Integral of ln u

Integral calculus is a branch of calculus that deals with integration. If f is a function of real variable x, then the definite integral is given by

int_a^bf(x)dx

Where a and b are intervals of a real line.

Integral of ln u means integration of natural logarithmic (ln) function with respect to the variable ' u '. In this article, we are going to see the list of natural logarithmic integral (ln) rules with few example problems.
List of Natural Logarithmic (ln) Integral Rules:

int 1/u dx = ln |u| + C

int  ln u du = uln u  - u + C

int uln u du = (u^2)/2 ln u  - 1/4 u2 + C

int u2 ln u du = (u^3)/3 ln u - 1/9 u3 + C

int (ln u)2 du = u(ln u)2 - 2uln u + 2u + C

int ((ln u)^n)/u dx = 1/(n + 1) (ln u)n+1 + C

int  (du)/(uln u) = ln (ln u) + C

Learning Natural Logarithmic (ln) Integral Rules with Example Problems:

Example problem 1:

Integrate the function f(u) = 5/u

Solution:

Step 1: Given function

f(u) = 5/u

int f(u) du = int5/u du

Step 2: Integrate the given function f(u) = 5/u with respect to ' u',

int5/u du = 5ln (u)+ C

Example problem 2:

Integrate the function f(u) = log (9u)

Solution:

Step 1: Given function

f(u) = log (9u)

int f(u) du = intlog (9u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let x = log (9u)    dv = du

dx = 1/u  du         v = u

int x dv = xv - int v dx

int log (9u) du = ulog (9u) - int u 1/u du

= ulog (9u) - int du

= ulog(9u) - u + C

My forthcoming post is on Mean Median Mode Calculator and Finding Interquartile Range will give you more understanding about Algebra.

Example problem 3:

Integrate the function f(u) = 10log (3u)

Solution:

Step 1: Given function

f(u) = 10log (3u)

int f(u) du = int10log (3u) du

The above function can be written as

int f(u) du = 10intlog (3u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let u = log (3u)    dv = du

du = 1/u  du         v = u

int x dv = xv - int v dx

10 int log (3u) du = 10[ulog (3u) - int u 1/u du]

= 10ulog (3u) - 10int du

= 10ulog(3u) - 10u + C

Equilateral Pentagon Angles

In geometry an equilateral pentagon is a polygon with five sides of equal length. Its five internal angles, in turn, can take several values, thus permitting to form a family of pentagons. In contrast, the regular pentagon is unique, because is equilateral but at the same time its five angles are equal. Four intersecting equal circles disposed in a closed chain, are sufficient to describe an equilateral pentagon. ( Source Wikipedia )


I like to share this Equilateral Polygon with you all through my article.

Examples

Problems 1

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 100, 58 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=100

Angle 3 a3=58

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+100+58+67+x=540

250+125+x=540

375+x=540

Subtract 375 on both sides

375-375+x=540-375

x=165

The missing angle is 165

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Problems 2

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 70, 158 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=70

Angle 3 a3=158

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+70+158+67+x=540

220+225+x=540

445+x=540

Subtract 375 on both sides

455-455+x=540-455

x=95

The missing angle is 95

Problems 3

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 100, 150, 90 and 40

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=100

Angle 2 a2=150

Angle 3 a3=90

Angle 4 a4=40

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

100+150+90+40+x=540

250+130+x=540

380+x=540

Subtract 375 on both sides

380-380+x=540-380

x=160

The missing angle is 160

Having problem with algebra questions keep reading my upcoming posts, i will try to help you.

Problems 4

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 155, 120, 60 and 70

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=155

Angle 2 a2=120

Angle 3 a3=60

Angle 4 a4 =70

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

155+120+60+70+x=540

275+130+x=540

405+x=540

Subtract 375 on both sides

405-405+x=540-405

x=135

The missing angle is 135

Distance Formula Proof

For distance formula proof statistics will be deals with the co-ordinates parameters.Formula for distance calculation by the co ordinates of the points given. For calculating distance statistics between the points we have the distance formula proof and we substitute the points in that formula we find the distance between the points.In this article we have the distance formula proof and the problems using the distance formula.
Distance Formula Proof:
Distance between two points co ordinates is a basic concept in geometry.Now, we give an algebraic expression for the same.

Let P1  (x1, y1) and P2 (x2, y2) be two points in a Cartesian plane and denotes the distance between P1 and P2 by d(P1, P2) or  by  P1P2. Draw the line segment ` bar(P_1P_2)`

distance formula proof

The segment ` bar(P_1P_2)` is parallel to the x axis  Then y1 = y2. Draw P1 L and P2 M, perpendicular to the x-axis. Then d(P1,P2) is equal to the distance between L and M. But L is (x1, 0) and M is (x2, 0).

So the length LM = |x1-x2| Hence d (P1, P2) = |x1-x2|.

therefore, [d(P1,P2)]2= |x1-x2|2+ |y1-y2|2

=(x1-x2)2+(y1-y2)2

=(x2-x1)2+(y2-y1)2
d(P1,P2) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Problems Using Distance Formula Proof:
Example 1:
Find the distance between the points A(-2,3) and B(3,7).

Solution:
Assume d be the distance between A and B.             (x1,y1) = (-2,3)

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                   (x2,y2) =  (3,7).

=`sqrt((3+2)^2 +(7-3)^2)`

=`sqrt(5^2+4^2)`

=`sqrt(25+16)`

=`sqrt41`
Example 2:
Find the distance between the points A(-1,1) and B (3,3).

Solution:
Assume d be the distance between A and B.                (x1,y1)= (-1,1)

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                                (x2-y2)= (3,3).

=`sqrt((3+1)^2 +(3-1)^2)`

=`sqrt(4^2+(2)^2)`

=`sqrt(16+4)`

=`sqrt20`
=2`sqrt 5`

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Example 3:
Find the co ordinates distance between the points A(-1,5) and B (1, 4).

Solution:
Assume d be the distance between A and B.                       (x1,y1)= (-1,5)

Then d (A, B) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                          (x2-y2)= (1,4).

=`sqrt((1+1)^2 +(4-5)^2)`

= `sqrt(2^2+(-1)^2)`

=`sqrt(4+1)`
=`sqrt5`

Multiply Formula

In this article we are going see about how to multiply the two variables and numbers. Multiplication is the process of find the product of the given values. We use the distributive formula for finding the multiplication values. Multiplication formulas are reducing the man power and easy for calculation. Now in this article we solve some problems using multiplying formula.

Please express your views of this topic Multiply Fraction by commenting on blog.

Distributive formula:

a * (b + c) = (a * b) + (a * c)

x * (y * z) = (x * y * z)
Example Problems for Multiply Formula

Multiply formula example problem 1:

Multiply the given two function f(x) = (2x + 4) and g(x) = (6x - 2)

Solution:

Given functions are f(x) = (2x + 4) and g(x) = (6x - 2)

Add each and every variables of the given two functions, we get

f(x) * g(x) = (2x + 4) * (6x - 2)

Using distributive property,

= 2x (6x - 2) + 4 (6x - 2)

Expand the above values, we get

= 12x2 - 4x + 24x - 8

After simplification, we get

f(x) * g(x) = 12x2 + 20x - 8

Answer:

The final multiplication value of the given function is 12x2 + 20x - 8

Multiply formula example problem 2:

Multiply the given two function f(x) = (10x + 3) and g(x) = (x - 3)

Solution:

Given functions are f(x) = (10x + 3) and g(x) = (x - 3)

Add each and every variables of the given two functions, we get

f(x) * g(x) = (10x + 3) * (x - 3)

Using distributive property,

= 10x (x - 3) + 3 (x - 3)

Expand the above values, we get

= 10x2 - 30x + 3x - 9

After simplification, we get

f(x) * g(x) = 10x2 - 27x - 9

Answer:

The final multiplication value of the given function is 10x2 - 27x - 9

I am planning to write more post on Log Calculator and Example of Histogram. Keep checking my blog.

Multiply Formula Example Problem 3:

Multiply the given two function f(x) = (3x2 + 1) and g(x) = (x + 2)

Solution:

Given functions are f(x) = (3x2 + 1) and g(x) = (x + 2)

Add each and every variables of the given two functions, we get

f(x) * g(x) = (3x2 + 1) * (x + 2)

Using distributive property,

= 3x2 (x + 2) + 1 (x + 2)

Expand the above values, we get

= 3x3 + 6x2 + x + 2

After simplification, we get

f(x) * g(x) = 3x3 + 6x2 + x + 2

Answer:

The final multiplication value of the given function is 3x3 + 6x2 + x + 2

Volume of Three Dimensional Shapes

Volume is how much three-dimensional space a substance or shape occupies or contains. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, can be easily calculated using arithmetic formulas. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space. (Source: From Wikipedia).

Here we are going to see the formulas to find the volume of three dimensional shapes and example problems.
Formulas to Find the Volume of three Dimensional Shapes

Here we are going to see some arithmetic formulas to find the volume of simple three dimensional shapes such as cube, cone, cylinder, and sphere.

Volume of cube = a3 cubic units. Where, a is the side of the cube.
Volume of cone = `1/3 pi r^2 h` cubic units. Where, r and h are the radius and height of the cone.
Volume of cylinder = `pi r^2 h` cubic units. Where, r and h are the radius and height of the cylinder.
Volume of sphere = `4/3 pi r^3` cubic units. Where, r is the radius of the sphere.

Example Problems to Find the Volume of three Dimensional Shapes

Example 1

Find the volume of a three dimensional shape with all sides equal to 3 feet.

Solution

A three dimensional shape with equal sides is a cube.

Volume of a cube = `a^3` cubic units

= 33

= 3 * 3 * 3

= 27

So, the volume of the given three dimensional shape is 27 cubic feet.

Example 2

Find the volume of a sphere, whose radius is 5 m.

Solution

Volume of a sphere = `4/3 pi r^3` cubic units

= `4/3` * 3.14 * `5^3`

= `4/3` * 3.14 * 5 * 5 * 5

= 523.33

The volume of the given sphere is 523.33 cubic meter.

Example 3

Find the volume of a cylinder with radius 3 cm and height 4 cm.

Solution

Volume of a cylinder = `pi r^2 h` cubic units

= 3.14 * 3 * 3 * 4

= 113.04

So, the volume of the given cylinder is 113.04 cubic cm.

Algebra is widely used in day to day activities watch out for my forthcoming posts on equation of line and Distance From a Point to a Line. I am sure they will be helpful.

Example 4

Find the volume of the cone, whose radius is 3 cm and height is 4 cm.

Solution

Volume of a cone = `1/3 pi r^2 h` cubic units

= `1/3` * 3.14 * 3 * 3 * 4

= 3.14 * 3 * 4

= 37.68

So, the volume of the given cone is 37.68 cubic cm.

Depreciation Value Formula

Depreciation is the reduction in the value of the asset year by year due to wear and tear. Depreciation can be calculated using the straight line depreciation method or the accelerated depreciation method.

The straight line method calculates the depreciation  by spreading the cost evenly over the life period of the fixed asset.
Accelerated depreciation method calculates the depreciation by expensing  a large part of the cost at the beginning of the life of the fixed asset.



Formula for Straight line depreciation method =  Cost / life.

Ex :- The cost of a machine is $ 100.  It is expected to last for 4 years. Calculate the annual depreciation.

Sol:  Cost of the machine = $ 100

Expected time         =       4

Therefore annual depreciation is 100 / 4 = $25

Every year $25 would be expensed as depreciation value.

In this method the salvage value is not taken into account.
Depreciation Value:

Another method used is called accelerated depreciation  method  or declining balance depreciation method. It uses a factor based on the life of the asset. The factor is the percentage of the asset that would be depreciated each year under straight line depreciation times the accelerator.

Let us take the cost of the machine that is $100. Under this system we double the depreciation period a 50% from 25%. This is called double declining balance

Ex:  Cost of a machine is $100.  Its life period is 4 years.  What is the depreciation factor?

Sol: The depreciation factor is calculated  by doubling  as 200% = 2 * (1/4) = 0.50

So the calculation runs like this:-

Year              Depreciable basis   Depreciation calculation    Depreciation expenses  Accumulated depreciation

1                     $100                       100 * 0.5                              50                                  50

2                     $  50                        50 * 0.5                              25                                   75

3                     $  25                        25 * 0.5                              12.50                              87.50

4                     $  12.50                 12.5 * 0.5                                6.25                              93.75

So over the  four years the depreciation has been $ 93.75

The salvage value is $100 - $93.75 = $6.25

This depreciation value formula is adopted by most of the manufacturing units where there is considerable wear and tear.

It is also used by  transport system such as lorries, goods carrier where the wear and tear is considerable.
Formula of Depreciation Value :

There is yet another formula for finding the depreciation value. It is  written as  A= P( 1 - i)n

A = the depreciated amount : P = the present value ;  i = rate of depreciation and n = number of years.

It is also written as FV = PV ( 1 - i)n  where FV is future value ; PV = present value ; i = rate of depreciation and n = number of years.

We use the subtraction sign (1 - i) because the value goes down.

Let us do a problem using this formula.

Ex: A machine depreciates  in value each year at the rate of 10% of its value at the beginning of a year.  The machine was purchased for $10,000.  Obtain its value at the end of 10th year.

Sol:

Present value = PV = $10,000

rate of depreciation =10% = 0.1

Number of years = 10                                                                                                                          _                               _

So   FV = 10,000 ( 1 -0.1)10  = 10,000 (0.9)10  = log  0f 10,000 + 10 log 0.9 = 4.0000 + 10 x  1.9542 = 4.0000 + 1.5420

=  3.5420

Antilog of  3.5420 = 3483

Hence the depreciated value of the machine whose purchase price was $10,000  at the depreciation rate of 10% for 10 years becomes $ 3,483

Ans = $ 3,483

Independent Random Vectors

The basic needs of vectors are to provide the relationship between the magnitude and direction. Let us take any two random vectors respectively s and t. If the directions of these random vectors are in the different direction and different magnitude we can say that these two vectors are independent random vectors.  If any of two vectors will be in the different places, we can say that the two are independent.

Independent Random Vectors:

Let us take {X1, X2, . . .  Xn} are the set of n random variables.

This is often convenient to consider these random variables set at a single object x = {X1, X2, . . .  Xn}. This will be called random vectors.  If it is particularly true true when this set of random variables submitted to the linear transformations.

If a random variable is described using its probability distribution the random vector is always described using its joint probability distribution of the n random variables which will make the random vector.
Expectation of Independent Random Vectors:

By the definition of independent random vectors the expectation is the vector whose components are the expectation of the independent random vectors.

If we make the vector

E(x) =` [[E[x_1]],[E[x_2]],[[....]],[E[x_n]]] `

Where the mean μ = E[x] and this is the center of gravity of the joint probability distribution of the random variables {X1, X2, . . .  Xn}

Linearity of the independent random vectors:

If the random vectors are immediately verified that the expectation is linear

If x and y are two independent random vectors, and if A and B are two constant matrices:

Then E [Ax + By] = AE[x] + BE[y]

And if b is a constant vector:

E [Ax + b] = AE[x] + b

My forthcoming post is on First Order Differential Equation and Definite Integrals will give you more understanding about Algebra.


Variance of the independent random vector:

Definition of the variance of the random vector and the random variable are equal. If we denote μ and the vector E[x], then by definition of the variance:

Var(x) = E[(x – μ)(x – μ)']