Let f(x) be a function satisfying f'(x) = f(x | vedclass

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(D) Given $f'(x) = f(x)$,we have $f(x) = Ce^x$. Since $f(0) = 1$,we get $C = 1$,so $f(x) = e^x$.Given $f(x) + g(x) = x^2$,we have $g(x) = x^2 - e^x$.W

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$e - \frac{1}{2}e^2 - \frac{3}{2}$

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(D) Given $f'(x) = f(x)$,we have $f(x) = Ce^x$. Since $f(0) = 1$,we get $C = 1$,so $f(x) = e^x$.Given $f(x) + g(x) = x^2$,we have $g(x) = x^2 - e^x$.We need to evaluate $I = \int_{0}^{1} f(x)g(x) \, dx = \int_{0}^{1} e^x(x^2 - e^x) \, dx = \int_{0}^{1} (x^2e^x - e^{2x}) \, dx$.Using integration by parts for $\int x^2e^x \, dx$:$\int x^2e^x \, dx = x^2e^x - \int 2xe^x \, dx = x^2e^x - 2(xe^x - e^x) = e^x(x^2 - 2x + 2)$.Evaluating the definite integral:$I = [e^x(x^2 - 2x + 2)]_{0}^{1} - [\frac{e^{2x}}{2}]_{0}^{1}$$I = (e(1 - 2 + 2) - e^0(0 - 0 + 2)) - \frac{1}{2}(e^2 - 1)$$I = (e - 2) - \frac{1}{2}e^2 + \frac{1}{2} = e - \frac{1}{2}e^2 - \frac{3}{2}$.

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