. A new treatise on the elements of the differential and integral calculus . •/ ( q)(x)dx = ^^ ^ , n and, since this is to hold for all values of a, we may differen-tiate with respect to a: hence , . (p(a) . (p(a)^^ ^ n n cpia) __n—l ^(p{a) a and by integration lq){a)=z(n—l)la+ from logarithms to numbers, cp{a)=: Ca-^: ., cp{x) = Cx-^; and the equation of the curve is y z= Cx^~^. Determine such a form for q)(x) that the integral w = / -~7^ r shall be independent of a. J 0 \/{a — x) Put x^=az ; then, since the limits a; =: 0, ic = a, correspondto ^ = 0, 2 == 1, cp(x)dx r*^ ^aq)(a
. A new treatise on the elements of the differential and integral calculus . •/ ( q)(x)dx = ^^ ^ , n and, since this is to hold for all values of a, we may differen-tiate with respect to a: hence , . (p(a) . (p(a)^^ ^ n n cpia) __n—l ^(p{a) a and by integration lq){a)=z(n—l)la+ from logarithms to numbers, cp{a)=: Ca-^: ., cp{x) = Cx-^; and the equation of the curve is y z= Cx^~^. Determine such a form for q)(x) that the integral w = / -~7^ r shall be independent of a. J 0 \/{a — x) Put x^=az ; then, since the limits a; =: 0, ic = a, correspondto ^ = 0, 2 == 1, cp(x)dx r*^ ^aq)(az)dz /^ cp[X)ax r* o \/(a — x)~J ^{a — x) J, A^{l—z^ INTEGRATION UNDER THE SIGN f. 451 By condition, u is to be independent of a: therefore the dif-ferential co-efficient of u, with respect to a, must be zero. But 1 duda cf (az) , / // N Wa + ^f ^^^ _ .« cp{x) + 2xcf\x) . V(i-^) ,\/(a — x) and, since this last integral is to be zero for all values of a,we must have q,(x) __ _ 1 q){x) -\- 2xq)^(x) z= 0 : (p{x) Therefore or l(p(^x) = —-Ix + 0,0 q)(x)^ \/x. Let A OB be a cycloid, with its vertex downwards ; and let itbe referred to the axisOx, and the tangentthrough its vertex, asco-ordinate axes. Pi Then, denoting theangle DCF by 0, we V f iT have for the co-ordinates 0F= y, OQ = x of the point P,X = FF = HL = a -f a cos.^^,y = OF-^ All - AE = AR - AD ^ FD=z ait — aO -\- a ^ 0 =z7t — (f, thou those values of .r and // becomeX = a ~a cos.(jp (\), y -— (uy -j- a (2).From (I) wo hndI (jT COS. X . 1 -, sni.(]r — I -ax — .r 452 INTEGRAL CALCULUS, and thus (2) becomes y=aco^r^- \- hax — x^, which is the equation of the cycloid. By differentiation, weget dx~~y a di^y ^\dx) ~y X and, by integration, s = \^Sax. We therefore conclude thatq){x), in Ex. 2, is the expression for the arc of a cycloid esti-mated from the vertex. This example is the solution of the problem in mechanicsfor finding the curve down which bodies, starting from dif-ferent p
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