ધારો કે x in R માટે,S_0(x) = x,S_k(x) = C_k x | vedclass

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(C) આપેલ છે કે $S_0(x) = x$ અને $C_0 = 1$.$k=1$ માટે: $C_1 = 1 - \int_0^1 S_0(x) dx = 1 - \int_0^1 x dx = 1 - \frac{1}{2} = \frac{1}{2}$.$S_1(x) = C_1

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$18$

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(C) આપેલ છે કે $S_0(x) = x$ અને $C_0 = 1$.$k=1$ માટે: $C_1 = 1 - \int_0^1 S_0(x) dx = 1 - \int_0^1 x dx = 1 - \frac{1}{2} = \frac{1}{2}$.$S_1(x) = C_1 x + 1 \int_0^x S_0(t) dt = \frac{1}{2}x + \int_0^x t dt = \frac{1}{2}x + \frac{x^2}{2}$.$k=2$ માટે: $C_2 = 1 - \int_0^1 S_1(x) dx = 1 - \int_0^1 (\frac{1}{2}x + \frac{x^2}{2}) dx = 1 - [\frac{x^2}{4} + \frac{x^3}{6}]_0^1 = 1 - (\frac{1}{4} + \frac{1}{6}) = 1 - \frac{5}{12} = \frac{7}{12}$.$S_2(x) = C_2 x + 2 \int_0^x S_1(t) dt = \frac{7}{12}x + 2 \int_0^x (\frac{1}{2}t + \frac{t^2}{2}) dt = \frac{7}{12}x + 2 [\frac{t^2}{4} + \frac{t^3}{6}]_0^x = \frac{7}{12}x + \frac{x^2}{2} + \frac{x^3}{3}$.$k=3$ માટે: $C_3 = 1 - \int_0^1 S_2(x) dx = 1 - \int_0^1 (\frac{7}{12}x + \frac{x^2}{2} + \frac{x^3}{3}) dx = 1 - [\frac{7x^2}{24} + \frac{x^3}{6} + \frac{x^4}{12}]_0^1 = 1 - (\frac{7}{24} + \frac{1}{6} + \frac{1}{12}) = 1 - \frac{7+4+2}{24} = 1 - \frac{13}{24} = \frac{11}{24}$.હવે,$S_2(3) = \frac{7}{12}(3) + \frac{3^2}{2} + \frac{3^3}{3} = \frac{7}{4} + \frac{9}{2} + 9 = \frac{7+18+36}{4} = \frac{61}{4}$.અંતે,$S_2(3) + 6C_3 = \frac{61}{4} + 6(\frac{11}{24}) = \frac{61}{4} + \frac{11}{4} = \frac{72}{4} = 18$.

ધારો કે $x \in R$ માટે,$S_0(x) = x$,$S_k(x) = C_k x + k \int_0^x S_{k-1}(t) dt$,જ્યાં $C_0 = 1$,$C_k = 1 - \int_0^1 S_{k-1}(x) dx$,$k = 1, 2, 3, \ldots$. તો $S_2(3) + 6C_3$ ની કિંમત $...........$ થાય.

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