What is Continuous Integration

In what is continuous integration, Let f (x) be continuous and single valued for a <= x <= b. Divide the interval (a, b) of the x axis into n parts at the points (a -= X_0), X_1, X_2… ( X_n -= b) as in below figure.

Define Delta x_i = x_i - x_(i-1) and let xi_i be a value of x in the interval x_(i-1) < xi_i <= x_i. Form the sum of the areas of each approximately rectangular area to get

sum_(i=1)^(n) f(xi_i) Delta x_i

continuous integration

If we take the limit as n ->oo and Delta x_i -> 0, the approximation of the area under the curve as a rectangle becomes more accurate and we obtain the definition of the definite integral of f (x) between the limits a, b; thus

^lim_((n -> oo)_(Delta x_i -> 0)) sum_(i=1)^n f(xi_i) Delta x_i -= int_a^b f(x)dx

The points a, b are referred to as the lower and upper limits of integration, respectively. The definite integral

int_a^b f(x) dx

equals the area under the curve f(x) in the interval (a, b) for what is continuous integration.

Fundamental Theorem - what is continuous integration:

In what is continuous integration, the connection between differential and integral calculus is through the fundamental theorem of calculus. In particular, if f(x) is continuous in the interval a <= x <=b and G(x) is a function such that (dG)/(dx) = f(x) for all values of x in (a, b), then

int_a^b f(x)dx = G(b) - G(a)

The definite integral acts like an antiderivative of f(x), or the inverse operation to differentiation. The concept of inverse differentiation can be made more explicit by introducing the notation

[G(x)]_(a)^(b) = G(b) - G(a)

and using (dG)/(dx) = f(x) in the definite integral above equation to give

int_a^b ((dG)/(dx))dx = [G(x)]_a^b

Example - what is continuous integration:

In what is continuous integration, let f(y) = ln y, u(alpha) = a, and v(alpha) = alpha. In this case, the function f does not depend on alpha. Consequently, we substitute u, v, and f into find.

(d)/(d alpha) int_a^(alpha) In y dy = f(alpha)(d a)/(d alpha) = In alpha

because (da)/(d alpha) = 0 and f(alpha) = ln alpha. This result is verified by first evaluating the integral

int_a^b In y dy = [y In y –y ]_a^alpha = alpha In alpha - alpha - (a In a - a)

Taking the derivative of above equation with respect to alpha gives

(d)/(d alpha)[alpha In alpha - alpha - (a In a –a)] =(d)/(d alpha) [alpha In alpha - alpha]

Answer is     = In alpha

Arithmetic Adding Fractions

A fraction is a part of a whole. A fraction can be altered to a decimal by dividing the upper number, or numerator, by the lower number, or denominator. Fractions are to indicate ratios, and to represent division which is one of the basic arithmetic operations. Thus the fraction 3/4 is also used to signify the ratio 3:4 (three to four) and the division 3 ÷ 4 as well.(three divided by four). Here in this topic we see about adding fractions.

Arithmetic adding Fractions with same denominator - Example problems:

Arithmetic adding Fractions - Example 1:

Add: 2/7   +    3/7

Solution

Here the denominator is equal

So we keep the denominator same as 7 and add the numerator alone

So the solution is   (2+3)/7 =  5/7

Arithmetic adding Fractions - Example 2:

Add:  5/8 +  3/8

Solution

Here the denominator is equal

So we keep the denominator same as 8 and add the numerator alone

So the solution is ,

(5+3) / 8 = 8 / 8  = 1.

Arithmetic adding Fractions with different denominator - Example problems:

Arithmetic adding Fractions - Example 1:

10/8   +    12/9


Answer:

Here the denominators are different that are 8 and 9. So take LCD of 8 and 9 first.

LCD of 8 and 9 is 72.

Therefore,
10 x 9        12 x 8          90      96       186
------------  +   ------------  =  -------- +  ------ = ----------
8 x 9          9 x 8          72       72        72


Simplifying this we get the answer as 31/12 .
Arithmetic adding Fractions - Example 2:

16/3     +   10/10

Answer:

Here the denominators are different that are 3 and 10. So take LCD of 3 and 10 first.

LCD of 3 and 10 is 30.

Therefore,
16 x 10       10 x 3       160       30       190
------------  +   ------------  =  -------- +  ------ = ----------
3 x 10          10 x 3       30       30        30


Simplifying this we get the answer as 190/30 .

Learning Tangent Line

If a line and a curve intersect at a point only then the line said to be tangent to the curve at that point. And the point is called the point of contact.

Tangents at (x, y) to y = f(x)

Lt y=f(x) be a given curve and P(x, y) and Q(x + δx, y + δy) be two neighbouring points on it

Equation of the line PQ is

Y – y = y + δy – y/x + δx – x(X – x)

Or       Y – y = δy/δx(X – x)                                             …(1)

Equation of line PQ

Equation of the line PQ



The line (1) will be a tagent to the give curve at P if Q → P which in turn means that δx → 0 and we know that

Lim δy/δx = dy/dx

δx → 0

Therefore the equation of the tangent is

Y – y = (dy/dx) (X – x)

Geometrical Intrepetation - slope and tangent

Geometrical meaning of dy/dx

From equation (1) we observe that dy/dx represents the slope of the tangent to the given curve y = f(x) at any point (x, y)

...    dy/dx = tan Ψ

Where Ψ is the angle which the tangent to the curve makes with +ive direction of x-axis. In case we are to find the tangent at any point (x1, y1) then (dy/dx)(x1, y1) i.e. the value of dy/dx at (x1, y1) will represent the slope of the tangent and hence its equation in this case will be

y – y1 = (dy/dx)(x1, y1) (x – x1)

Algorithm to draw tangent of a circle

Algorithm to obtain the equations of tangents draw from a given point to a given circle.

Step i. Obtain the point, say (x1, y1)

Step ii. Write a line passing through (x1, y1) having slow m i.e., y – y1 = m (x – x1)

Step iii. Equate the length of the perpendicular from the centre of the circle to the line in step ll to the radius of  the circle

Step iv. Obtain the value of m from the equation in step iii

Step v. Substitute m in the equation in step ll

Normal to the tangent:

The normal to a curve at a given point is a straight line passing through the given point, perpendicular to the tangent at this point. From the definition of a normal it is clear that the slope of the normal m′ and that of the tangent m are connected by the equation m′ = –`1/m .`

i.e., m′ = –1/f ′(x1)

= − 1/  dx/dy    (x1, y1)

Hence the equation of a normal to a curve y = f(x) at a point P(x1,y1) is of the form y – y1= –1/f ′(x1)( x – x1).

The equation of the normal at (x1,y1) is

(i)                x = x1 if the tangent is horizontal

(ii)               y = y1 if the tangent is vertical and

(iii)              y – y1 = –1 / m (x – x1) otherwise.

Example problems:

Example 1 :  Find the equations of the tangents through (7, 1) to the circle x2 + y2 = 25

Solution :  The equation of any line through (7, 1) is

y – 1 = m (x – 7)

mx – y – 7m + 1 = 0             …(i)

The coordinates of the centre and radius of the given circle are (0, 0) and 5 respectively

The line (i) will touch the given circle if,

Length of the perpendicular from the centre = radius

| (1 - 7m)/√(m2 + (-1)2 ) | = 5

│(1 – 7m)/√(m2 + 1)│ = 5

(1 – 7m)2/(m2 + 1) = 25

24m2 – 14m – 24 = 0

12m2 – 7m – 12 = 0

(4m + 3) (3m – 4) = 0

m = –3/4, 4/3

Substituting the values of m in (i), we obtain

3/4x – y + 21/4 + 1 = 0 and 4/3x – y – 28/3 + 1 = 0

3x + 4y – 25 = 0 and 4x – 3y – 25 = 0 which are the required equations.


Example 2: Find the equation of the tangent and normal to the curve y = x3  at the point (1,1).

Solution:

We have y = x3 ; slope y′= 3x2.

At the point (1,1), x = 1 and m = 3(1)2 = 3.

Therefore equation of the tangent is y − y1 = m(x − x1)

y – 1 = 3(x – 1) or y = 3x – 2

The equation of the normal is y − y1 = −`1/m `  (x − x1)

y – 1 =–`1/3` (x – 1) or y = –`1/3` x +4/3 .

normal tangent line graph

Probability Simple Event

Probability is a method of expressing information or principle that an event will occur or has occurred. In mathematics the ideas has given a correct meaning in probability theory  is used widely in the areas of study as mathematics, statistics, science, and finance to depict conclusions about the possibility of potential events .An event is a group of possible outcomes.

A simple event in a probability is the event of a single outcome.

Simple Event

The simple event is as follows:

If we bring together a deck of 52 playing cards and no jokers, and pick a single card from the deck, then the sample space is a 52-element set, as each individual card is a possible outcome.

An event is any subset of the sample space, as well as any single-element set, the empty set and the sample space itself which is defined to have chance one.

Further events are correct subsets of the sample space that surround multiple elements. So, for example, potential events include:

Black and red at the same time without being a joker (0 elements),
The five of Diamond(1 element),
A Queen(4 elements),
A Face card(12 elements),
A Spade (13 elements),
A card(52 elements).

Seeing as all events are sets, they are generally written as sets (e.g. {1, 2, 3}), and symbolized graphically using Venn diagrams. Venn diagrams are mainly useful for symbolizing events because the probability of the event can be identified with the ratio of the area of the event and the area of the sample space.



Example:

The event that 1 roll a 7 with two dice is an event.

But the event that 1 roll snake-eyes (1 and 1) is a simple event.

A complementary event is just the opposite of an event.

The balancing event to rolling a 7 is the event that 1 do not roll a seven. It consists of all the outcomes not in the unique event.

Hypothesis Test Power

A statistical hypothesis test is a technique of making decisions using experimental information. In statistics, an effect is called statistically significant if it is doubtful to have happen by possibility. Critical examination of this category may be called tests of significance, and when such examination is available we may determine whether a next sample is or is not significantly dissimilar from the initial. Let us study about the topic hypothesis test power given below content with some example problems.


Example problems for hypothesis test power:

Example 1:

A sample of 900 members is establish to have a mean of 3.4 cm and standard deviation 2.61 cm. is the example in use from a great population of mean 3.25 cm and standard deviation 2.61 cm. If the population is usual and its mean is unknown, Get 95% self-confidence limits of exact mean.

Solution:

Given:

H0 : µ = 3.25
H1 : µ ? 3.25
L.O.S  a = 0.05

Test Statistic:

Z = (barx-mu)/(sigma)/(sqrt(n))

Given bar x = 3.4, µ=3.25, n=900, s=2.61

Z = (3.4- 3.25)/(2.61)/(sqrt(900))

= 1.724

Critical value:

At 5% level, the table  value of Z = 1.96

Conclusion:

Since |Z| <1 .96="" 5="" accepted="" at="" h0="" is="" level="" of="" p="" significance.="">
The sample is taken from population whose mean is 3.25 cm

Mean µ is unknown, 95% confidence limits of µ are = barx+-1.96 sigma/(sqrt(n))

=3.4+- 1.96*(2.61)/(sqrt(900))

= 3.4+- 0.17

=3.23,3.57

Example 2:-using hypothesis test power

A construct maintains that his artificial fishing line has a mean contravention strength of 8 kg and standard deviation 0.5 kg . Can we consider his state if a random sample of 50 lines yield a denote contravention power of 7.8kg?

Solution:

Given:

H0 : µ = 8
H1 : µ ? 8
L.O.S: a = 0.01

Test Stastistic

Z =(barx-mu)/((sigma)/(sqrt(n)))

Given:

barx = 7.8, mu = 8, n = 50, sigma = 0.5

Z = ((7.8)-8)/((0.5)/(sqrt(50)))

= - 2.828

|Z| = 2.828

Critical value:

At 1% level of significance the table value of Z = 2.58

Conclusion:

Since |Z|>2.58,H0 is rejected at 1% level.

The manufacture’s claim is not accepted.


Practice problem for hypothesis test power:

Problem 1: using hypothesis test power

On a test given 60 students at a huge number of several colleges, the mean position was 74.5 and S.D was 8.0. at one exacting college, where 200 students took test, the mean position was 75.9%. Discuss the meaning of this answer at 5% level.

Solution:

Z = 2.45, significant at 5% level.

Problem 2: using hypothesis test power

A random sample of 400 items is pinched from a normal population whose mean is 5 and whose variation is 4. if the sample mean is 4.45, can the sample be regard as truthfully random sample?

Solution:

Z= 5.5; H0 is rejected , No.

Standard Deviation

Standard Deviation is defined as "the measure of the variability". If the spread is more then the standard deviation will also be more. Standard deviation can also be stated as the "square root of the variance". Standard deviation has its application in computing the the annual returns of the banks

Arithmetic Mean:

A mathematical representation of the series of the number which is totalled as the sum of all the number in series and it is divided by the total number of values in the set.

Variability :

When the  standard deviation is raised to the power of two , the value which we get is the variance i.e Variance is equal to the square of the standard deviation. It is the measure of the degree of the spread among certain set of the values. Variance is also stated in the other way as follows"Variance is the measure of tendancy of individual values that vary from the arithmetic mean value

Formula to determine the Standard deviation :

[Formula of Standard deviation]

Where  ,     Σ = sum of

xi = Individual scores

m = arithmetic mean of all scores

n = Size of the number of scores

Variance :

Variance = σ2

Example

Example 1: To find the standard deviation of  2 , 4 , 6 , 8 , 10

Step-1 :Calculate the mean and the deviation

xi       m        xi-m         (xi-m)2

2        6        -4              16

4         6        -2              4

6        6          0              0

8        6          2              4

10       6         4             16

Step-2 :Find the sum of  (x-m)2

16 + 4 + 0 + 4 + 16 = 40

Step-3 :n = 5 , the total number of values , find n-1

5-1 = 4

Step-4 :Now find standard deviation

√40 / √4 = 3.162

3.162 is the standard deviation

To find the variance just square the standard deviation value

Step-5 : Variance = (3.162)2 = 9.998244

Example 2

Example 2 :To find the standard deviation of 3 , 6 , 9 , 12 , 15

Step-1 : Calculate the mean deviation

xi          m          xi-m           (xi-m)2

3           9           -6               36

6           9           -3                 9

9           9            0                 0

12          9            3                 9

15          9           6                36

Step-2 :

calculate the sum of (x-m)2

36+9+0+9+36 = 90

Step-3 :

n=5 ,the total number of values , find n-1

5-1 = 4

Step-4 :

Now let us find the standard deviation

√90/√4 = 4.743

The standard deviation is 4.743

Step-5 :

Variance = (4.743)2

= 22.496

Non Terminating Decimals

In this page we are going to discuss about Non terminating decimals.Below we will get introduction to Non terminating decimals. A non terminating decimal is one of the type of decimals which are divided into two types. They are,

Terminating Decimals
Non terminating Decimals

Non terminating Decimals is the continuous decimal number. It has been going on till the value infinitive. It is the very big value sometimes these non terminating decimals are rounded to its nearest values. But the Terminating decimal number is the fixed value of the decimal value.

Calculating Non Terminating Decimals

Below you can see the steps of calculating non terminating decimals:

Divide 5/3

Step 1: This is the type of an improper fraction. because here numerator takes the large value than the denominator.

Step 2: One time 3 is 3. So the remainder is 2 and the quotient value is 1

Step 3: Now 2 is not divided by 3

Step 4: So we can add zero to the remainder 2 and put a point the quotient value 1

Step 5: Now the value is 20. Again we can do the same procedure

Step 6: Now 6 times 3 is 18 again the remainder is 2 and the quotient value is 6

Step 7: Again we can add zero to 2 then the value is 20

Step 8: Again 6 times 3 is 18. Again the remainder is 2 and the quotient value is 6

Step 9: Now the decimal value is 1.66 …….its going on

Step 10: This decimal part 1.66……is known as the non terminating decimal.

Step 11: It can be rounded as 1.67

Step 12: Because it’s decimal part value is greater than or equal to 5. So the value of the last decimal value is consider as one and that value is added to the previous decimal value.

Step 13: Otherwise the last decimal part value is less than five. So the value of the last decimal value is simply leave. And the previous decimal value is considered as the last decimal digit value.

Pictorial Representation of this problem is shown below:

Non- terminating decimals

Practice problems

Here are some practice problems on non terminating Decimals:

Identify which of the fractions gives the non terminating decimals:

(4)/(2)
(2)/(3)
(6)/(3)

Answer:

Terminating decimals (4)/(2) =2 and (6)/(3) =2

Non terminating decimal (2)/(3) = 0.6666.....