# 95548117-MATH22-Module-3 - APPLICATIONS Area by Integration...

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APPLICATIONS Area by Integration DISCUSSION OF THE TOPICS Geometric Interpretation of Area by Integration. Consider ) x ( f y = The area will be divided into n th rectangles with equal width. Such that, Let, S = area of 1 rectangle A = summation of the area of n-rectangles Where, ) x x ( y S a 1 1 1 - = ) x x ( y S 1 2 2 2 - = ) x x ( y S 2 3 3 3 - = …….. ) x x ( y S 1 b b b b - - = b 3 2 1 S , ..... . S S S A + + + + = ) x x ( y ..... ) x x ( y ) x x ( y ) x x ( y A 1 b b b 2 3 3 1 2 2 a 1 1 - - + + - + - + - = Since there widths are equal, then: 1 b b 2 3 1 2 a 1 x x ....... x x x x x x x - - = = - = - = - = ) x ( y ..... ) x ( y ) x ( y ) x ( y A b 3 2 1 + + + + = O x = a x = b y = f(x) x y y 1 y 2 y 3 y b x 1 x 2 x 3 x b-1
x ) y ..... y y y ( A b 3 2 1 + + + + = , if ) x ( f y = [ ] x ) x ( f ..... ) x ( f ) x ( f ) x ( f ( A b 3 2 1 + + + + = = = n 1 i i x ) x ( f A Apply the limits on both sides as 0 x : = = n 1 i i 0 x 0 x x ) x ( f lim lim Since limit of a constant is equal to the constant itself, if A is a constant, then: dx ) x ( f A b a = [ ] [ ] [ ] C ) a ( F C ) b ( F C ) x ( F A b a + - + = + = ) a ( F ) b ( F A - = ) a ( F ) b ( F - is the definite integral of f(x) if f(x) is a continuous function at the closed interval from a to b . That is at the point [ ] b , a , f(x) is defined. Properties of the Definite Integral 1. If a > b, then - = b a a b dx ) x ( f dx ) x ( f provided f(x) is defined at the point [ ] b , a 2. If a = b, then = = - = a a b a b b 0 dx ) x ( f dx ) x ( f dx ) x ( f provided f(a) or f(b) exists 3. [ ] ± b a dx ) x ( g ) x ( f ± = b a b a dx ) x ( g dx ) x ( f ] b a ) x ( G ) x ( F ± = [ ] [ ] ) a ( G ) a ( F ) b ( G ) b ( F ± - ± =
4. If f(x) is a continuous function at the point [ ] b , a , where a < c < b, then: + = b c c a b a dx ) x ( f dx ) x ( f dx ) x ( f Suggested Steps to Determine the Area of a Plane Figure by Integration: 1. Plot the given function, and other given data. 2. Shade the area to be determined. 3. Get a representative strip (in the form of a very thin rectangle) anywhere along the region using the stripping. (Either horizontal or vertical) 4. Find the equation of the area using the formula of a rectangle with the strip as the basis for computation. 5. Apply the integral using the extreme points as the limit of integration to determine the total area. 6. Use properties of the definite integral to get the actual area. Solution 1: Using Vertical Stripping O x y (0,4) (2,0) (x,y) y = 4 - 2x L = y w = dx
Solution 2: Using Horizontal Stripping Example 1: Find the area of the region bounded by the line x 2 4 y - = and the coordinate axes. Graphing the given, we have: Line: x 2 4 y - = O x y (0,4) (2,0) (x,y) y = 4 - 2x L = x w = dy O x y (0,4) (2,0) (x,y) y = 4 - 2x L = y w = dx
a. Using Vertical Stripping: The rectangular strip has a partial area: = ydx dA = ydx A , since x 2 4 y - = = = - = 2 x 0 x dx ) x 2 4 ( A - - = 2 0 dx ) x 2 ( 2 A 2 0 2 2 ) x 2 ( 2 A - - = [ ] 2 2 ) 2 0 ( ) 2 2 ( A - - - - = ) 4 ( A - - = 4 A = sq. units b. Using Horizontal Stripping: O x y (0,4) (2,0) (x,y) y = 4 - 2x L = x w = dy
The rectangular strip has a partial area: = xdy dA = xdy A , since ) y 4 ( 2 1 x - = = = - = 4 y 0 y dy ) y 4 ( A - - = 2 0 dy ) y 4 ( 2 1 A 4 0 2 2 ) y 4 ( 2 1 A - - = ( 29 ( 29 [ ] 2 2 0 4 4 4 4 1 A - - - - = [ ] 16 4 1 A - - = 4 A = sq. units CHECK: The figure formed is a right triangle.
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