Online Tutoring Standard Definite Integrals

The expression of definite integral with lower and upper limit is `int_a^b`f(x) dx . The definite integral is also known as Riemann integrals. In definite integral, `int_a^b`f(x) dx . where, a, b and x is the complex numbers and path of integration from a to b. The calculus fundamental theorem allows the definite integrals to be computed in terms of indefinite integrals. Where, F is the indefinite

integral and f(x) is continuous function. The standard form of indefinite integral is `int` f(x) dx. The standard form of definite integral with an interval a to b `int_a^b`f(x) dx = F(b) - F(a).

Standard integral formulas:

1.Standard integral formula for nth power ` int` xn dx  = `x^(n+1)/(n+1)` + c

2.Standard integral formula for exponential function `int` ex dx = ex + c

3.Standard integral formula for logarithmic function ` int ` log x dx = x (log x - 1) + c

4.Standard integral formula for trigonometric function ` int ` sec2x.dx = tan x + c

These formulas are helps to solve the definite integrals. Trained tutors are available in online to assist the learners for learning standard definite integrals at every time. The online tutors will solve every problem in step by step manner with the correct and neat explanation. Tutors are real person, who is working as a teaching staff in online with the help of Internet.

Online tutoring standard definite integrals problems:

Online tutoring standard definite integrals problem 1:

Find the value of definite integral sin x with the limits 0 to `pi`

Solution:

Given definite integral function is, `int_0^(pi) ` sin x dx

So, `int_0^(pi) ` sin x dx =  `[-cosx]_0^(pi)` .

Substitute the upper limit and lower limit value. So, we get

= - cos `(pi)` - (-cos 0)

= - (-1) - (-1)       (Because the trigonometric value of cos 0 = 1 and cos`pi` = -1)

= 1+ 1

= 2

Answer:    `int_0^(pi) ` sin x dx = 2.

Online tutoring standard definite integrals problem 2:

Find the value of definite integral : e-x with the limits 1 to 2

Solution:

The integral of e-x can be expressed as `int_1^2` e-x dx

Already we know the integral formula `int` eax dx = `1/a` eax + c

here,    a = -1

So, apply the above condition  ` int` e-x dx = `(1/(-1))` e(-1)x + c

`int_1^2`e-x dx   = `(1/(-1))` e(-1)x

= [-e-x ]`_1^2`

= [-e-2 ] - [-e-1]

= [-(0.135)] - [-(0.3678)]

= - 0.135 + 0.3678

we know  e-1 = 0.3678     and e-2 = 0.135

= 0.2328

Answer: 0.2328

Odd are Even Numbers and Math

Let us study about odd and even numbers in math. Odd numbers are defined as the numbers that are not divisible by 2 or any other even numbers but they are divisible by some other odd numbers.
Even numbers are defined as the numbers that are divisible by 2 and some other even numbers but not divisible by any odd numbers.
Examples for odd and even numbers are below.


Odd are even numbers and math:

Odd are even numbers and math example 1:

List out all the possible odd numbers from 1 to 50.



Solution:

All the possible odd numbers that are not divisible by 2 or any other even numbers between 1 and 50 are as follows:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47 and 49



Odd are even numbers and math example 2:

List out all the possible even numbers from 1 to 50.



Solution:

All the possible even numbers that are divisible by 2 or any other even numbers between 1 and 50 are as follows:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46 and 48



Odd are even numbers and math example 3:

Find the odd numbers from the following number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700



Solution:

The odd numbers present in the given number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700 are found to be as follows:
21, 55, 81, 95, 111, 311, 431 and 673 since these numbers cannot be divided by 2 or any other even numbers but can be divided by some other odd numbers.



Odd are even numbers and math example 4:

Find the even numbers from the following number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700



Solution:

The even numbers present in the given number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700 are found to be as follows:
34, 76, 512, 224 and 700 since these numbers cannot be divided by any odd numbers but can be divided by 2 or some other even numbers.


Odd are even numbers and math exercises:

List out all the possible odd numbers from 70 to 100. (Answer: 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97 and 99)
List out all the possible even numbers from 100 to 130. (Answer: 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128 and 130)
Find the odd numbers from the following number series 423, 322, 333, 129 and 112. (Answer: 423, 333 and 129)
Find the even numbers from the following number series 423, 322, 333, 129 and 112. (Answer: 322 and 112)

Synthetic Methods

Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials.

Let's look at the quadratic equation: y = x2 + 5x + 6. From the rational root testwe  know that ± 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring, we know that the zeroes are, in fact, –3 and –2.) How would we use synthetic division to check the potential zeroes?  If we guess that x = 1 is a zero, then this means that x – 1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide x2 + 5x + 6 by x – 1, we would get a zero remainder.

we divide x2 + 5x + 6 by x – 1 by synthetic method as follows:

1. write the coefficients ONLY inside an upside-down division symbol:

write coefficients in upside-down division symbol

Make sure you leave room inside, underneath the row of coefficients, to write another row of numbers later.


2.Put the test zero, x = 1, at the left:

write test zero at left

3. Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the division symbol:

carry down leading coefficient


4.Multiply this carry-down value by the test zero, and carry the result up into the next column:

multiply by test zero, and carry result up into next column


5.Add down the column:

add down the column

6. Multiply the previous carry-down value by the test zero, and carry the new result up into the last column:

multiply result by test zero, and carry result into next column


7.Add down the column:

This last carry-down value is the remainder.

that means (x2 + 5x + 6) ÷ (x – 1) =12

Hence x-1 is not a factor of the polynomial x2 + 5x + 6

Hermite Polynomials

The Hermite polynomials which are a classical orthogonal polynomial sequence arises in mathematics in probability, such as the Edgeworth series; in combinatorials, as an example of an Appell sequence, obeying the umbral calculus; and in physics, where they give rise to the eigenstates of the quantum harmonic oscillator. They are named in honor of Charles Hermite.

Definition

Hn(x) = (-1)nex2/2 [`d/dx`(e-x2/2)]n

(the "probabilists' Hermite polynomials"), or sometimes by

Hn(x) =(-1)nex2 [`d/dx`(e-x2)]n

(the "physicists' Hermite polynomials"). The above two definitions are not exactly equivalent; either is a rescaling of the other, to wit

Hnphys(x) = 2n/2Hnprob(`sqrt(2)`x )

Usually follow the first convention is followed. That convention is often preferred by probabilists because

`1/sqrt(2Pi)` e-x2/2

is the probability density function for the normal distribution in which expected value 0 and standard deviation 1.

Properties of Hermite polynomials

Hn is a polynomial of degree n. According to probabilists' definition it has leading coefficient 1, while according to physicists' definition it has leading coefficient 2n.

Orthogonality

Hn(x) is an nth-degree polynomial for n = 0, 1, 2, 3, .... These polynomials show orthogonality with respect to the weight function     (measure)

W(x) = e-x2/2   (probabilist)

or

W(x) = e-x2   (physicist)

Completeness

The Hermite polynomials (probabilist or physicist) form the basis of orthogonality of the Hilbert space of functions satisfying
`int_-oo^oo|f(x)|^2w(x)dx` < `oo`


Hermite's differential equation

The probabilists' version of Hermite polynomials are solutions of the differential equation

(e-x2/2 u')' + `lambda`e-x2/2 u = 0

Recursion relation

Hermite polynomials' sequences also satisfies the recursion

Hn+1(x) = xHn(x) - H'n(x) (probabilist)

Hn+1(x) = 2xHn(x) - H'n(x) (physicist)


Applications of Hermite polynomials

Hermite functions

One can define the Hermite functions from the physicists' version of polynomials:

`psi`n(x)  =   `1/sqrt(n!2^nsqrt(Pi))`  e-x2/2 Hn(x)


Recursion relation

Following recursion relations of Hermite polynomials, the Hermite functions satisfy

`psi`'n(x) = `sqrt(n/2)``psi`n-1(x) - `sqrt((n+1)/2)``psi`n+1(x)

Cramér's inequality

The Hermite functions satisfy the following inequality due to Harald Cramér

|`psi`n(x)|   `<=` K

for x real, where the constant K is less than 1.086435.

Hermite functions act as eigenfunctions of the Fourier transform

Area using Perimeter

Area:
Area is a measure of  the outermost surface of an object. Area is a bounded space with its surrounding. Area is the  measure of  plane region or  amount at surface which any object occupies is called the area for that object. Area for any object is a measure ofbounded surface to its bounded space

Perimeter:
The length of the boundary of any closed figure is called its perimeter. The distance around a two-dimensional shape,for a circle perimeter is a circumference.closed surface for inner space occupied it said to be the perimeter for an object.

Finding Area using Perimeter For Square:

Area of Square =a2
Perimeter of Square =4a
Example 1:
For square board the perimeter is 84cm find the area for square board.

Solution:
we know perimeter is 84cm
for square perimeter,4a=84
a=84/4
a=21
area of square =a2
=212
area of square =441cm

Example 2:
The perimeter of a square playground is 1200m. Find its area using perimeter.

Solution :
Perimeter of the square ground, p = 4a
4a = p
a =p/4

Hence, a = 1200
4 m [since p = 1200 m ]
∴ a = 300 m
Area of the square ground A = a2
= 300 m × 300 m
Area = 90000 sq.m.

Finding Area using Perimeter For Rectangle:

Area of Rectangle =Length `xx` Breadth
Perimeter of rectangle =2(Length+Breadth)

Example 1:
If perimeter of rectangle is 18 m and its breadth 4m. find its area using perimeter.

Solution:
2 length + 2 breadth = perimeter
2l + 2`xx` 4 = 18
2l + 8 = 18
2l = 18– 8
2 l = 10
l =5
l=5,b=4
Now the area of the rectangle A = l `xx` b
= 5 m × 4 m
= 20 sq.m.

Example 2:
The perimeter of the floor of a rectangular room is 24 m and the  length is 7 m. Find its area.

Solution :
2 length + 2 breadth = perimeter
2 × 7 + 2b = 24
14 + 2b = 24
2b = 24 – 14
2 b = 10
b =`10/2`
= 5
Now the area using perimeter of  rectangle A = l` xx` b
= 7 m × 5 m
= 35 sq.m.

Dividing Fractions Solver

As division of whole numbers shows how often one integer is contained in another integer, to division of fraction shows how often one fraction is contained in another. If the numerator of any fraction be made a denominator, and the denominator a numerator, the fraction of made is called the reciprocal of the former. Thus 4 / 3 is the reciprocal of 3 / 4.

In division * as the divisor is to the dividend, of is a unit of the quotient (both in whole numbers and fraction).

Dividing fractions solver – Steps and methods:

To dividing algebraic fraction, follow these steps:

Write the given fractions.
Change the division sign to a multiplication sign and invert the second fraction.
Write the given fraction and cancel any common factors.
Multiply the numerators.
Multiply the denominators.

Dividing fractions solver – Methods:

Dividing common fraction
Dividing a fraction and a whole number
Dividing mixed number


Dividing fractions solver – Three methods and example problem:

Dividing common fractions:

To divide common fraction, follow these steps:

Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

(5 / 6) / (3 / 4)

Solution:

(5 / 6) / (3 / 4) (divisor)

= (5 /6) * (4 / 3)

= (5 * 4) / (6 * 3)

= 20 / 18

= 10 / 9

Dividing a fractions and a whole number:

To divide a fraction and a whole number, follow these steps:

Change the whole number to a fraction by the placing the whole number over one.
Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

16 / (5 / 8)

Solution:

16 / (5 / 8)

= (16 / 1) / (5 / 8)

= (16 / 1) * (8 / 5)

= (16 * 8) / (1 * 5)

= 128 / 5

= 25 3/5

Dividing mixed number fractions:

To divide mixed numbers, follow these steps:

Change the mixed number to an improper fraction
Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

(2 7/8) / (4 / 5)

Solution:

(2 7/8) / (4 / 5)

= (2 7/8) = ((2 * 8 + 7) / 8) = 23 / 5

= (23 / 8) / (4 / 5)

= (23 / 8) * (5 / 4)

= (23 * 5) / (8 * 4)

= (115 / 32)

= 3 19/ 32.

Hard Subtraction Problems

Subtraction is one of the arithmetic operations; it is the inverse of addition, meaning that if we start with any number and add any number and then subtract the same number we added, we return to the number we started with. Subtraction is denoted by a minus sign in infix notation. The traditional names for the parts of the formula

c − b = a

(Source: Wikipedia)

Hard subtraction example problems:

Hard subtraction problem 1:

Dylan made five hundred twenty-three cupcakes. Two hundred ten of them have already been put into packages. How many cupcakes are left to be packaged?

Solution:

Dylan made five hundred twenty-three cupcakes.

Two hundred ten of them have already been put into packages.

Cupcakes are left to be packaged = 523 – 210

= 313 cupcakes.

Answer: 313 cupcakes are left to be packaged.

Hard subtraction problem 2:

There are four hundred fifty employees working in an office building. Hundred an sixteen of them are about to leave to go out to lunch. How many employees will be left in the building?

Solution:

There are four hundred fifty employees working in an office building.

Hundred an Sixteen of them are about to leave to go out to lunch.

= 450 – 116

= 334

Answer: 334 employees will be left in the building.

Hard subtraction problem 3:

The pet store had three hundred twenty-eight bags of bird feed. Two hundred six will be eaten today. How many bags will be left?

Solution:

The pet store had three hundred twenty-eight bags of bird feed.

Two hundred six will be eaten today.

= 328 – 206

= 122 bags.

Answer: 122 bags will be left.

Hard subtraction practice problem:

Practice problem 1:

There were twenty-three magazines on a bookshelf this morning. Students came and borrowed ten of them. How many magazines are left?

Answer: 13 magazines are left

Practice problem 2:

There are fifty-two colored pens in a desk. There are twenty students and each will need one. How many colored pens will be left?

Answer: 32 colored pens will be left.

Practice problem 3:

There are forty-two birds sitting in two trees. One tree has twenty-two birds in it. How many birds are in the other tree?

Answer: 20 birds are in the other tree.