Â  Â  Â  Â
Â  Â  Â  Â  Â  Â

# Application Of Integration

Anti derivative of function f is the function F whose derivative is function f. We can understand it by an equation as F'=f. This process is also known as anti differentiation. This term is related to the definite integrals by using the Functions of Calculus. It can be understand by an example as the function F(x)=x3/3 is an anti derivative of the function f=x2 means x2 have indefinite number of anti Derivatives because the derivative of a constant is zero so x3/3+3,x3/3+78,x3/3+453 and so on. By it we can understand that when the value of the constant is changed then the new anti derivative is obtained as F(x)=x3/3+c here c is a arbitrary constant.

The Applications of Antiderivatives is understand with an example of acceleration and velocity in physics as v=u+at ,where v and u are velocity and a is acceleration as anti derivative of this equation means Integration of the acceleration yields the velocity and a constant as:

a=dv / dt + c and ∫t1t2 a(t) dt = v (t2) -v (t1) .This same pattern is applied to all other parts of equation as Position ,velocity, acceleration and so on .These are some essential application of anti derivative. These anti derivative are important to compute the definite integrals and using the formulas of calculus. If a function F is anti derivative of function f then it is shown as ∫ f(x) dx = F(b)-F(a).

We also define the anti derivative F of a function of f that have interval as then each anti derivative is different from other because of the value of the constant such as G (x) = F (x) + c here constant c has the different values on that basis the value of anti derivatives so c is known as arbitrary constant of integration.

We can find the anti derivative of a function F(a) if F' (a) = 4 – 3 ( 1 + a2-1

and f (1) = 0.

The general anti derivative of function is F ( a) = 4a - 3 arc tan ( a ) + c

For an arbitrary constant c the derivative of arc tan ( x) is ( 1 + x) . For find the specific Antiderivative, we evaluate:

F(1)=4 - 3 arc tan ( 1 ) + c = 0 so c = 3 arc tan(1) -4 = π / 4 - 4

Thus we have F (a ) =4a - 3 arc tan( a ) + π / 4 - 4

if we find the G ( a) then G ' (a) = a ( 6 + 5a) 1/2 and G ( 1 ) = 10.

So this is the way of generate the anti derivative of the Functions.

We can take another example for the finding the anti derivative as a ball dropped from a roof and hits the ground at 120 fts / s .What is the height of the roof ? (Note that the acceleration due to gravity is

a(T) =- 32 ft / s2)

Solution of it as follows:

we have a (t) = - 32 ft / s2

v (t) =- 32 t + c here c is an arbitrary constant ,to find c we evaluate getting

v (0) = c = 0

But the initial velocity is 0.Thus we have v(t)=-32t.

Next we calculate height

d (t) = -16 t+ c

Note that when t = 0 we have d ( 0 ) = c ,and so c is the height of the roof . In particular for getting the answer, we need to decide c . As we know that when the ball hits the ground, we have

v (t) = -32t = -120

And so it follows that the ball hits the ground when t 120 / 32 = 15 / 4 .

Since the ball gets the height 0 when it hits the ground, we have

d ( 15 / 4 ) = -16.225 / 16 + c = 0 and so c = 225.

So the height of the roof is 250 ft.

For defining the application of anti derivative we take some more example of it:-

The most general form of anti derivative as follows f(a)=a2

The most general anti derivative is F(a) =a3+c for an arbitrary constant c. g(a) = 5 – 4a+ 2a/ a

When initially we solve it we get g(a) = 5 / a- 4 / a+ 2 = 5a-6 – 4a-3 + 2

Anti derivative of the most general type is g(a) = -a-5 + 2a-2 + 2a + c

For an arbitrary constant c. So these are some examples that show the application of anti derivative.

There is also some other application of anti derivatives as

Anti derivatives and differential equation: anti derivative can be used in finding the general solution of the differential equation

dp / dq = f ( q ).

Another example of application of anti derivative is as:-

A brick is dropped from an apartment whose window is 20 meters higher from the ground. Suppose that the acceleration of the brick is −16 m / secthen we have to calculate some of the answers for the following questions:

(1) The velocity of the brick based on the function of time t;

(2) The position d ( t) ;

(3) Time of the brick when hits the ground and the corresponding velocity.

Solution of the above questions as follows :-(1) The motion is described by the equation dv / dt = −16 m / sec2 . The general solution of this equation is given by v (t) = −16 t + C. Since the initial velocity is v (0) = 0, we find C = 0 and therefore v ( t ) = −16 t.

(2) The solution to the equation dh / dt= −16 the anti derivative

h ( t) = −8 t+ C .

But h (0) is the initial distance of the brick from the ground, that is h (0) = 20 .

Thus, 20 = C and therefore h (t) = −8 t2 + 20.

(3) The brick hits the ground when h (t) = 0 that is, −8 t2 + 20 = 0. Solving for t > 0 we find t ≈ 1.6 sec. At that time, v = −16 (1.6) = −25.6 m / sec. (The velocity is negative because we are decide up as positive and down as negative.)

We can take another example of anti derivative as we have a function ∫ cos (3a) da=? So as the solution of this function as the derivative of the sine is cosine so anti derivative of cosine is sine.

For solution of it we can take the bottom up approach that the anti derivative of the function most probably is sin (3x).So if we take the derivative sin (3x) then what do:

d / da sin (3a)=cos (3a) d/ da (3a) = 3 cos (3a)

So the anti derivative of it is almost sin (3x) but there is some other factor of 3 which comes at differentiation. It is because of Chain Rule of differentiation. Then we calculate little bit of changing function of sine as:-

d / da sin (3a) / 3 = 1 /3 cos (3a) d / da (3a)=1 / 3 .3 cos (3a) = cos (3a).

So the derivative of sin (3a) / 3 equals to cos (3a), that is as follows:

∫cos (3a) da= sin(3a)/3 + c,

The problem of initial value that is defined by an example as

da / db =4x3-2x+1 and b=3 when a=0 then find b as a function of a.

b=∫ [4a3- 2a + 1] da =4 ∫a3 da -2 ∫a da + ∫ da,

c = 4 . a4/ 4 - 2.a2/2 + a + c,

y = a- a+ a + c,

Initial condition that is b=3 when a=0 we use for solving the constant of integration:-

b = a- a+ a + c.

3=0-0+0 +c

3=c

so,

b=a- a+ a +3.

## Velocity and Distance Problems Involving Motion Along a Line

In daily life, motion is common thing to us, we walk, run and ride is a called as motion. General meaning of motion is change with respect to time like earth rotates once in every 24 hours.
In real life, when we change our Position in straight line, then this type of motion is called as a Straight Line motion and difference b...Read More

## Area of region

Suppose a function is given by y = g (x) represents a curve in the closed interval [a, b]. The area of the region enclosed by the curve of g (x), the vertical lines x = p and x = q and the x axis is given by:
Area A = pq g (x) dx,
For the area problems some properties of definite integrals is here:
If a function g (x) is defined at x = p then,
pp g (x) dx = 0,

## Area between curves

In Calculus when two curves intersects each other then area is also intersected by these two curves. Suppose we have two curve where first curve is  f(x) and second curve is g(x) ,so area between two curve is   A = f(x) – g(x). That is area between two curves if we take the interval between to two curve is [a,b] then formula would below:

A = ab (f(x) – g(x)...Read More

## Volumes of Solid of Revolutions

If a plane area is revolved about a fixed line in its own plane then the body or Solid that is formed by the revolution of the plane area is called as solid of revolutions and the fixed line about which the area revolves is known as the axis of revolution, it can be used to find the area of the region enclosed by the curves.
In other words we can say th...Read More

## Area of Region Bounded by Parametrically Defined or Polar Curves

Area of region bounded by polar curve, in this curve are lies between an interval [a ,b] and function is continuous at this interval, and f(x) ≥ 0 then the area of the region between the function and the x-axis is given by Area = ab f(x) dx.
If y-axis curve f(y) is continuous and another y-axis curve g(y) is also continu...Read More

## Arc Length

The topic "Arc length" is generally related to the geometries but sometimes we use it in the Calculus Geometry also. As the name suggests “Arc Length”, means a part of a curve in any region. To understand the arc length we first have to understand that what is arc and then arc length, so Arc is curved line made by two distinct points on the curve, and the arc length is the...Read More

## Velocity and Acceleration Vectors of Planar curves

Velocity is defined as speed with direction. Speed is equals to the distance covered per unit time.  Speed is defined in Scalar terms without direction and if we consider the direction, it is said to be the velocity of the object.  Velocity and acceleration vectors of planar curves are formed in the 2 dimensional planes. Velocity always...Read More

## Other Applications involving the use of Integral of Rates

Other applications involving the use of integral of rates can be understood with the help of various examples from our day to day life. Some of the examples for better understanding of the implementation of the above property are:
We will start with a small example from ancient civilization: The art of modern pottery which is st...Read More