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PLANE AREA The problem of obtaining the area bounded by a curve y-fix), the x-axis and two ordinates xarandx=b is one problem that led to the basic ideas and techniques which later on form the elementary calculus A simple method in solving this area is to construct set of rectangles as shown in figure Sa Using the formula length times width the area of the rectangles are computed. Adding these areas will approximately give us the required area. ¹y = f(x) If we let the width of the rectangles approach zero the number of rectangles will correspondingly become farge, then the sum of the areas of the rectangles will be equal to the area bounded by the curve and the lines. This is the definite integral of the function y=f(x) from a to b. It is also called Riemann Sum or Riemann integral in honor of Bernhard Riemann a German Mathematician 1826- 1866. The formula us to solve area bounded by any curves or lines by taking an element (rectangle) parallel to the coordinate axes. This will mean that either the element is vertical or horizontal Make a sketch of the area to be found. Draw an element vertical or horizontal, and then use any of the following formulas. For Vertical element _4-10₂-9₂dx xxx-3 For Horizontal element f(x,-x, )dy Example 1. Find the area bounded by y=2x-3, y=0.x-2)x=5 Solution 1. We use the vertical element A=f(x,-₁)dx a=2 bas y₁=2x-3 y₁=0 -0)dx= 4-[5¹-3(5)]-[2-3(2)]-25-15-4+6-12 CEMEREECE, College of Engine
Son no.2 Using the horizontal element y=2(2)-3=1 P(2.1) 4 = 1 y* 2x-3 Ya b=7 2x-3 4 (42-4)=(18)=9sq.units P(2.1) x-5 ing (SY 2008-2009) xaz X A₂ 42 = $(xx- 4₂ a₂ = 0 X₁ = 5 P(5,1) XR X+5 -x)dy 4₂ = f(5-2)dy-f3dy=[3y)-3(1)-3(0)=3 ^ - (-2³) - 1¹0-2-³-1²-²4-v-41--2] ---- A 9 sq units A A, A • 9+5=12 sq. units Solution No. 3 b₂ = 1 x₁ = 2 Engr. Myma Mama X Looking at the figure A is a trapezoid Area of trapezoid is equal to the product of one-half of the sum of a and b times the height. 4= (a + b)(h) = (1+7)(3)=(8)(3)=(4)(3) = 12 sq.unis
Example No.2 Solution 1. a=0 Find the area in the first quadrant bounded by y=x' and y=4x *888* A- jou- 46 y,)dx Equate yx and y-4x for the points of intersection. When x 2 P(2,8) When y=-8 -yax Solution 2. Horizontal Element 9 Vertical Element P(-2,-8) -Yox² b=2 LIMITS -x¹)x= [2x¹-x]-[2(2)-(2)]-101-8-4-4sq units XL Xa -yax³
Math 221 Example No. 3 Find the area between the parabola yx and y² =2-x. y² =2-x y² = -x+2 (y-0)²=-1(x-2) Parabola (0,0) 4a=1 2a Opening to the right -43)dy -[-]-[(8) -(8)]-[01-12-8-4squnits x=2-x 2x=2 x = 1 When A= V(2.0) 4a=1 a=-1 x = 2-y² 4-12-²-² s 2a- opening to the left For the points of intersection equate y² =2-x, y² = x x = 1 P,(1,1) y² = 1 y=±1 ao Solution 1. Using the horizontal element A = [(x₂-x₂) (10) b=1 2,0 Engr. Myma Mama P(1.-1)
Applying the theorem or even function A=2 (2–2y^²)dx = 4 [(1 – 3 −2y³)dx = 4 [(1-y¹³ )dx = 4[y-+y³] =4[1-+(¹1)' ]— 4[0] = 4(3) = ³, sq units Solution No.2 A₂ Looking at the figure 4, is identical to A A=4₂+4₂ A = 2 [(y₁ - y₂)dx 11 4,4 8 A==+= 3-3 3 a=0 b=1 4, = 2 [(x¹² - 0) dx = 2 [3x¹¹] =4[³¹]-[0] = 2√(x¹ 90 Solution 3. 4 is a parabola and identical to 4 [d Area of parabola = 4 = (2x1) = s sq units im 6,1) sq.units == sq.units 2.0. Engr. Myma Mame 4 = 4₂
Example No.4 Find the area bounded by y-In x, y=0andx=e SERBER x-1(x,-v₁dx a=1 b=e y₁ = ln x y₁ = 0 vja X x-0)dx = in xdx u = ln x dv=dx dx X du Example No.5 Find the area under one arch of the curve y=cos.x O %%%%%%% 1 0.9 0.5 0 -0.5-0.91-1 品 Tyscosx V=X 4-2 cos xax =2[sin x] -2[sin-sino-2 a=0 y₁ = cos.x -2(1)=2sq units A=2(x,-₁)d Engr Myma Mama *E A=1sq.unit y₁ = 0 4=[xln x] - 4=[xinx-x] A=(elne-e)-(1in1-1) y.Inx x=e