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,x^3/3+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 ∫_t1^t2 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 + x² ) . 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 t² + 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 – 4a³ + 2a⁶ / a

When initially we solve it we get g(a) = 5 / a⁶ - 4 / a³ + 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 / sec² then 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 t² + 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 = a⁴ - a² + a + c,

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

b = a⁴ - a² + a + c.

3=0⁴ -0² +0 +c

3=c

so,

b=a⁴ - a² + a +3.