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APPLICATIONS Area by IntegrationDISCUSSION OF THE TOPICSGeometric Interpretation of Area by Integration.Consider )x(fy=The area will be divided into nthrectangles with equal width. Such that,Let,S=area of 1 rectangleA=summation of the area of n-rectanglesWhere,)xx(ySa111-=)xx(yS1222-=)xx(yS2333-=……..)xx(yS1bbbb--=b321S,......SSSA++++=)xx(y.....)xx(y)xx(y)xx(yA1bbb233122a11--++-+-+-=Since there widths are equal, then:1bb2312a1xx.......xxxxxxx--==-=-=-=∆)x(y.....)x(y)x(y)x(yAb321∆++∆+∆+∆=Ox = ax = by = f(x)xyy1y2y3ybx1x2x3xb-1
x)y.....yyy(Ab321∆++++=,if )x(fy=x)x(f.....)x(f)x(f)x(f(Ab321∆++++=∑=∆=n1iix)x(fAApply the limits on both sides as 0x→∆:∑=→∆→∆∆=n1ii0x0xx)x(flimlimSince limit of a constant is equal to the constant itself, if A is a constant, then:dx)x(fAba∫=C)a(FC)b(FC)x(FAba+-+=+=)a(F)b(FA-=)a(F)b(F-is the definite integral of f(x) if f(x) is a continuous function at the closed interval from ato b. That is at the pointb,a, f(x) is defined.Properties of the Definite Integral1.If a > b, then∫∫-=baabdx)x(fdx)x(fprovided f(x) is defined at the point b,a2.If a = b, then∫∫∫==-=aababb0dx)x(fdx)x(fdx)x(fprovided f(a) or f(b) exists3. ∫±badx)x(g)x(f∫∫±=babadx)x(gdx)x(f]ba)x(G)x(F±=)a(G)a(F)b(G)b(F±-±=
4.If f(x) is a continuous function at the pointb,a, where a < c < b, then:∫∫∫+=bccabadx)x(fdx)x(fdx)x(fSuggested 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 StrippingOxy(0,4)(2,0)(x,y)y = 4 - 2xL = yw = dx
Solution 2: Using Horizontal StrippingExample 1:Find the area of the region bounded by the line x24y-=and the coordinate axes.Graphing the given, we have:Line: x24y-=Oxy(0,4)(2,0)(x,y)y = 4 - 2xL = xw = dyOxy(0,4)(2,0)(x,y)y = 4 - 2xL = yw = dx
a.Using Vertical Stripping:The rectangular strip has a partial area:∫∫=ydxdA∫=ydxA,since x24y-=∫==-=2x0xdx)x24(A∫--=20dx)x2(2A2022)x2(2A-⋅-=22)20()22(A----=)4(A--=4A=sq. unitsb.Using Horizontal Stripping:Oxy(0,4)(2,0)(x,y)y = 4 - 2xL = xw = dy
The rectangular strip has a partial area:∫∫=xdydA∫=xdyA,since )y4(21x-=∫==-=4y0ydy)y4(A∫--=20dy)y4(21A4022)y4(21A-⋅-=(29(2922044441A----=1641A--=4A=sq. unitsCHECK: The figure formed is a right triangle.