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

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(D) Given $f'(x) = f(x)$,we have $\frac{f'(x)}{f(x)} = 1$. Integrating both sides,we get $\ln|f(x)| = x + C$,which implies $f(x) = ce^x$. Since $f(0)

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(D) Given $f'(x) = f(x)$,we have $\frac{f'(x)}{f(x)} = 1$. Integrating both sides,we get $\ln|f(x)| = x + C$,which implies $f(x) = ce^x$. Since $f(0) = 1$,we have $1 = ce^0$,so $c = 1$. Thus,$f(x) = e^x$. Given $f(x) + g(x) = x^2$,we have $g(x) = x^2 - e^x$. Now,we calculate the integral: $\int_0^1 f(x)g(x) dx = \int_0^1 e^x(x^2 - e^x) dx = \int_0^1 (x^2 e^x - e^{2x}) dx$. Using integration by parts for $\int x^2 e^x dx$: $\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(x e^x - e^x) = e^x(x^2 - 2x + 2)$. Evaluating the definite integral: $\int_0^1 x^2 e^x dx = [e^x(x^2 - 2x + 2)]_0^1 = e(1 - 2 + 2) - e^0(0 - 0 + 2) = e - 2$. Evaluating $\int_0^1 e^{2x} dx$: $\int_0^1 e^{2x} dx = [\frac{1}{2} e^{2x}]_0^1 = \frac{1}{2}(e^2 - 1)$. Thus,the integral is $(e - 2) - \frac{1}{2}(e^2 - 1) = e - 2 - \frac{1}{2}e^2 + \frac{1}{2} = e - \frac{1}{2}e^2 - \frac{3}{2}$.

Column $I$	Column $II$
$(A) \int_{-1}^1 \frac{dx}{1+x^2}$	$(p) \frac{1}{2} \log \left(\frac{2}{3}\right)$
$(B) \int_0^1 \frac{dx}{\sqrt{1-x^2}}$	$(q) 2 \log \left(\frac{2}{3}\right)$
$(C) \int_2^3 \frac{dx}{1-x^2}$	$(r) \frac{\pi}{3}$
$(D) \int_1^2 \frac{dx}{x \sqrt{x^2-1}}$	$(s) \frac{\pi}{2}$

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be the function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_0^1 f(x)g(x) dx$ is equal to

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