 1 Indefinite Integrals, page 4 Simple Differential Equations

Fundamentals

Area under a curve

The problem of calculating the area of different geometric forms preocupied mankind for a very long time, as seen in ancient written documents. Analytical geometry uses presentations in a Cartesian coordinate system, which facilitates more precise answers. In a metric system we can ask the question what is the area between a curve and the x axis in a given interval of x. The area is commonly denoted by the capital letters A (for area) or S (for surface).

Let's take the simple example of a single valued, continuous and bounded non-negative function in a closed interval of x :

$y = f (x) {\begin{cases} 0 \leq y \leq bound \\ for a \leq x \leq b \end{cases}}$	(2.2.1.1)

If we denote by y_m and y_M the corresponding minimal and maximal values of y in the interval, then it is obvious that the required area S is bounded by

$y_{m} (b - a) \leq S \leq y_{M} (b - a)$	(2.2.1.2)

One can divide the interval in two parts and add up separetely, the minimal and the maximal contributions of the parts. The continuation of this procedure with more and more subdivided intervals, leads to closing the bounds, and the lower and upper bounds form monotonic convergent sequences to the same limit S . In view of the sandwich rule we could also randomly choose any value of the function for each particular interval (not necessarily the minimal or the maximal value) and obtain the same limit.

We are going to illustrate this behaviour with the following example. The function within the interval (2.2.1.3)

$\begin{array}{l} y = 1 - {(\frac{x}{3})}^{2} \\ 0 \leq x \leq 3 \end{array}}$	(2.2.1.3)

is positive and monotonic decreasing, therefore the minimal and maximal values corresponding to any sub interval are on its right and left edge accordingly. The area under the curve is S = 2 (we will later learn how to calculate it) and we will obtain graphically the convergence to this value by consecutive subdivision of the intervals in two equal parts as seen in Fig. Area of parabola(1).

Instead of using the area of the rectangles corresponding to the left-most and right-most values of y for each interval, we could use only the arithmetic average of both. This won't affect the result of the convergence, because of the sandwich rule previously mentioned. On the other hand the area of this rectangle is equivalent to the area of a trapezoid, with vertices at the left-most and the right-most values of y, as seen in Fig. Area of a trapezoid.

The comparison of two methods for convergence: maximal rectangles and trapezoids is illustrated in Fig. Area of parabola(2).

In the case of a monotonic function (as shown in the examples so far), the end points of an interval represent the minimal and the maximal values of the function in the interval, and therefore the use of trapezoids takes the average of these values. In the case of a non-monotonic function, there are intervals in which the end points are not anymore the corresponding minimal and the maximal values, but never the less the use of trapezoids converges to the area under the function due to the sandwich rule.

As an example of a function that is not monotonic in all the interval, we'll use

$\begin{array}{l} y = 4 - x^{2} \\ - 1 \leq x \leq 2 \end{array}}$	(2.2.1.4)

There is a maximum at x = 0 , but of course at each consecutive approximation only one of the intervals could contain it, making the behaviour of the function monotonic in all the others. The maximum is at x = 0 and the area under the curve is S = 9 . The use of the trapezoid method for this case is shown in Fig. The trapezoid method.

The definite integral

After we have studied how the area under a curve can be calculated, we are ready to introduce the Riemann's definition of a definite integral.

If a function f(x) in an interval of x = [a,b] is defined, bounded and continuous except for a countable number of points, then the limit of the sum over n sub-intervals Δ_kx

$\lim_{n \to \infty} \sum_{k = 1}^{n} f (ξ_{k}) Δ_{k} x$	(2.2.1.5)

where all the sub-intervals converge to zero and ξ_k is any value of x in the appropriate sub-interval, converges to a finite value. The limit is called the definite integral of f(x) for the interval of x = [a,b] and is denoted by

$\int_{a}^{b} f (x) d x = \lim_{n \to \infty} \sum_{k = 1}^{n} f (ξ_{k}) Δ_{k} x$	(2.2.1.6)

The values of a and b are called in this notation, the limits of the integral. Such a function is called integrable (could be integrated) in the interval. Later on we'll enlarge the field of integrable functions beyond the Riemann's definition.

As we'll see, the similarity in notation and name with the indefinite integrals is not accidental. Without making the formal proof of (2.2.1.6), we'll study this definition more carefully.

One difference with the case of an area under a curve is that there is no restriction on functions with non-negative values now. It means that the integral can be also negative. The sign of the integral is not dependent only on the sign of the function, but also on the direction of integration, from a to b or from b to a, since the direction inverts the sign of the sub-intervals in the summation. Formally it can be written as

$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$	(2.2.1.7)

which translated into words is: the inversion of the limits of an integral inverts its sign. As a direct result of the definition it follows that

$\int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x = \int_{a}^{c} f (x) d x$	(2.2.1.8)

which combined with (2.2.1.7) gives

$\int_{a}^{a} f (x) d x = 0$	(2.2.1.9)

From the definition of definite integrals it follows also that

$\begin{array}{l} for {\begin{cases} a \leq x \leq b \\ m \leq f (x) \leq M \end{cases} \\ m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a) \end{array}}$	(2.2.1.10)

where m and M are the bounding values of the function f(x) in the interval.

The trapezoid method could also be applied when calculating a definite integral, but always be careful of the sign.

In the case of calculating the area under a curve, we did not see an example of a discontinuous function. We'll use the following example with one point of discontinuity, but with an easily calculated definite integral.

$\begin{array}{l} f (x) = \frac{x}{\| x \|} = {\begin{cases} - 1 (x < 0) \\ 1 (x > 0) \end{cases} \\ f (0) = b (b is a constant) \\ \int_{- 1}^{3} f (x) d x = - \int_{- 1}^{0} d x + \int_{0}^{3} d x = \\ = - 1 + 3 = 2 \end{array}}$	(2.2.1.11)

The function should be defined and therefore the constant b was introduced. This value at the point of discontinuity cannot affect the calculation of an area, so we should obtain the same result for any choice of b . By using the trapezoid method, if the point of discontinuity does not coincide with the border of any subinterval, the value of b is irrelevant. If it coincides, the sequence leading to the limit could be b dependent, but not the final result. This example is shown graphically and animated by the use of the trapezoid method in Fig. Integration with discontinuity.

In addition to the trapezoid method, there are faster ways for numerical calculations of definite integrals such as: the Simpson rule and the Gauss integration, which make interpolation of the function based on the information from the neighbouring intervals. Those methods are very often used in scientific analysis and computer programs exist for their implementation.

Relation with indefinite integrals

Although the definite integrals were known a long time before the invention of calculus in the 17^th century, only at this time the use of the indefinite integrals made the definite integrals analytically (and not only numerically) solvable.

Let's try to differentiate with respect of x , the following expression of a definite integral

$\frac{d}{d x} \int_{a}^{x} f (u) d u$	(2.2.1.12)

By using the properties of derivatives and of definite integrals one obtains

$\begin{array}{l} \frac{d}{d x} \int_{a}^{x} f (u) d u = \lim_{Δ x \to 0} \frac{\int_{a}^{x + Δ x} f (u) d u - \int_{a}^{x} f (u) d u}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{\int_{x}^{a} f (u) d u + \int_{a}^{x + Δ x} f (u) d u}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{\int_{x}^{x + Δ x} f (u) d u}{Δ x} = \lim_{Δ x \to 0} \frac{f (x) Δ x}{Δ x} = f (x) \end{array}}$	(2.2.1.13)

On the other hand we know that by definition the derivative of the indefinite integral I(x) gives the same result:

$\frac{d}{d x} I (x) = \frac{d}{d x} \int f (x) d x = f (x)$	(2.2.1.14)

We obtained therefore that up to an additive constant

$\int_{a}^{x} f (u) d u = \int f (x) d x = I (x)$	(2.2.1.15)

An additive constant comes with the indefinite integral which should be adjusted in order to balance the contribution of the lower limit a of the definite integral.

Now we are ready to express any definite integral in terms of indefinite integrals:

$\begin{array}{l} \int_{a}^{b} f (u) d u = \int_{a}^{b} f (x) d x = \\ = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x = \\ = \int_{c}^{b} f (x) d x - \int_{c}^{a} f (x) d x = \\ = I (x = b) - I (x = a) = \\ = I (x) \|_{a}^{b} \end{array}}$	(2.2.1.16)

which was obtained by means of (2.2.1.15). The contribution of the additive constants of the indefinite integrals vanishes, because of the subtraction. The new symbol $I (x) |_{a}^{b}$ expresses the subtraction of the indefinite integrals.

Finally we can summarize the important result:

$\begin{array}{l} \int_{a}^{b} f (x) d x = I (x = b) - I (x = a) = I (x) \|_{a}^{b} \\ where I (x) = \int f (x) d x \end{array}}$	(2.2.1.17)

The relation (2.2.1.17) offers a simple way to calculate definite integrals, if the indefinite integral is known analytically. On the other hand if the analytical expression of an indefinite integral does not exist, it can be obtained from (2.2.1.15) by the use of the known methods for numerical calculation of definite integrals.

We already used the trapezoid method, for calculating two definite integrals representing the area, which we can obtain now with the help of (2.2.1.17):

example (2.2.1.3): $\int_{0}^{3} [1 - {(\frac{x}{3})}^{2}] d x = x - {(\frac{x}{3})}^{3} |_{0}^{3} = (3 - 1) - 0 = 2$
example (2.2.1.4): $\int_{- 1}^{2} (4 - x^{2}) d x = 4 x - \frac{x^{3}}{3} |_{- 1}^{2} = (8 - \frac{8}{3}) - (- 4 + \frac{1}{3}) = 9$

Few facts about definite integrals

Here are some facts about definite integrals, which are obvious or very easy to prove, but it is important to know them in order to benefit fully from their use.

The formulae (2.2.1.1-2) become for definite integrals.

$\begin{array}{l} y = f (x) {\begin{cases} y_{m} \leq y \leq y_{M} \\ for a \leq x \leq b \end{cases} \\ y_{m} (b - a) \leq \int_{a}^{b} y d x \leq y_{M} (b - a) \end{array}}$	(2.2.1.18)

For an odd function

$\begin{array}{l} f (- x) = - f (x) \\ \int_{- a}^{a} f (x) d x = 0 \end{array}}$	(2.2.1.19)

For an even function

$\begin{array}{l} f (- x) = f (x) \\ \int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x \end{array}}$	(2.2.1.20)

A periodic function with period Δx holds

$\begin{array}{l} f (x + Δ x) = f (x) \\ \int_{a}^{a + Δ x} f (x) d x = \int_{b}^{b + Δ x} f (x) d x \\ {\begin{cases} \int_{a}^{a + n Δ x} f (x) d x = n \int_{a}^{a + Δ x} f (x) d x \\ for n = integer \end{cases} \end{array}}$	(2.2.1.21)

The trigonometric functions

$\begin{array}{l} {\begin{cases} \sin (x + 2 π) = \sin x \\ \int_{a}^{a + 2 π} \sin x d x = 0 \end{cases} \\ {\begin{cases} \cos (x + 2 π) = \cos x \\ \int_{a}^{a + 2 π} \cos x d x = 0 \end{cases} \end{array}}$	(2.2.1.22)

The most encountered definite integrals are usually listed in the mathematical tables. Have a close look.

Learning by examples

This example is a very often used integral and it is worth remembering it

$\int_{0}^{π} \sin x d x = - \cos x \|_{0}^{π} = - (- 1) + 1 = 2$	(2.2.1.23)

From this result one can also deduce

$\int_{0}^{π} \sin x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos x d x = 2$	(2.2.1.24)

The following example is also a very common integral. We'll use a simple trick to calculate it without actually doing any integration. The functions $\sin^{2} x$ and $\cos^{2} x$ have a period of π and are equal, but just translated on x by π ⁄2 . According to (2.2.1.21)

$\int_{0}^{π} \sin^{2} x d x = \int_{0}^{π} \cos^{2} x d x$	(2.2.1.25)

but on the other hand

$\int_{0}^{π} \sin^{2} x d x + \int_{0}^{π} \cos^{2} x d x = \int_{0}^{π} d x = π$	(2.2.1.26)

From (2.2.1.25-26) one obtains

$\int_{0}^{π} \sin^{2} x d x = \int_{0}^{π} \cos^{2} x d x = \frac{π}{2}$	(2.2.1.27)

The following example uses a parametric presentation. As already known, for solving such indefinite integral one needs to change (substitute) the variables. In case of a definite integrals, this also requires an appropriate change of the limits of the integral.
The cycloid was introduced in the previous chapter and its parametric presentation (1.2.5.9) is reproduced here:

${\begin{cases} x = r (φ - \sin φ) \\ y = r (1 - \cos φ) \end{cases}}$

We want to calculate the area under the curve for one period. Since the period is 2πr we need to calculate $\int_{0}^{2 π r} y d x$ . For this period of x , the span of the angular parameter is 2π and therefore we write:

${\begin{cases} x = r (φ - \sin φ) \\ y = r (1 - \cos φ) \end{cases}}$

$\begin{array}{l} \int_{0}^{2 π r} y d x = \int_{0}^{2 π} y (φ) \frac{d x}{d φ} d φ = \\ = r^{2} \int_{0}^{2 π} {(1 - \cos φ)}^{2} d φ = \\ = r^{2} \int_{0}^{2 π} (1 - 2 \cos φ + \cos^{2} φ) d φ = \\ = r^{2} (2 π + 0 + 2 \frac{π}{2}) = 3 π r^{2} \end{array}}$	(2.2.1.28)

where we used (2.2.1.22) and (2.2.1.27).

Exercises

Exercise 1. Give the general solution of the following derivatives

$\frac{d}{d x} \int_{a}^{v (x)} f (u) d u where a = constant$

$\frac{d}{d x} \int_{w (x)}^{b} f (u) d u where b = constant$

Exercise 2. Calculate the area of the ellipse

${(\frac{x}{a})}^{2} + {(\frac{y}{b})}^{2} = 1$

by use of

the Cartesian coordinates
the parametric representation $x = a \cos β and y = b \sin β$

Exercise 3. Calculate the area enclosed between the hyperbola

${\begin{cases} x = a \cosh u \\ y = b \sinh u \end{cases}$

where a is a positive constant, and the straight line

$x = 2 a$

Exercise 4. Calculate the following integrals

$\int_{0}^{1} \frac{atan x}{1 + x^{2}} d x$

$\int_{1}^{2} \frac{x d x}{{(1 + x^{2})}^{2}}$

$\int_{0}^{\sqrt{π}} x \sin (x^{2}) d x$

Exercise 5. Calculate the following integrals where n is a natural number.

$\int_{0}^{\frac{π}{2}} \sin^{n} x \cos x d x$

$\int_{- π}^{π} x^{2} \sin (n x) d x$

$\int_{- π}^{π} x^{2} \cos (n x) d x$

Previous topic: Section 1 Indefinite Integrals, page 4 Simple Differential Equations

Next topic: Page 2 Applications

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 2: Definite Integrals; page 1

Home | Table of Contents | A-Z Index | Help

Fundamentals

Area under a curve

The definite integral

Relation with indefinite integrals

Few facts about definite integrals

Learning by examples

Exercises

Home | Table of Contents | A-Z Index | Help

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)