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(C) Given $f^{\prime}(x)=f(x)$.Dividing by $f(x)$,we get $\frac{f^{\prime}(x)}{f(x)}=1$.Integrating both sides with respect to $x$,we get $\ln|f(x)| =

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$e-\frac{e^2}{2}-\frac{3}{2}$

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(C) Given $f^{\prime}(x)=f(x)$.Dividing by $f(x)$,we get $\frac{f^{\prime}(x)}{f(x)}=1$.Integrating both sides with respect to $x$,we get $\ln|f(x)| = x + C$.Since $f(0)=1$,we have $\ln(1) = 0 + C$,which implies $C=0$.Thus,$f(x) = e^x$.Given $f(x)+g(x)=x^2$,we have $g(x) = x^2 - e^x$.Now,we need to evaluate the integral $I = \int_0^1 f(x)g(x) dx = \int_0^1 e^x(x^2 - e^x) dx$.$I = \int_0^1 (x^2 e^x - e^{2x}) dx = \int_0^1 x^2 e^x dx - \int_0^1 e^{2x} dx$.Using integration by parts for $\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(x e^x - e^x) = e^x(x^2 - 2x + 2)$.Evaluating the definite integral: $[e^x(x^2 - 2x + 2)]_0^1 = (e(1-2+2)) - (e^0(0-0+2)) = e - 2$.Evaluating the second part: $\int_0^1 e^{2x} dx = [\frac{1}{2} e^{2x}]_0^1 = \frac{1}{2}(e^2 - 1)$.Therefore,$I = (e - 2) - \frac{1}{2}(e^2 - 1) = e - 2 - \frac{e^2}{2} + \frac{1}{2} = e - \frac{e^2}{2} - \frac{3}{2}$.

If $f(x)$ is a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ is a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$ is

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