Prevención y Control

Consider the curve f (x) in figure 9.6. It shows a number of turning points, known as local maxima and local minima.

Figure 9.6 Graph of f (x)

At these turning points, the tangents are parallel to the x-axis, that is, their slopes are zero. So at a maximum or minimum point, the gradient of the curve, and hence

dx

dy , is zero.

To the left of every local maxima, the slope is positive.

To the right of every maxima, the slope is negative.

To the left of every minima, the slope is negative.

To the right of every minima, the slope is positive.

This is an important result that leads to the solution of many practical problems.

Example

Find the maxima and minima of the following function:

y = 3x⁵ – 20x³

(1) Find the first derivative of y   x y



 using the power rule

x

 3x⁵ – 20x³

= 15x⁴ – 60x²

(2) Set the derivative equal to zero and solve 15x⁴ – 60x² = 0

15x² (x² - 4) = 0

15x² (x + 2) (x – 2) = 0 15x² = 0 or x = 2 or x = -2 x = 0, -2 & 2

(3) Put the values of x (0, -2, 2) into the original function y = 3x⁵ – 20x³ which gives y = 0, -64 & 64

We now have the values for the local maxima and minima:

(0, 0) (-2, 64) (2, -64)

(4) Determine whether the coordinates represent maxima or minima by calculating the second derivative

2 2

x

 15x⁴ – 60x²

= 60x³ - 120x

(5) Input values of x into the second derivative to determine the turning points For x = 0, the value is 0 a point of inflection

x = -2, the value is = -240, which is negative x = 2, the value is = 240, which is positive Using the rules given above:

the coordinate (-2, 64) is a maxima the coordinate (2, -64) is a minima 9.4 INTEGRATION

The previous sections have shown that using differentiation it is possible to find solutions to the problem:

given y = f(x), find dx dy

It is also possible to find solutions to the reverse problem:

given dx

dy = f (x), find y

This process is called integration.

9.4.1 AREA UNDER A GRAPH

Suppose that you have to solve the problem of calculating the area A bounded by the curve y = (x), the x-axis, and lines x = a, x = b as in figure 9.7

Figure 9.7 curve of y = (x)

Such an area cannot be found directly as sum of rectangular or triangular areas, but we could find an approximation to its value.

Figure 9.8 curve of y = (x)

As can be seen from figure 9.8 the first attempt using rectangular approximation will not get a result very close to the true value.

Figure 9.9 curve of y = (x)

By using smaller and smaller rectangles it can be seen that the accuracy of approximation is greatly increased.

Integration then may be considered as the process of summing up an ‘infinite number of rectangles’ to give an exact result.

(An algebraic limiting process can be used to evaluate the area A using rectangles but is beyond the scope of this course).

9.4.2 INTEGRALS

It would be a very tiresome business if you had to evaluate the limit of a sum in order to find an area.

It would be advantageous to develop a technique for finding the areas bounded by the graphs of a wide variety of functions.

If we apply a method of upper and lower sums to find the area of the region under the graph y = x, from x = a to x = b (figure 9.10).

Figure 9.10 graph of y = x

Knowing that the area of a triangle is ½ base × height and by applying the limiting process mentioned earlier it is found that the required area is:

^{2 2}

2 1 2

1b  a

The same process can be used to find the area A bounded by the curve y = x², the x-axis, and the lines x= a and x = b (figure 9.11).

Figure 9.11 Graph of y = x²

The results show that from the x, y intersect (we will call 0), the area from 0 to a is

3

3

1a and, similarly, the area from 0 to b is, ³ 3

1b thus the required area is just the difference between the two:

³ 3

1b - ³ 3 1a

Although it is not very wise to guess a general result from only two special cases, it is tempting in this case to do so as the results have striking similarities.

(a) Both answers are the difference of two terms of the same form.

(b) The first term involves b and the second term involves a.

(c) In this form, the exponent is one more than the exponent in the original function.

(d) The exponent is the same as the number in the denominator.

From previous discussions if we have a function

xn (1)

the standard derivative of the function will be

1

applying the logic of the previous page to the integral a first guess of integrating could be

1

xn (2)

if (2) is differentiated with respect to x the result is:

n if (4) is differentiated with respect to x the result is:

n

This gives us a general rule of integration:



^{xndx } _n^xn_^₁¹

When integration is to be carried out the notation



^{xndxis used.}

The symbol



is the mathematical notation for integration.

xⁿ represents the variable about to be integrated.

dx is what is to be integrated with respect to x.

9.4.3 INDEFINITE INTEGRALS If we consider the functions:

y = x³, y = x³ + 6 and y = x³ + 10 all have the derivative:

3x2

dx y 



Looking at the graph of y = x³ (figure 9.12) it can be seen that are an infinite number of possible places that x³ may be placed on the graph.

Figure 9.12 Graph of x³ + c

Thus the knowledge of the gradient is insufficient to describe uniquely the solution for x³. So when a function is integrated an arbitrary constant must be included to take account of the infinite number of ‘parallel’ functions.

Using the earlier examples:

y =



^3x2

= ³ 3

1x adding the arbitrary constant gives

= ^x3^c 3 1

Where c could represent 0, 6 or 10 (or in fact any constant).

Thus for indefinite integrals the general rule is:



^{xndx } _n^xn_^₁^{1 c}

9.4.4 DEFINITE INTEGRALS

When carrying out integration with definite values the method is as follows:

(a) Integrate the function, omitting the constant of integration.

(b) Substitute the value of the upper limit for x: repeat the process for the value of the lower limit.

(c) Subtract the lower limit from the upper limit.

Example

Consider the graph of y = x², x = 2 and x = 5 (figure 9.13)

Figure 9.13 Graph of y = x²

Integrate y = x² with respect to x, for x = 5 and x = 2 This is written as:



⁵₂ ^x2dx

(i) Integrate the function



⁵₂ ^{x2dx =} 3 ^{3 5}₂

 1x 

(ii) Substitute the value of the upper limit for x: repeat the process for the value of the lower limit and subtract the lower limit from the upper limit

 

 _^

 

_^

For definite integrals the basic rule is:



^{ }_^ _^{  }_^

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In document Tesis de Maestría en Ciencias en el Uso, Manejo y Preservación de los Recursos Naturales con Orientación en Acuacultura. Camilo Pohlenz Castillo (página 28-0)