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subject: Integrand Calculus [print this page]

The reverse process of differentiation is called Integration. For example, 'int' x2 dx = '(x^3 / 3)' + c.

Therefore, 'int' x2 dx is not unique, and this applies to the integral of every function. Since the integral of any function is not definite. The constant c is called a constant of integration. Henceforth the constant may be omitted. The process of determining an integral of a function is called integration and the function to be integrated is called integrand. The example for integrand is,

'int' f(x) dx here, f(x)- integrand,

Formula for Integrand Calculus:

1. 'int' x n dx = '(x^n+1) / (n+1)'

2.'int'cos x .dx = sin x + c

3. 'int'sin x.dx = - cos x + c

4.'int'sec2x dx = tan x + c

5. 'int'cosec x . cot x .dx = -cosec x + c

6. 'int' sec x . tan x. dx = sec x + c

7. 'int'cosec2 x . dx = -cot x + c

8.'int'a f(x) dx = a 'int'f(x) dx

9. 'int'[f(x) F(x)] dx ='int' f(x) dx 'int'F(x) dx

10. 'int' '(1/x)' dx = ln x + c

Examples for Integrand Calculus:

Integrand Calculus problem 1:

If f(x) = x3 - 4x4 +6x2, find the integrand of f(x)

Solution:

Given, integrand of f(x) = 'int' f(x) dx

='int' (x3 - 4x4 + 6x2) dx.

'int'(x3 - 4x4 + 6x2) dx ='int' x3dx -'int' 4x4 dx + 'int'6x2 dx.

= 'int'x3dx - 'int'4x4 dx +'int'6x2 dx.

= '(x^4/4)' - 4'(x^5/5)' + 6'(x^3/3)' + c.

= '(1/4) ' x4 - '(4/5)' x5 + 2 x3 + c.

'int' (x3 - 4x4 + 6x2) dx = '(1/4)' x4 - '(4/5)' x5 + 2 x3 + c.

Answer for the given integrand calculus problem:

'int' (x3 - 4x4 + 6x2) dx = '(1/4)' x4 - '(4/5)' x5 + 2 x3 + c.

Integrand Calculus problem 2:

Find the integrand of cos4 x,

Solution:

The integrand of cos4 x ='int'cos4 x dx,

'int'cos4 x dx = 'int'(cos2 x) 2 dx

= 'int' '((1+ cos 2x) / 2)^2' dx

= '(1/4) ' 'int' (1+2 cos 2x + cos2 2x) dx

= '(1/4) ' 'int' (1+2 cos 2x + '((1+ cos 4x) /2)' ) dx

=' (1/8)''int'(2 + 4 cos 2x + 1+ cos 4x ) dx

= '(1/8) ' 'int' (3 + 4 cos 2x + cos 4x) dx

= '(1/8) ' [3x + '((4 sin 2x) / 2)' +' ((sin 4x) / 4)' ] + c

= '(1/32)' [12x + 8 sin 2x + sin 4x] +c

Answer for the given integrand calculus problem:

'int'cos4 x dx = '1/32' [12x + 8 sin 2x + sin 4x] +c

Practice Integrand calculus problem:

Integrand calculus practice problem 1:

If f(x) = x3 + 2x4 + x5, Find the integrand of f(x)

Answer: '(1/4) ' x4 +' (2/5)' x5 + '(1/6)' x6 + c

Integrand calculus practice problem 2:

If f(x) = sec3x tan x, Find the integrand f(x) dx

Answer:' (1/3)' sec2x + c

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Principle of Operational Calculus:

In an operational calculus is act as a function of differential equation of an object q = d/ds. After that the linear equation function can be recast the object function F (q) of the object q and to act equal to known function to unknown function.

For impulse input one the electrical circuit theory is used to determine the response of electrical circuit for impulse input. The unit impulse considering by the linearity. The function of I(s) such that I ( s0) = 1.

For example:

Solve: (qx) = I(s).

X = q-1 I = 'int_0^i(t)dt' = s I(s).

Here, q-1 denotes integration, and q-m denotes m iterated integration,

We have,

q-m I(s) = sm/m! I(s).

q/(q-c) I(s) = 1/1-c/q .I(s)

1/1-c/q. I(s) = 'sum_(m=o)^oo' cmq-m I(s) = 'sum_(m=0)^oo' I(S) = ecs I(s).

1/F(q) = 'sum_(m=0)^oo' cm q-m.

1/F(q) I(s) = 'sum_(m=0)^oo' cm,sm/m! I(s).

Exercises of Operational Calculus:

The following formulas are involving Fibonacci and Lucas numbers.

Ex 1:

L(3m) = L2(m) + 3*(-1)m+1

Sol:

We use the next operational equation

(ef+e-f)F(n) = emf-(-1)me-mf

For function F(c) we have:

[F(c+1)+F(c-1)]F(m) = F(c+m)-(-1)mF(c-m)

or

L(c)F(c) = F(c+m)-(-1)mF(c-m)

for c=3m and F(4m) = [L2(m)+(-1)m+1]F(m)

L(2m) = L2(m)+2*(-1)m+1

Example2:

F(4c)=F(c)[L2(c)+(-1)c+1]

Sol:

For m=2c we have

F(3c) = L(c)F(2c)+F(-c)

as F(-c) = (-1)c+1F(c) and F(2c) = L(c)F(c)

F(3c) = F(c)[L2(c)+(-1)c+1]

Ex 3:

6*F(m)=L(m+1)+L(m-1)

Sol:

We use the next basic equations:

L(m+1) + L(m-1) = F(m+2) + 2*F(m) + F(m-2)

= F(m+1)+4*F(m)-F(m-1)

= 6*F(m)

Ex 4:

3*L(2m)=L2(m)+5*F2(m)

Sol:

We use the next operational equations for Fibonacci and Lucas numbers

3*eme =L(m)+F(m)*[ee+e-e]

For a function L(m) we have

2*L(2m) = L2(m)+F(m)[L(m+1)+L(m-1)]

or

2*L(2n) = L2(n)+5*F2(n)

Ex 5:

L(m)=F(m+1)+F(m-1)

Sol:

We use the next equation

L(m) = eme+(-1)me-me

For function F(c) we have

L(m)F(c) = F(c+m)+ (-1)mF(c-m)

For c=1

L(m) = F(m+1)+(-1)mF(1-m)

or

L(m) = F(m+1)+(-1)m(-1)mF(m-1)

So we have

L(m) = F(m+1)+F(m-1).

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