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Integration Made Easy
By Masood Amir
Calculus is a tool used to measure change or variation of a function with respect to the independent variable.
It is the branch of calculus used to measure change or variation of a function in a very small interval of time, the techniques use to measure such changes is called “differentiation”.
It is the branch of calculus used to measure changes or variation over an interval of independent variable, e.g to find length of curve, the area of region and the volume of a solid in a specified period of time.
The technique used to measure such changes or variation is called “Integration” or “Antiderivatives”. It a reverse process of differentiation.
Mathematically, Integration is defined as “ If f’(x) represents the differential coefficient of f(x), then the problem of integration is given f’(x), find f(x) or dy/dx, find y.
Notation: ”∫” is used to show the integration, it is a symbol of “S” derived from the word “Sum”. i.e. Integration is a process in which we have to sum up the derivatives over a specified interval and to find the function.
Techniques of Integration:
As we know that integration is the reverse process of differentiation, our problem is to find the function f(x) or Y, when f’(X) or dy/dx is given.
dy/dx = f’(X)
?dy = ?f ’(X)dx
Y= f(x) is our solution
Ist Formula of Integration (1st Rule of Integration)
Ist Formula of Integration (Ist Rule of Integration):
Let ? dy = ? xndx
y = xn+1/n+1 + C
In the process of differentiation, we eliminate constant, as the derivative of a constant is “zero”.
So, In functions like Xn, Xn+ 6, Xn + 3 , Xn – K, the derivatives of all of them is Xn-1, in finding the anti derivative of Xn-1, we put a constant “C”, as we don’t know which constant was present in the original function, untill and unless we have initial boundary values (Definite Integral).
Example: Solve ∫x3dx
Solution: x3+1/3+1 + C = x4/4 + C
Example: Solve ∫(x3 + x2 + 5x + 6)dx
Solution: ∫x3dx + ∫x2dx + ∫5xdx +∫6dx
X4/4 + x3/3 + 5×2/2+ 6x + C
Worksheet # 1
Find the Integral of the following:
∫(4v3x2 -2x)dx ∫(3x + 5×2 –x3/2-0.4×4)dx
∫(x-3 + x-4)dx ∫(2×3-3)3x4dx
∫(4×7 + 3×12 -5×8 + 2x -1)dx ∫(ax3 – bx2 + cx –d)dx
∫(1/x3 + 2/x2 -6)dx ∫(-3x-8 + 2vx)dx
∫((x3 -5)(2x + 5)dx ∫(7x-6 + 5vx)dx