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Câu hỏi: Cho hàm số $y=f\left( x \right)$ biết $f\left( 0 \right)=\dfrac{1}{2}$ và $f'\left( x \right)=x.{{e}^{{{x}^{2}}}}$ với mọi $x\in \mathbb{R}$. Khi đó tích phân $I=\int\limits_{0}^{1}{xf\left( x \right)\text{d}}x$ bằng
A. $\dfrac{e+1}{4}$.
B. $\dfrac{e-1}{4}$.
C. $\dfrac{e-1}{2}$.
D. $\dfrac{e+1}{2}$.
A. $\dfrac{e+1}{4}$.
B. $\dfrac{e-1}{4}$.
C. $\dfrac{e-1}{2}$.
D. $\dfrac{e+1}{2}$.
Ta có $\int{x.{{e}^{{{x}^{2}}}}\text{d}}x=\dfrac{1}{2}\int{{{e}^{{{x}^{2}}}}\text{d}\left( {{x}^{2}} \right)}=\dfrac{1}{2}{{e}^{{{x}^{2}}}}+C$
Mặt khác $f\left( 0 \right)=\dfrac{1}{2}\Rightarrow C=0$
Do đó $f\left( x \right)=\dfrac{1}{2}{{e}^{{{x}^{2}}}}$.
$I=\int\limits_{0}^{1}{xf\left( x \right)\text{d}}x=\dfrac{1}{2}\int\limits_{0}^{1}{x.{{e}^{{{x}^{2}}}}\text{d}}x=\dfrac{1}{4}\int\limits_{0}^{1}{{{e}^{{{x}^{2}}}}\text{d}\left( {{x}^{2}} \right)}=\left. \dfrac{1}{4}{{e}^{{{x}^{2}}}} \right|_{0}^{1}=\dfrac{e-1}{4}$.
Mặt khác $f\left( 0 \right)=\dfrac{1}{2}\Rightarrow C=0$
Do đó $f\left( x \right)=\dfrac{1}{2}{{e}^{{{x}^{2}}}}$.
$I=\int\limits_{0}^{1}{xf\left( x \right)\text{d}}x=\dfrac{1}{2}\int\limits_{0}^{1}{x.{{e}^{{{x}^{2}}}}\text{d}}x=\dfrac{1}{4}\int\limits_{0}^{1}{{{e}^{{{x}^{2}}}}\text{d}\left( {{x}^{2}} \right)}=\left. \dfrac{1}{4}{{e}^{{{x}^{2}}}} \right|_{0}^{1}=\dfrac{e-1}{4}$.
Đáp án B.