Finding the Areas With Integrals

We use inegral method to evolve the area finding problems for the continous curves which bound on some points normally we said as the area. Sonetimes the integral functions are given on the which satisfying some lines the graph. These are all considered here. Now we are going to deal with finding the area on the continous curves and lines.

Explanation "finding areas with integrals"

On the first step here we going to separate the different areas according to the axis and the graphs present in the bounds which is drawn according to our problem.

By the way, we have to separate the integral limits.

And then we prosecute the execution.


Examples:

Find the area of the region bounded y = x2 − 5x + 4, x = 2, x = 3 and the x-axis.

solution:

For all x, 2 ≤ x ≤ 3 the curve lies below the x-axis.Integral area

Required area = int_2^3(-y)dx

=   int_2^3 -(x^2 - 5x + 4)dx

= - [x^3/3 - 5x^2/2 + 4x]_2^3

= - [(9 - 45/2 + 12) - (8/3 - 20/2 + 8)]

= - [-13/6 ] = 13/6   sq. units

This is the required area represented in the square units.

Find the regional area for the region for x = 1, x = 4 bounded by the 3x − 15y − 15 = 0 line in  x-axis.

Solution:

The line 3x − 15y − 15 = 0 appears in  the x-axis (below)  in the interval x = 1 and x = 4Integral area

∴Required area  = int_1^4(-y)dx

=   int_2^3 -1/15(3x - 15)dx

= 3/15int_2^3 (5 - x )dx

=  3/15[5x - x^2/2 ]_1^4

= 3/15 [5(4 - 1) -1/2 (16 - 1)]

= 3/15 [15 - 15/2]

= 3/2   sq. units.

This is the required area represented in the square units.

problems to explain "finding areas with integrals"

Find the area between the curves y = x2 − x − 2, x-axis and the lines x = − 2 and x = 4

Solution:

y = x2 − x − 2Integral area

= (x + 1) (x − 2)

This curve intersects x-axis at x = − 1 and x = 2

Required area = A + B + C

The part B lies below x-axis.

∴ B = = -int_-1^2(y)dx

Hence required area = int_-2^-1(y)dx  +  int_-1^2(-y)dx  + int_2^4(y)dx

= int_-2^-1(x^2 - x - 2)dx  +   int_-1^2(-(x^2 - x - 2))dx   +  int_2^4(x^2 - x - 2)dx

= [x^3/3 - x^2/2 - 2]_-2^-1 + [-[x^3/3 - x^2/2 - 2]]_-1^2 + [x^3/3 - x^2/2 - 2]_2^4

= 11/6 + 9/2 + 26/3  = 15 square units

This is the required area represented in the square units.

Mixed or Whole Numbers

A fraction is defined as the relation of a whole digit. For case 7/4 is a fraction. What does “4” position for? It is the integer part into which the total number division. What does “7” position for? It is the number of equivalent parts which contain taken out. A mixed fraction is said to be the whole integer and a fraction number joint into one number known as “mixed number”.

Example for mixed or whole numbers:

Change Mixed numbers to improper numbers.

To modify mixed numbers into an improper numbers.

Steps to be followed:

Multiply the complete number part by fraction’s denominator.
Calculation up to the numerator.
Then reminder to the on top of the denominator.

Example: Convert 4 2/6 to an improper fraction.

Multiply the complete number by the denominator: 4 × 6 = 24
Calculation up the numerator to that: 24 + 2 = 26.
Then note down that on top of the denominator, like this: 26/6 or 13/3.


Rounding Whole Numbers

Rounding whole number means to the closest to the means to evaluate the nearest integers ,having the zeros to the right position of the tens position. Note: when the number 5, 6, 7, 8, or 9 shows in the ones position, round up; when the numbers 0, 1, 2, 3, or 4 shows in the ones position, round down.

Examples:

Rounding 219 to the closest ten gives 220.
Rounding 355 to the closest ten gives 360.
Rounding 105 to the closest ten gives 100.

Similarly, to round a numeral to any position value, to find the digit with zeros in all of the positions to the right of the position value being rounded to is nearest in value to the new numbers.

Whole Number Portion:

The decimal number to the whole number portion is which of the numbers to the remaining of the decimal position.

Example:

In the numeral 48.65, the whole number portion is 48.

In the numeral 0.004, the whole number portion is 0.

Common Monomial Factor

Monomial means a single term of a polynomial. Ex : 2xy or 8mn or 3x2. Monomial factor is a factor which has a variable and a term with some exponents. Common monomial factor means finding the common factor from the given set of monomials.  The common monomial factor can be a combination of number and variable, a variable, a number, two or more numbers and two or more variables.

Steps to find the common monomial factor:

Step 1: write down all the given monomials.

Step 2: write the factors for each given monomial separately.

Step 3: Highlight the common factor from all the factors.

Step 4: The common factors are the common monomial factor.

Example based on Common monomial factor:

Ex 1: Find the common monomial factor for the monomials 9pq and 12pq.

Sol:

The given monomials are 9pq and 12pq

The factors of 9pq are 3, 3, p and q

The factors of 12pq are 2, 2, 3, p and q.

The common factors are 3, p and q.

The common monomial factor is 3pq.

Ex 2: Find the common monomial factor for the monomials 9x3y2 and 15xyz.

Sol:

The given monomials are 9x3y2 and 15xyz.

The factors of the monomial 9x3y2 is 3, 3, x, x, x, y and y and the factors 15xyz is 3, 5, x, y and z

The common factors are 3, x and y.

The common monomial factor is 3, x and y. It can be written as 3*x*y=> 3xy.

Ex 3:Find the common monomial factor for the monomials 25a2bc, 252a2b and 20 a2b2.

Sol:

The given monomials are 25a2bc, 252a2b and 20 ab2.

The factors of 25a2bc is 5, 5, a, a, b and c.

The factors of 252a2b is 5, 5, 5, 5, a, a and b

The factors of 20 a2b2  is 2, 2, 5, a, a, b and b.

The common factors are 5, a, a and b.

The common monomial factor is 5a^2b.



Ex 4:Find the common monomial factor for x, yz and xyz.

Sol:

The given monomials are x, yz and xyz.

The factors of x is x.

The factors of yz are y and z.

The factors of xyz are x, y and z.

The common monomial factor is 0 because there is no common factor from the given monomials.

Multiple Value Function

During mathematics, a multivalued function is an overall relation; specifically all input is related with one or else further outputs. Exactingly words, a well-defined function connections one, also only one, output toward at all particular input. The word "multiple value function" is, consequently, a misnomer while functions be single-valued.Multiple value functions frequently occur as of functions are not injective. Such functions perform not contain an inverse function, except they perform include an inverse relation. It is also called the set-valued function.

Examples for multiple value function

 

This figure do not symbolize a right function since the element 3 within x be related through two elements b with c within y.

Each real number larger than zero or else each complex numbers except for 0 have two square roots. The square roots of 9 be in the set {+3,−3}.

The square roots of 0 be describe through the multiset {0,0}, since 0 be a root of multiplicity 2 of the polynomial x².

Every complex number include three cube roots

Complex logarithm functions are multiple-valued. The values implicit with log (1) be 2πni for every integers n.

Inverse trigonometric function is multiple-valued since trigonometric function is periodic.

tan(π/4)=tan((5π/4)=tan((-3π/4)=tan(2(n+1)π/4)=...1

Therefore arctan(1) might be consideration of since contain multiple values such as π/4, 5π/4, −3π/4, and rapidly. We know how to treat arctan since a single-valued function through restrict the domain of to -π/2 < x < π/2. Therefore, the range of arctan(x) becomes -π/2 < x < π/2. These values as of a limited field are call principal values.

Example problems

Problem 1
Given function f(x)=7x-5, what is the value of f(1) and(f(2)?
solution:

Given function f(x)=7x-5)

We can  find the value of f(1), to substitute 1 for the given function,

f(1)=7x1-5

      =7-5

f(1)=2

We can  find the value of f(2), to substitute 2 for the given function,

f(2)=7x2-5

      =14-5

f(2)=9

Problem 2
Given function f(x)=5x²+6x-5, what is the value of f(5)?
solution:

Given function f(x)=5x²+6x-5

We can  find the value of f(5), to substitute 5 for the given function,

f(5)=5x5²+6x5-5

      =5x25+6x5-5

     =125+30-5

  f(5)=150

Rectangular Objects

Rectangular objects are one of the basis of mathematics. Rectangular objects are having six sides. One of the rectangular objects is cuboid. Cuboid are also having six sides. These rectangular cuboid are having same faces like the rectangular objects. Rectangular prism is also included in the rectangular objects. Length of the rectangular prism is equal to the length of the rectangular.

Explanation for rectangular objects

There are many rectangular objects are present. They are defined as,

1. Rectangular cuboid

2.Rectangular prism

Rectangular cuboid:

The volume of the rectangular cuboid are find using the formula that are shown below,

Volume = Height `xx` Width `xx` Length

This can also be written as,

V = h `xx` w `xx` l.

Surface area of the rectangular cuboid are find using the formula,

Surface area = 2wl `xx` 2lh `xx` 2hw

The diagrammatic representation of the cuboid are shown below,



Example problem for rectangular objects

Example problem 1: Find the surface area and volume of the cuboid by using the formula, where the height = 4, length = 3, width = 2.

Solution:

Step 1: The volume of the rectangular cuboid are,

Volume = Height `xx` Width `xx` Length

from given h= 4, l =3, w =2

Step 2: Therefore by substituting the given information, we get,

V = 4 `xx` 3 `xx` 2

= 24.

Therefore, the volume of the cuboid is 24units.

Step 3: The surface area of the cuboid is given by,

Surface area = 2wl `xx` 2lh `xx` 2hw

Step 4: by substituting the above information given, we get,

Surface area = 2 ( 3 `xx` 2 ) `xx` 2( 4 `xx` 3 ) `xx` 2 ( 3 `xx` 2 )

= 2(6) `xx` 2(12) `xx` 2(6)

= 12 `xx` 24 `xx` 12

= 3456

This is the required area for cuboid.

Example problem 2: Find the surface area and volume of the cuboid by using the formula, where the height = 3, length = 4, width = 5.

Solution:

Step 1: The volume of the rectangular cuboid are,

Volume = Height `xx` Width `xx` Length

from given h= 3, l =4, w =5

Step 2: Therefore by substituting the given information, we get,

V = 3 `xx` 4 `xx` 5

= 60

Therefore, the volume of the cuboid is 60 units.

Step 3: The surface area of the cuboid is given by,

Surface area = 2wl `xx` 2lh `xx` 2hw

Step 4: by substituting the above information given, we get,

Surface area = 2 ( 6 `xx` 4 ) `xx` 2( 4 `xx` 3 ) `xx` 2 ( 3 `xx` 6 )

= 2( 24 ) `xx` 2( 12 ) `xx` 2( 18 )

= 288 `xx` 24 `xx` 36

= 248832

This is the required area for cuboid.

Adding Fractions Grade Six

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.(Source: Wikipedia)

This article will help for grade six students. Through this article grade six students learn about adding fractions.We are going to see some solved problems on adding fractions and some practice problems on adding fractions.

Solved problems on adding fractions grade six:

Problem 1:

Add the fractions 1/8 and 1/8

Solution:

Given, 1/8 + 1/8

Here both fractions have the same common denominator.

So we Can add the numerator normally and keep the denominator as it is.

1/8 + 1/8 = (1+1)/8

= 2/8

We can also simplify it further,

(2÷2)/(8÷2) = 1/4

Answer: 1/8 + 1/8 = 1/4

Problem 2:

Adding the fractions 1/8 + 1/16

Solution:

Given, 1/8  + 1/16

Both fractions have different denominator,

So we need to find least common denominator

Multiple of 8 = 8, 16, 24 ,32...

Multiple of 16 = 16 , 32 ....

Least common multiple = 16.

So to make a common denominator, multiply 1/8 by 2 on both  numerator and denominator,

(1 * 2) / (8 * 2) = 2/16

Now we can subtract,

1/8 + 1/16 = 2/16 + 1/16

= (2 + 1) / 16

= 3/ 16

Answer: 1/8 + 1/16 =3/16

Problem 3:

Adding the fraction 12 / 32 + 6 / 18

Solution:

Given , 12 / 32 + 6 / 18

Both fractions have different denominator,

So we need to find least common denominator

Multiple of  32 = 32, 64, 96, 128, 160, 192, 224 32, 64, 96, 128, 160, 192, 224, 256, 288, 320.........

Multiple of  18 = 18 36 54 72 90 108 126 144 162 180 198 216 234 252 270 288 306 324........

The least common multiple of 32 and 18 is 288.

So we need to make the denominator as 288.

(12 * 9) / ( 32 * 9 ) = 108 / 288

( 6 * 16) / ( 18 * 16) = 96 / 288

12 / 32  + 6 / 18 = 108 / 288 + 96 / 288

= ( 108 + 96) / 288

= 204 / 288

= 51 / 72

Practice problems on adding fractions grade six :

Problems:

1. Adding the fractions  6/7 + 15 /7

2. Adding the fractions 3/4 + 5/6

Answer:

1.3

2.19 / 12

Probability and Chance

Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory, which is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems. In this article we shall discuss about probability and chance problems.(Source: wikipedia)

Probability and chance example problem

Example 1:

If the odds in favor of an event be 3/5 find the chance of the occurrence of the event.

Solution:

Let the given event be E and let P(E)=x then

Odds in favor of E =`(P(E))/(1-P(E))`

= `(P(E))/(1-P(E))` =`3/5`

= `(x)/(1-x)` =`3/5`

5x=3-3x

8x=3

x= `3/8`

Therefore required probability = `3/8`

Example 2:

There dice are thrown together. Find the chance of getting a total of at least 6.

Solution:

In throwing 3 dice together the number of all possible outcomes is (6x6x6)=216

Let E = event of getting a total of at least 6.

Then, E = event of getting a total of less than 6.

= event of getting a total of 3 or or 5.

={(1, 1, 1),(1,1,2),(1,2,1),(2,1,1)(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2)(2,2,1)}

Now , n(`barE` )=10= P(not E)=P(`barE` )=`(n(barE))/(n(S))` =`10/216` =`5/108`

=P(E)=1-P(not E)=1-`5/108` =`103/108`

Hence the required probability is `103/108`

Example 3:

A bag contains 12 red and 15 white balloon. One ball is drawn at random. Find the chance that the ball drawn is red

Solution

Total number of balls=(12+15)=27

Let S be the sample space. Then

n(S)= number of ways of selecting 1 balloon out of 27=27

Let E be the event of drawing a red balloon. Then

n(E)= number of ways of selecting 1 red balloon out of 12=12

Therefore P(getting a /red balloon)=P(E)=`(n(E))/(n(S))` =`12/27` = `4/9`

Probability and chance practice problem

Problem 1:

The odds in favor of the outcomes of an event are 8:13 find the chance that the event will occur

Answer:

`8/21`

Problem 2:

If the odds against the occurrences of an event be 4:7 find the chance of the occurrences of the event?

Answer:

`7/11`