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Câu hỏi: Cho hàm số $f\left( x \right)$ thỏa mãn $f\left( 0 \right)=4$ và $f'\left( x \right)={{e}^{x}}+x,\forall x\in \mathbb{R}$. Khi đó $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{6e+13}{6}$
B. $\dfrac{6e+25}{6}$
C. $\dfrac{6e+25}{3}$
D. $\dfrac{6e+19}{6}$
A. $\dfrac{6e+13}{6}$
B. $\dfrac{6e+25}{6}$
C. $\dfrac{6e+25}{3}$
D. $\dfrac{6e+19}{6}$
Theo giả thiết $f'\left( x \right)={{e}^{x}}+x,\forall x\in \mathbb{R}$ nên:
$f\left( x \right)=\int\limits_{{}}^{{}}{f'\left( x \right)dx}=\int\limits_{{}}^{{}}{\left( {{e}^{x}}+x \right)dx}={{e}^{x}}+\dfrac{1}{2}{{x}^{2}}+C$
Mà $f\left( 0 \right)=4$ nên ${{e}^{0}}+\dfrac{1}{2}{{0}^{2}}+C=4\Leftrightarrow C=3$
Suy ra $f\left( x \right)={{e}^{x}}+\dfrac{1}{2}{{x}^{2}}+3$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( {{e}^{x}}+\dfrac{1}{2}{{x}^{2}}+3 \right)dx}=\dfrac{6e+13}{6}$
$f\left( x \right)=\int\limits_{{}}^{{}}{f'\left( x \right)dx}=\int\limits_{{}}^{{}}{\left( {{e}^{x}}+x \right)dx}={{e}^{x}}+\dfrac{1}{2}{{x}^{2}}+C$
Mà $f\left( 0 \right)=4$ nên ${{e}^{0}}+\dfrac{1}{2}{{0}^{2}}+C=4\Leftrightarrow C=3$
Suy ra $f\left( x \right)={{e}^{x}}+\dfrac{1}{2}{{x}^{2}}+3$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( {{e}^{x}}+\dfrac{1}{2}{{x}^{2}}+3 \right)dx}=\dfrac{6e+13}{6}$
Đáp án A.