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Câu hỏi: Cho hàm số $f(x)=\left\{ \begin{aligned}
& x+1\text{ khi x}\ge \text{0} \\
& {{e}^{2\text{x}}}\text{ khi x}\le \text{0} \\
\end{aligned} \right. $. Tích phân $ I=\int\limits_{-1}^{2}{f(x)d\text{x}}$ có giá trị bằng bao nhiêu?
A. $I=\dfrac{7{{\text{e}}^{2}}+1}{2{{\text{e}}^{2}}}$
B. $I=\dfrac{\text{11}{{\text{e}}^{2}}-11}{2{{\text{e}}^{2}}}$
C. $I=\dfrac{\text{3}{{\text{e}}^{2}}-1}{{{\text{e}}^{2}}}$
D. $I=\dfrac{\text{9}{{\text{e}}^{2}}-1}{2{{\text{e}}^{2}}}$
& x+1\text{ khi x}\ge \text{0} \\
& {{e}^{2\text{x}}}\text{ khi x}\le \text{0} \\
\end{aligned} \right. $. Tích phân $ I=\int\limits_{-1}^{2}{f(x)d\text{x}}$ có giá trị bằng bao nhiêu?
A. $I=\dfrac{7{{\text{e}}^{2}}+1}{2{{\text{e}}^{2}}}$
B. $I=\dfrac{\text{11}{{\text{e}}^{2}}-11}{2{{\text{e}}^{2}}}$
C. $I=\dfrac{\text{3}{{\text{e}}^{2}}-1}{{{\text{e}}^{2}}}$
D. $I=\dfrac{\text{9}{{\text{e}}^{2}}-1}{2{{\text{e}}^{2}}}$
$I=\int\limits_{-1}^{2}{f(x)d\text{x}}=\int\limits_{-1}^{0}{f(x)d\text{x}}+\int\limits_{0}^{2}{f(x)d\text{x}}=\int\limits_{-1}^{0}{{{e}^{2\text{x}}}d\text{x}}+\int\limits_{0}^{2}{(x+1)d\text{x}}=\left. \dfrac{1}{2}{{e}^{2\text{x}}} \right|_{-1}^{0}+\left. \dfrac{{{(x+1)}^{2}}}{2} \right|_{0}^{2}=\dfrac{9{{\text{e}}^{2}}-1}{2{{\text{e}}^{2}}}$.
Đáp án D.