A) \[e-\frac{{{e}^{2}}}{2}-\frac{5}{2}\]
B) \[e+\frac{{{e}^{2}}}{2}-\frac{3}{2}\]
C) \[-\frac{{{e}^{2}}}{2}-\frac{3}{2}\]
D) \[e+\frac{{{e}^{2}}}{2}+\frac{5}{2}\]
Correct Answer: C
Solution :
Given, \[f'(x)=f(x)\] and \[f(0)=1\] Then, let \[f(x)={{e}^{x}}\] ?.. (i) Also, \[f(x)+g(x)={{x}^{2}}\] \[g(x)={{x}^{2}}-{{e}^{x}}\] ... (ii) Now, \[\int_{0}^{1}{f(x)g(x)dx=\int_{0}^{1}{{{e}^{x}}({{x}^{2}}-{{e}^{x}})dx}}\] [from Eqs. (i) and (ii)] \[=\int_{0}^{1}{({{x}^{2}}{{e}^{x}}-{{e}^{2x}})dx}\] \[=\int_{0}^{1}{{{x}^{2}}{{e}^{x}}dx-\int_{0}^{1}{{{e}^{2x}}dx}}\] \[=[{{x}^{2}}{{e}^{x}}-\int{2x{{e}^{x}}dx]_{0}^{1}-\frac{1}{2}[{{e}^{2x}}]_{0}^{1}}\] \[=[{{x}^{2}}{{e}^{x}}-2\,(x{{e}^{x}}-{{e}^{x}})]_{0}^{1}-\frac{1}{2}[{{e}^{2}}-{{e}^{o}})\] \[=[{{x}^{2}}{{e}^{x}}-2\,x{{e}^{x}}+2{{e}^{x}})]_{0}^{1}-\frac{1}{2}({{e}^{2}}1)\] \[=[({{x}^{2}}-2x+2){{e}^{x}}]_{0}^{1}-\frac{1}{2}{{e}^{2}}+\frac{1}{2}\] \[=[(1-2+2){{e}^{1}}-(0-0+2){{e}^{0}}]-\frac{1}{2}{{e}^{2}}+\frac{1}{2}\] \[=(e-2)-\frac{1}{2}{{e}^{2}}+\frac{1}{2}=e-\frac{1}{2}{{e}^{2}}-\frac{3}{2}\]
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