Anti Derivative of Log X

Introduction to anti derivative of log x:

A function `Phi (x) ` is called a primitive or an antiderivative of a function f(x) if `Phi '(x)`  = f(x).

For example , `(x^4)/4` is an antiderivative of x 3, because `d/dx (x^4)/4`  = x 3.

To find the antiderivative of log x we use a special form of antiderivatives known as uv form (antiderivative by parts).

Theorem : If u and v are two fucntions of x, then

`int` u v dx  = u `int` vdx - `int { du/dx int v dx } `  dx

Finding Antiderivative of Logx

We use the above said formula of uv form to find the antiderivative of logx.

We need two functions for that logx can be written as 1 * logx.

We take logx = u , as differentiation of log x is easy to find and v = 1.

`(du)/dx = (d(logx))/dx`  = `1/x`

and `int` v dx =  `int` 1 dx = x

We use all these values and plug it in the formula

The uv form is `int` uv dx = u `int` v dx - `int { (du)/dx int v dx }` dx

`int` 1 logx dx = logx `int` 1 dx  -  `int { (du)/dx int 1 dx }` dx

= logx * x -  `int 1/x xx x dx`  dx

= x logx - `int 1 dx`

= xlogx - x  +c

= x(logx -1) +c

The antiderivative of logx is x(logx-1) +c

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Solved Problem on Antiderivative of Logx

Find the antiderivative of xlogx   

Solution :  The function is x logx

Let v =x and u = logx, as log x is easily differentiable and the integral of x is easy to find.

Now `int` v dx  = `int` x dx = `(x^2)/2`

and `(d)/dx = (d(logx))/dx = 1/x`

Plugging in all the values in the formula

`int` uv dx  =  u `int` vdx - `int { (du)/dx int vdx }` dx

`int` x logx dx = logx `int` x dx - `int {(d(logx ))/dx int x dx}`  dx

= logx `x^2/2`  - `int` `1/x xx x^2/2` dx

= `x^2/2`  logx - `1/2 int x` dx

= `x^2/2` logx - `1/2 * x^2/2` +c

= `x^2/2`    ( logx - `1/2` ) +c

Antiderivative of xlogx is  `x^2/2`    ( logx - `1/2` ) +c

Odd Number Theorem

Introduction to odd number theorem:

The strong gravitational lensing is use the odd number theorem and it is derived from differential topology. The odd number theorem derive the odd numbers by using formula. The integer is determined by normal form of odd number theorem and converse the given integer for odd number theorem.

Explanation for Odd Number Theorem

Statement of odd number theorem:

The odd number theorem illustrated as the square number is equal to the sum of first ‘n’ odd numbers. This statement is expressed as `sum_(k=1)^n` (2k -1) = n2.

The converse is performed in this theorem and defined as the nth odd number is equal to subtraction of  squares of nth and (n – 1)st value. This can be expressed as n2 – (n – 1)2 = 2n – 1.

Odd number theorem output:

The odd numbers are derived from the odd number theorem. The odd number means if the number is divided by two,the number have a remainder. The odd number is an integer and the form that integer is n = 2k + 1. The series of odd numbers are 1, 3, 5, 7, 9 and so on. The odd numbers are also called as gnomonic numbers by odd number theorem.

The parity value of odd number is 1 and the odd  number’s generating function is `(x(1 +x))/(x-1)^2` .

More about Odd Number Theorem

Examples for finding the odd numbers by using odd number theorem:

Problem 1: Find t out the odd numbers from given set of data by using function:

12, 14, 6, 21, 16, 24, 30

Answer:

The odd number of  given set of data is 21. Because it give the remainder as 1 when it divided by 2.

Problem 2 : Choose the odd number from given set of numbers.

7, 4, 2, 8, 10, 12, 14

Answer:

The odd number of given set of number  is 7.Because the remainder of  that number is 1 when divided by 2.

Exercise problem for finding the odd number by using odd number theorem:

1. Find out the odd number from given set of number:

54, 26, 38, 41, 52, 46

Answer: The odd number is 41.

2. Choose the odd number from series of data:

25, 24, 32, 44, 58, 60

Answer: The odd number is 25.

Graph Parabola Equation

Introduction to graph parabola equation:

The parabola is a member of conic section which means the intersection of a cone with a plane.
The general equation of a parabola is given as  (Ax + By)2 + Cx + Dy + E = 0.
It is very simple to graph parabola equation. To graph parabola equation, we have to find vertex, x  and y intercepts. The steps for graphing parabola equations are given below with an example problem.

Steps to Graph Parabola Equation:

Step 1: Allocate the variables a, b and c from the given equation

Step 2: Check and determine whether the parabola opens upwards or downwards.

If a > 0, then the parabola opens upwards (U-shaped).

If a < 0, then the parabola opens downwards (n-shaped)

Step 3: To find vertex:

The next step is to find vertex. To find vertex, we have to find the x-coordinate of the maximum point (or minimum point) by

X = - `b/(2a)`

By substituting this x-value into the given quadratic function i.e., the y expression, we obtain y-coordinate.

The obtained coordinate (X, Y) is the vertex of the parabola.

Step 4: To find the coordinates of the y-intercept.

By substituting x = 0 in the expression, we can find the y-intercept coordinate.

Step 5: To find the coordinates of the x-intercept.

By substituting y = 0 in the expression and solving the quadratic equation, we can find x-intercept coordinate.

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Example Problem to Graph Parabola Equation:

Graph the equation y = x2 - 7x +12.

Step 1:  Allocate the variables a, b and c from the given equation

a = 1

b = - 7

c = 12

Step 2: In the given equation, a is greater than 0 ( a = 1), therefore the graph of given equation is a parabola opens upwards (U-shaped) i.e., it has minimum point.

Step 3:  To find the vertex

x = - `b/(2a)` .

=  -`(-7)/(2(1))` .

= `7/2` .

= 3.5 

Y has the minimum point. So,

y = (3.5)2 - 7(3.5) + 12

= 12.25 - 24.5 + 12

= -0.25

Therefore, the minimum point obtained is (3.5, - 0.25)

Step 4: To find y-intercept.

By substituting x = 0 in the given equation y = x2 - 7x + 12, we get

y = (0)2 - 7(0) + 12 = 12

So, y intercept = (0, 12)

Step 5: To find x-intercept.

By substituting y = 0 in the given equation y = x2 - 7x + 12, we get

0 = x2 - 7x + 12

By factoring the above equation, we get

x = 3,  x = 4

The graph is drawn below with the help of the above information.  

   

Properties of Absolute Value

Introduction to properties of absolute value:

Normally absolute value is nothing but if any value regards to its sign in math. For example let us consider the numbers 8 and -8. Absolute value of these numbers is 8. |8| = +-8. Here we are going to learn about the properties of the absolute value. If we know the properties of the absolute value it is easy to do the operations on the absolute vales.

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Properties of Absolute Value:

Non negativity property:

The absolute value of any numbers is greater than 0. There is no negative numbers in this.

For example take any value -9. So the absolute value is 9. Because|9| = +-9. So for -9 and +9 the absolute value is +9.

|x| >= 0

Positive definiteness:

The absolute value of 0 is always 0. |x| = 0 then x = 0 (always)

Normally for an absolute value we have two values. Here the absolute value is 0. So value of this is 0. There is no sign for the value 0.

Multiplicative property:

Multiplication of any two absolute values same as the individual Absolute Value Equations. This mean

|x `xx` y | = |x| `xx` |y|

Example:

|-2 X 3| = |-6| = + 6

|-2| `xx` |3| = +2 `xx` +3 = + 6

So both the values are equal.

Subtraction addition property:

Addition of any two absolute values is always less than its individual addition.

|x + y| `lt=` |x| + |y|

Example:

|5 + -3| = |2| = 2

|5| + |-3| = 5 + 3 = 8

2 `lt` 8

Other Properties of Absolute Value:

Symmetry property:

Symmetry property is nothing but |-x| = |x|

Absolute value of –x and absolute value of x is always equal.

Identity of indiscernible property:

If the subtraction of any two absolute values is 0 then these two absolute values are equal.

|x - y| = 0 then x = y

Preservation of division:

The division of any two absolute values same as individual absolute value division.

|x / y | = |x| / |y| (where y `!=` 0)

These are the some basic properties of the absolute values.

Least Common Multiple and Greatest Common Divisor

Introduction to least common multiple and greatest common divisor
Least common multiple

In mathematics, the least common multiple (lcm), of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder.

Greatest common divisor

In mathematics, the greatest common divisor (gcd), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. (Source: From Wikipedia).

Example Problems to Find the least Common Multiple and Greatest Common Divisor

Example problems to find the least common multiple of two numbers

Example 1

Find the least common multiple of 3 and 7.

Solution

To find the least common multiples of 3 and 7, first we have to find the multiples of 3 and 7.

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Multiples of 7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70

Here, 21 is the lowest common number in the multiples of 3 and 7.

So, 21 is the least common multiple of 3 and 7.

Example 2

Find the least common multiple of  5, 11

Solution

The multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70.

The multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132.

Here, 55 is the lowest common number in the multiples of 5 and 11.

So, 55 is the least common multiple of 5 and 11.

Example problems to find the greatest common divisor of two numbers.

Example 1

Find the greatest common divisor of  54, and 40

Solution

The greatest common divisor of 54 and 40 can be found by the prime factors of the given numbers.

Prime factorization of 54 = 2 * 3 * 3 * 3

Prime factorization of 40 = 2 * 2 * 2 * 5

The common factors in the prime factorization is 2

So, 2 is the greatest common divisor of 54 and 40.

Example 2

Find the greatest common divisor of 21 and 35

Solution

Prime factorization of 21 = 3 * 7

Prime factorization of 35 = 5 * 7

Here, 7 is the common term.

So, 7 is the greatest common divisor of 21 and 35.

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Practice Problems to Find the least Common Multiple and Greatest Common Divisor

Problem 1

Find the greatest common divisor of 3, 5, 7

Answer: 1

Problem 2

Find the greatest common divisor of 54, 27

Answer: 27

Problem 3

Find the least common multiple of 21 and 3

Answer: 21

Problem 4

Find the least common multiple of 51 and 16.

Answer: 816

Finding Percentages

Introduction :
Percentages are numerators of fractions with denominator 100 and have been used in comparing results. Per cent is derived from Latin word ‘per centum’ meaning ‘per hundred’. Per cent is represented by the symbol % and means hundredths too. That is 1% means 1 out of hundred or one hundredth. It can be written as: 1% =1/100 =0.01. We saw how percentages were helpful in comparison. We have also learn to convert fractional numbers and decimals to percentages. Now, we shall learn how percentages can be used in real life.

Finding Percentages for Fractions

Fractional numbers can have different denominator. To compare fractional numbers, we need a common denominator and we have seen that it is more convenient to compare if our denominator is 100. That is, we are converting the fractions to Percentages. Let us try converting different fractional numbers to Percentages.

Example: Write 1/3 as per cent.

Solution:

We have, 1/3 = 1/3 x 100/100 = 100 %/3=33 and 1/3 %

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Finding Percentages for Decimals

We have seen how fractions can be converted to percents. Let us now find how decimals can be converted to percents.

Example1: Convert the given decimals to per cents:

(a)    0.75 (b) 0.09 (c) 0.2

Solution:

(a)    0.75 = 0.75 × 100 % =(75/100)× 100 % = 75%

(b)   0.09 =9/100= 9 %

(c)    0.2 =(2/10) × 100% = 20 %

Example2: The monthly salary of Meena is Rs. 4000. She spends 80% of her salary every month. How much does she save every month?

Solution:

Meena's monthly salary = Rs. 4000

Expenditure = 80% of 4000 = 80/100 × 4000 = Rs. 3200

Therefore, Savings = 4000 – 3200 = Rs. 800

Prime Numbers to 1000

Introduction to prime numbers to 1000:

A natural number other than 1, divisible only by itself and 1. Or any natural number which has precisely two divisors. The following numbers are prime numbers are prime 2, 3, 5, 7, 11, 13 … 37… every natural numbers greater than 1 may be resolved uniquely into a product of prime numbers. For example 180 = 2 * 2 * 3 * 3 * 5. In the case of prime number p, the product has to be interpreted as p itself.

Finding Prime Number to 1000:

Here we will see how to find the prime numbers. Which two types of prime, Both Fermat and Mersenne have presented us a few techniques to identify the prime numbers. Neither always gets up a prime number, but in some time of Mersenne, his formula leads to finding exact numbers.

Fermat’s formula: 2n + 1

Messene’s formula: 2n – 1

We do not know whether there are a countless number of Mersenne primes.

Or whether we can get the infinite number of primes from Fermat’s Formula.

Some facts:

The even prime number is 2 only. Remaining even numbers can be divided by 2.
If the total of a number's digits are multiple by 3, which number can be divided by 3.
No prime number < 5 ends in a 5. Several numbers < 5 which ends in a 5 can be divided by 5.
0 and 1 are not measured as a prime numbers.
Apart from for 0 and 1, a number is whether a prime number or a composite number. A composite number is distinct as any number, < 1, which is not a prime.
To demonstrate either a number is a prime number, initially try to dividing it by 2 and look if we get a whole number. If we do, it cannot be a prime number. If we do not get a whole number, then try to dividing it by prime numbers: 3, 5, 7, and 11 (9 are divisible by 3) etc, Let us see some prime numbers from 1 to 1000 in the below table.

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Example for Prime Numbers to 1000:

The following is the list of prime numbers below 1000.

2  3  5  7  11  13  17  19  23  29  31  37  41  43

47  53  59  61  67  71  73  79  83  89  97  101  103  107

109   113  127   131   137   139   149   151   157   163   167   173   179   181

191   193   197   199   211   223   227   229   233   239   241   251   257   263

269   271   277   281   283   293   307   311   313   317   331   337   347   349

353   359   367   373   379   383   389   397   401   409   419   421   431   433

439    443  449   457   461   463   467   479   487   491   499   503   509   521

523    541    547    557    563    569   571   577   587   593   599   601   607   613

617    619     631    641    643    647   653   659    661    673    677    683    691   701

709   719    727    733    739    743    751    757    761    769     773     787    797    809

811    821    823    827     829     839    853    857    859    863    877    881    883    887

907    911    919    929    937   941   947   953   967   971   977   983    991   997