## Integration – Integration Formulas

February 6, 2012 in Additional Mathematics, Integration, Integration Formulas I by Shivakkumar Vadiveyl

## Copyright by MathsAchiever.com

In this lesson, we will be learning some of the basic Formulas of Integration.

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We have learnt in Differentiation the formula for Differentiation as follows,

$begin{array}{ccccc}& y & = & x^{m}& dfrac{dy}{dx} & = & mx^{m-1}end{array}$

The derivative is derived by reducing the power of the original x term by one and multiplying the resulting term with the original power of the x term.

Since the Integration is the reverse of Differentiation, the formula for Integration of $y=x^{n}$ can be written as

$begin{array}{ccccc}& int y, dx & = & dfrac{x^{n+1}}{n+1}+C & nneq-1,,, n=integerend{array}$

The Integral is derived by increasing the power of the x term by one and dividing the resulting term by n+1. The Integral is then added with the Constant of Integration, C.

We have also seen that when a constant multiplier exists in the function, it is maintained in the results of the Differentiation.

$begin{array}{ccccc}& y & = & ax^{m}& dfrac{dy}{dx} & = & amx^{m-1}end{array}$

Similarly, in the case of Integration, the constant multiplier is treated in the same way for a function, $y=ax^{n}$

$begin{array}{ccccc}& int y, dx & = & int ax^{n}dx& & = & aint x^{n}dx& int y, dx & = & adfrac{x^{n+1}}{n+1} & nneq-1,,, n=integerend{array}$

## Integration Example 4B

Integrate this function, $y=4x^{3}+2x^{2}+5x+1$

$begin{array}{ccccc}& y & = & 4x^{3}+2x^{2}+5x+1& int ydx & = & int(4x^{3}+2x^{2}+5x+1)dx+Cend{array}$

We have learnt in the previous lesson that for a function y = u + v, the Integration is $begin{array}{ccccc}& int y & = & int u, dx+int v, dxend{array}$

$begin{array}{ccccc}& int ydx & = & int(4x^{3}+2x^{2}+5x+1)dx+Cvspace{2mm}& & = & int4x^{3}dx+int2x^{2}dx+int5xdx+int1dx+Cvspace{2mm}& & = & 4dfrac{x^{4}}{4}+2dfrac{x^{3}}{3}+5dfrac{x^{2}}{2}+x+Cvspace{2mm}& int ydx & = & x^{4}+dfrac{2}{3}x^{3}+dfrac{5}{2}x^{2}+x+Cend{array}$

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