Definite Integrals

February 6, 2012 in Additional Mathematics, Definite Integrals, Integration by Shivakkumar Vadiveyl

Section E Definite Integrals

So far we have seen the Indefinite Integration whereby the equation of curve is derived for a continous function of each value of x and y. In this section, we will be learning the Definite Integral in which the exact value of the Integration will be obtained.

Firstly, let us take a look at the formulas for the Definite Integrals.

The most basic Definite Integral is written in this form, $begin{array}{ccc}& int_{a}^{b}f(x)dxend{array}$ where a is called the lower limit, b is called the upper limit. Let us say that the result of Integration of f(x) is F(x). The formula can then be expanded as follows.

$begin{array}{ccccc}& int_{a}^{b}f(x)dx & = & [F(x)]_{a}^{b}& & = & F(b)-F(a)end{array}$

Definite Integral Example

Find the solution to $int_{1}^{2}(x+5)dx$

First, we integrate for $int(2x+5)dx$ ,

$begin{array}{ccccc}& y & = & int(x+5)dx& & = & dfrac{x^{2}}{2}+5x+Cend{array}$

$begin{array}{ccccc}& int_{1}^{2}(x+5)dx & = & left[dfrac{x^{2}}{2}+5xright]_{1}^{2}vspace{2mm}& & = & left[dfrac{(2)^{2}}{2}+5(2)right]-left[dfrac{1^{2}}{2}+5(1)right]vspace{2mm}& & = & left[dfrac{4}{2}+10-dfrac{1}{2}-5right]vspace{2mm}& & = & dfrac{3}{2}+5vspace{2mm}& int_{1}^{2}(x+5)dx & = & 6dfrac{1}{2}end{array}$

There are a few important properties of Definite Integral as listed here.

$begin{array}{ccccc}& int_{a}^{a}f(x)dx & = & 0vspace{2mm}& int_{a}^{b}kf(x)dx & = & kint_{a}^{b}f(x)dxvspace{2mm}& int_{a}^{b}f(x)dx & = & -int_{b}^{a}f(x)dxvspace{2mm}& int_{a}^{b}(f(x)+g(x))dx & = & int_{a}^{b}f(x)dx+int_{a}^{b}g(x)dxvspace{2mm}& int_{a}^{c}f(x)dx & = & int_{a}^{b}f(x)dx+int_{b}^{c}f(x)dxend{array}$

Definite Integrals Exercise E

1) Show that $int_{a}^{a}f(x)dx=0$
Let’s write the integration results of f(x) as F(x).

$begin{array}{ccccc}& int_{a}^{a}f(x)dx & = & left[F(x)right]_{a}^{a}& & = & left[F(a)right]-left[F(a)right]& int_{a}^{a}f(x)dx & = & 0end{array}$

2) Find $int_{1}^{2}2(x+1)dx$
First, we integrate for $2(x+1)$,

$begin{array}{ccccc}& y & = & int(x+1)dx& & = & dfrac{x^{2}}{2}+x+Cend{array}$

$begin{array}{ccccc}& int_{1}^{2}2(x+1)dx & = & left[dfrac{x^{2}}{2}+xright]_{1}^{2}vspace{2mm}& & = & 2left[dfrac{2^{2}}{2}+2right]-2left[dfrac{1^{2}}{2}+1right]vspace{2mm}& & = & 8-1-2vspace{2mm}& int_{1}^{2}2(x+1)dx & = & 5end{array}$