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Integration Made Easy

By Masood Amir

Calculus:

Calculus used to measure change or variation of a function with respect to the independent variable.

### Differential Calculus:

The branch of calculus used to measure change or variation of a function in a very small interval of time,“differentiation” mesure such changes.

### Integral Calculus:

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.

“Integration” or “Antiderivatives” measure such changes, It is reverse process of differentiation.

Mathematically, Integration defined as “ If f’(x) represents the differential coefficient of f(x), then the problem of integration is to find f(x) or dy/dx, such that we have f’(x).

Notation: ”∫” used to show the integration, 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, find the function f(x) or Y, when we have f’(X) or dy/dx .

dy/dx = f’(X)

∫dy =∫f ’(X)dx

Y= f(x) is our solution

Ist Formula of Integration (1st Rule of Integration)

### Indefinite Integration:

Ist Formula of Integration (Ist Rule of Integration):

Let ∫dy =∫xndx

y = xn+1/n+1 + C

Why “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:

SET 1

- ∫(x3-4×2+5x-6)dx
- ∫(3×5-4×3+3×2)dx
- ∫(x3-4×2+5x-6)dx
- ∫(3×5-4×3+3×2)dx
- ∫(ax5-bx4)dx
- ∫(x3/2-5×4/3+3×2)dx
- ∫(4v3x2 -2x)dx
- ∫(3x + 5×2 –x3/2-0.4×4)dx
- ∫(x(8x-1/2)dx
- ∫(2-x)(4+3x)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

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