View source for Math 22 Area and the Fundamental Theorem of Calculus

==Definition of a Definite Integral==
  Let <math>f</math> be nonnegative and continuous on the closed interval <math>[a,b]</math>. 
  The area of the region bounded by the graph of <math>f</math>, the x-axis, 
  and the lines <math>x=a</math> and <math>x=b</math> is denoted by
  <math>\text{Area}=\int_a^b f(x)dx</math>
  
  The expression <math>\int_a^b f(x)dx</math> is called the definite integral from a to b, 
  where a is the lower limit of integration and b is the upper limit of integration.
==The Fundamental Theorem of Calculus==
[[File:Fundametal_theorem_of_calculus.png]]
  If <math>f</math> is nonnegative and continuous on the closed interval [a,b], then
  
  <math>\int_a^b f(x)dx=F(b)-F(a)</math>
  
  where <math>F(x)</math> is any function such that <math>F'(x)=f(x)</math> for all <math>x</math> in [a,b]

Notation <math>\int_a^b  f(x)dx=F(x)\Biggr |_a^b =F(b)-F(a)</math>

==Properties of Definite Integrals==
  Let <math>f</math> and g be continuous on the closed interval [a,b].

  1.<math>\int_a^b k f(x)dx=f\int_a^b f(x)dx</math> for <math>k</math> is constant.
  
  2.<math>\int_a^b [f(x)\pm g(x)]dx=\int_a^b f(x)dx \pm \int_a^b g(x)dx</math>
  
  3.<math>\int_a^b f(x)d=\int_a^c f(x)dx+\int_c^b f(x)dx</math> for <math>a<c<b</math>
  
  4.<math>\int_a^a f(x)dx=0</math>
  
  5.<math>\int_a^b f(x)dx=-\int_b^a f(x)dx</math>

'''Exercises''' 

'''1)''' Find the area of the region bounded by the x-axis and the graph of <math>y=5x</math> when <math>0\le x\le 3</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|<math>\int_0^3 5xdx=\int e^x=[\frac{5}{2}x^2]\Biggr |_0^3=[\frac{5}{2}(3)^2-\frac{5}{2}(0)^2]=\frac{45}{2}</math>
|}

'''2)''' Evaluate <math>\int_1^2 \frac{1}{x^2}dx</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|Let <math>\int_1^2 \frac{1}{x^2}dx=\int_1^2 x^{-2}=[\frac{x^{-2+1}}{-2+1}]\Biggr|_1^2=[x^{-2}]\Biggr|_1^2=[\frac{-1}{x}]\Biggr|_1^2=[\frac{-1}{2}-\frac{-1}{1}]=\frac{1}{2}</math>
|}



[[Math_22| '''Return to Topics Page''']]
'''This page were made by [[Contributors|Tri Phan]]'''