Applications of Integration Work

Integration is applied on a given function within certain limits to get various output.

Here let us see three areas of integration application work. In the normal sense, the integrations are applied to get the area under the given curve between certain limits.

1. Area under the curve f (x) between the limits ‘a’ and ‘b’ is given by`int_a^bf(x)dx`

2. The arc length of a curve f (x) between the limits ‘a’ and ‘b’ is given by `int_a^bsqrt[ 1 + [dy/dx] ^2] dx` .

3. The volume of a curve f(x) between the limits ‘a’ and ‘b’ is given by `pi int_ a^b y ^2 dx`

Now let us see problems on this topic.


Example Problems on Application of Integration Work.

Ex 1: Find the area under the curve f (x) = 2

4x 2 + 3x + 2 between 2 and 4.

Soln: Given: f (x) = 4x 2 + 3x + 2

The limits are ‘2’ and ‘4’.

Therefore Area =` int_2^4(4x ^2 + 3x + 2) dx` = `[4x ^3 / 3 + 3x ^2 / 2 + 2x] ` , the limit is between 2 to 4.

= `[(4 (4 ^3) / 3 + 3 (4) ^2 / 2 + 2 (4) ) ** (4 (2) ^3 / 3 + 3 (2) ^2 / 2 + 2 (2) )]`

Therefore Area = `96 2 / 3 ` sq units.

Ex 2: Find the arc length of a curve f (x) = 4x + 3 between 0 and 1.

Soln: Given: f (x) = 4x + 3

Therefore f ‘ (x) = 4.

Therefore arc length =` int_0^1 sqrt [(4) ^2 + 1] dx ` = `sqrt 17` `int_0^11 dx`

= sqrt 17 (x) between 0 to 1 = sqrt 17 (1 – 0) = sqrt 17 units.


More Example Problem on Application of Integration Work.

Ex 3: Find the volume generated when the finite region enclosed by the curve y = x ^2 and the lines x = 2, x = 4 and the x axis.

Soln: Volume = `pi int_2^4 y ^2 dx`

=` pi int_2^4 (x ^2) ^2 dx`

= `pi` [`x ^5 / 5` ] 2 to 4

= `pi / 5 [ 4 ^5** 2 ^5] = 992 / 5` cubic units.

Direct Inverse Relationship

Chemistry is study of physical and chemical properties of elements and compounds. While studying the same, various observations are made and inference derived from it. As such it involves discovery of correlations. Correlations means dependency. Dependency may be direct or inverse.

Establishing the dependencies scientifically is  very important as this can be converted to standard laws. These standard laws are useful in study of chemistry and chemical calculations
Direct Dependencies

Direct dependency means a parameter changes as the another changes. If one increases the another too increases.

e.g. An observation is made that the volume of a gas increases as the temperature increases. How can this direct dependendeny be presented mathematically?

V `alpha ` T

This equation is further rationalised as

V   = k T

where k is the constant for the corelation

OR

V / T =Constant

This is the established equation for Charles' Law

The direct dependencies may be proportional to 1st or 2nd or 3rd or even  1/2 or 1/4 etc. power too.

As in case of famous Einstein equation

E= mc2

lt denotes that the energy released is directly proportional not to the speed of light but, to the square of  the speed of light
Inverse Dependencies

Inverse dependency means a parameter changes as the another changes.If one increases the another decreases.

e.g. An observation is made that the volume of a gas increases as the temperature decreases. How can this inverse dependendeny be presented mathematically?

V `alpha ` 1/P

This equation is further rationalised as

V   =k / P

where k is the constant

for the corelation between volume and pressure..

or PV=Constant

This is the established equation for Boyles' Law

The inverse dependencies may be proportional to 1st or 2nd or 3rd or even  1/2 or 1/4 etc. power too

Rhombus Perimeter Formula

In geometry, a rhombus or rhomb is a quadrilateral whose four sides all have the same length. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the latter sometimes refers specifically to a rhombus with a 45° angle. (Source: Wikipedia)

Formula for perimeter:

Perimeter of a rhombus = 4 * Side length


Example problem 1:

A Rhombus has the side length is 17 m. Find the perimeter of the rhombus.

Solution:

Given side length of the rhombus is 17 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 17 m

= 68 m

Answer:

The final answer is 68 m

Example problem 2:

Find the perimeter of the given rhombus in centimeter.

Rhombus perimeter formula - Example problem 1

Solution:

Given side length of the rhombus is 12 cm

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 12 cm

= 48 cm

Answer:

The final answer is 48 cm

Example problem 3:

Find the perimeter of the given rhombus in meter.


Rhombus perimeter formula - Example problem 2

Solution:

Given side length of the rhombus is 9.1 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 9.1 m

= 36.4 m

Answer:

The final answer is 36.4 m

Example problem 4:

Find the perimeter of the given rhombus in meter.

Rhombus perimeter formula - Example problem 4

Solution:

Given side length of the rhombus is 7.7 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 7.7 m

= 30.8 m

Answer:

The final answer is 30.8 m


Practice Problems for Rhombus Perimeter Formula

Practice problem 1:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 1

Answer:

The final answer is 128 m

Practice problem 2:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 2

Answer:

The perimeter of the rhombus is 14.4m

Practice problem 3:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 3

Answer:

The final answer is 34 m

Revenue Formula Accounting

Revenue formula is defined as one of the important formula in accounting. Revenue is defined as the income of a company.  The income of revenue is mainly obtained from the sales.  The multiplication of selling price and the income is called as the revenue formula. The revenue is directly proportional to the quantity and it always from the linear manner. In this article, we are going to study to study about the revenue formula in accounting.


Explanation to Revenue Formula Accounting

The explanation to revenue formula accounting is given as follows,

Formula:

Revenue Formula = P `xx` Q

where,

P = Price per unit
Q = Quantity

Example Problems to Revenue Formula Accounting

Problem 1: Find the revenue by using the formula, where, P = 12500, Q = 200.

Solution:

Step 1: Given: The given values for finding the revenue is as follows,

P = 12500

Q = 200

Step 2: To find:

Revenue

Step 3: Formula: The formula given for finding the revenue is as follows,

Revenue Formula = P `xx` Q

Step 4: Solve:

Revenue Formula = P `xx` Q

= 12500   `xx`   200

= 2500000

Result: Revenue = 2500000

Therefore, this is the answer for solving the revenue formula in accounting.

Problem 2: Find the revenue by using the formula, where, P = 35000, Q = 250.

Solution:

Step 1: Given: The given values for finding the revenue is as follows,

P = 35000

Q = 250

Step 2: To find:

Revenue

Step 3: Formula: The formula given for finding the revenue is as follows,

Revenue Formula = P `xx` Q

Step 4: Solve:

Revenue Formula = P `xx` Q

= 35000   `xx`   250

= 875000

Result: Revenue = 875000

Therefore, this is the answer for solving the revenue formula in accounting.


Practice Problems to Revenue Formula Accounting

Problem 1: Find the revenue by using the formula, where, P = 45000, Q = 650.

Answer: 29250000

Problem 2: Find the revenue by using the formula, where, P = 55000, Q = 65.

Answer: 3575000

Problem 3: Find the revenue by using the formula, where, P = 43000, Q = 35.

Answer: 1505000

Algebra Geometry Practice

Algebra is a branch of math that deals with properties of lines, curves and surfaces generally divided into two types, Algebraic and geometry in agree with the mathematical techniques. The relationship between Algebra and geometry was introduced by Descartes.

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The ‘Geometry’ means study of properties. A point is used to denote a position in space. such that all points lying on the line joining any two points on the surface. The algebra geometry example problems and practice problems are given below.
Example Problems for Algebra Geometry

Example problem 1:

Find the values of a and b if 3x4 + x3 + ax2 + 5x + b is exactly divisible by x + 2 and x – 1.

Let P(x) = 3x4 + x3 + ax2 + 5x + b

(x+2) and (x–1) are factors of P(x) and hence both P(–2) and P(1) = 0

P(–2) = 3(–2)4 + (–2)3 + a(–2)2 + 5(–2) + b = 30 + 4a + b

Since P(–2) = 0 we get

4a + b = –30 (1)

P(1) = 3(1)4 + (1)3 + a(1)2 + 5(1) + b = 9 + a + b

Since P(1) = 0 we get

9 + a + b = 0 or a + b = –9 (2)

(1) – (2) gives

4a + b = –30

`(a+b=-9)/(3a = 21)`

` After solving this, We get,`

` a = 7`

Substituting a = –7 in equation (2) we get

–7 + b = –9 (or) ∴ b = –2

a = –7, b = –2.

Example problem 2:

Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1).

Solution:

P(x) = x3 – 5x2 + 7x – 4. By using the remainder theorem when P(x) is divided by (x – 1), the remainder is P(1).

The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1.

Example problem 3:

If all the sides of a parallelogram touch a circle prove that the parallelogram is a rhombus.

Solution:


Given: All the sides of a parallelogram ABCD touch a circle with centre O.

To prove: ABCD is a rhombus.

Proof: Since the lengths of the tangents from an external point to a given circle are equal (Fig)

AP = AS; BP = BQ; CR = CQ; DR = DS

Adding (AP+BP)+ (CR+DR) = (AS+DS) + (BQ+CQ)

AB + CD = AD + BC (Fig)

AB + AB = AD+AD (ABCD is a parallelogram CD= AB, BC=AD)

2AB = 2AD AB= AD

But AB= CD and AD = BC ( opposite sides are equal)

AB = BC = CD = AD. Hence ABCD is a rhombus.
Practice Problems for Algebra Geometry:

Practice problem 1:

Find the quotient and the remainder using synthetic division

1.  x3 + x2 – 3x + 5 ÷ x – 1

2. 4x3 – 3x2 + 2x – 4 ÷ x – 3

Answer: 1. Remainder = 4, Quotient = x2 + 2x + 1

2. Remainder = 83, Quotient = 4x2 + 9x + 29

Practice problem 2:

In the Fig AB is the diameter of the circle, ∠BAC = 42°. Find ∠ACD

Answer: m∠ACD = 48°

Inequalities on a Coordinate Plane

In mathematics, an inequality is a statement about the relative size or order of two objects or about whether they are the same or not.

• The notation a < b means that a is less than b.

• The notation a > b means that a is greater than b.

• The notation a ≠ b means that a is not equal to b, but does not say that one is greater than the other or even that they can be compared in size. Source – wikipedia.


Inequalities on a Coordinate Plane : Definition of Coordinate Plane:

The give figure is known as a coordinate plane. It has two axis and four quadrants. The two number lines form the axis. The horizontal number line is known as the x-axis and the vertical number line is known as the y-axis.

coordinate plane

Concepts for inequalities on a coordinate plane:

• The shaded area of a graph symbolizes all of the coordinates that will work in a specified equation.

• A solid frame of the shaded area specifies the edge is part of the answers to the equation.

• A dashed edge of the shaded area indicates that the edge of the graph is not element of the answers.



Steps to inequalities on a coordinate plane:

Ex: Graph the equation y`>=` x

Step1: Sketch the graph y = x. This equations in slope intercept form would appear like this y = 1/1 x + 0. The 0 indicates that you go through the origin, put a point there. Now use the slope to sketch the break of the line. From the origin go up one and to the right one and place another point. Do again until you have more than a few points on an inequalities on a coordinate plane.

example 1

In an inequalities on a polar coordinates now sketch a solid line since the equation to be charted is larger than or equivalent to. Your diagram must now appear like this:

example 1

Step 2: Then shade all over the place above the line because the equation indicates that the y values are larger than or equal to the line for any certain x value.

example 1

Now check your solution by inserting a pair of points from the shaded area and non-shaded area.
Example Problems for Inequalities on a Coordinate Plane:

Ex 1: Graph y > 2, in an inequalities on a coordinate plane.

Sol:

Keep in mind that is now a horizontal line. This is just a parallel line that is shaded above the line and dashed since it is not equivalent to the line it is only bigger than the line.

example 2

Ex 2: Graph x < -3, in an inequalities on a coordinate plane.

Sol:

Keep in mind that is just a perpendicular line. This is just a perpendicular line that is shaded to the left of the line and dashed since it is not identical to the line it is only fewer than the line. The x values on the left are fewer than the line.

Euclidean Geometry

Geo means “earth” and metron  means “measurement”. ”Euclid, a distinguished Greek mathematician, called the father of geometry. A point is used to represent a position in space. A plane to be a surface extending infinitely in every directions such that all points lying on the line joining any two points on the surface. The euclidean geometry example problems and practice problems are given below.
Example Problems - Euclidean Geometry:

Example problem 1:

Can the following angles are triangle or not?

(a) 50° , 70°, 60° (b) 40°, 80°, 70°

Solution :
(a) The sum of the measure of the three angles is
50° + 70° + 60° = 180°
Therefore 50°, 70°, 60° can be the measure of the angles of a triangle.
(b) The sum of the measure of the three angles is
40° + 80° + 70° = 190°.
But the sum of the measure of the angles of a triangle is 180°.
Therefore 40°, 80°, 70° cannot be the measures of the angles of a triangle.

Example problem 2:

Find the supplementary angle of 95°

Solution:

The supplementary angle of 95° = 180° – 95° = 85°.

Example problem 3:

The lengths of two sides of right triangles are 6cm and 8cm. Find its hypotenuses.

Solution:

AC = 6 cm
BC = 8 cm
AB =?
AB2 = 62+ 82
= 36 + 64
AB2 = 100
AB = √100 = 10

Thus, the hypotenuses are 10 cms in length.
Practice Problems - Euclidean Geometry:

Practice problem 1:

Two angles of a triangle are of measures 52°and 78°. Find the measure of the third angle.

Ans: 50

Practice problem 2:

Find the angles in each of the following:
(i) The angles are supplementary and the larger angles are twice the small.
(ii) The angles are complementary and the larger is 20° more than the other

Ans:    1. smaller angle = 60°, larger angle = 120°.

2. smaller angle = 35°, larger angle =  55°.