. Differential and integral calculus, an introductory course for colleges and engineering schools. nite integral, dx we shall have ydx=A -B + C,notA + B+C. f U a f %J a The process of finding the area undera curve is termed quadrature.^ Example 1. To find the area boundedby the cubical parabola y = x*, the z-axis,and the ordinates whose abscissas are aand b. * On a later page it will be shown why this definite integral is negative wheny is negative. f The term quadrature is also made to include any integration withoutreference to the geometrical interpretation of the process. Thus, the integra
. Differential and integral calculus, an introductory course for colleges and engineering schools. nite integral, dx we shall have ydx=A -B + C,notA + B+C. f U a f %J a The process of finding the area undera curve is termed quadrature.^ Example 1. To find the area boundedby the cubical parabola y = x*, the z-axis,and the ordinates whose abscissas are aand b. * On a later page it will be shown why this definite integral is negative wheny is negative. f The term quadrature is also made to include any integration withoutreference to the geometrical interpretation of the process. Thus, the integra-tion of J f(x) dx is termed a quadrature. 228 INTEGRAL CALCULUS 160 By formula (3) we have U = I ydx = ( x3dx = PJM Ja Ja L4Jo If a = 0, we have area OAS6 = *7 = b4 ■jH-v b4 C~° r~x?~\ ~ ° Again, U = | x? dx =\- \ J -a L4J-a * and this area is negative because b4 < a4. Example 2. To find the area bounded by the upper branch of theparabola y2 = 2 mx, the z-axis, and the ordinate whose abscissa is b. In this case the lower limit is 0. Byformula (3) we have U = £ydx = V2mfo a*dx 9 k a/9^. 3 *- _„ 2 6 V2m6 If c be the ordinate of B, then c = v2 mb, and therefore t/ = f 6c. Now6c = area of rectangle OMBC. Therefore U = f OMBC. From thesymmetry of the figure it follows that Parabolic segment SOS7 =%BMMB\ or, the area of the segment of the parabola cut off by a double ordinateis two-thirds the area of the circumscribing rectangle. Example 3. Let us find the area of the ellipse t. _l £. = 1a2 ^ 62 It will be simpler to find first the area of one quadrant. In this case y = -V a2 — x2. Then by formula (3) U= Caydx = - fV-z2)*J0 a Jn b dx b r /- j , 2 . .an- 2 a L a J 0 = —sm 11 =2 4 ab /- §160 AREAS 229 (We might of course integrate from—a to -\-a and so get half the areaof the E.)Or we may proceed as follows: Let z = asin0; then Va2 — x2 = a cos 0, dx = a cos 0 for the limits of integration, when x = 0, 0 = 0, and when x = a, sin 0 = lt 0 = - • Consequently6
Size: 1787px × 1399px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1912
