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Câu hỏi: Tính tích phân $I=\int\limits_{0}^{2019}{{{\text{e}}^{2x}}\text{d}x}$.
A. $I={{\text{e}}^{4038}}-1$.
B. $I=\dfrac{1}{2}\left( {{\text{e}}^{4038}}-1 \right)$.
C. $I=\dfrac{1}{2}{{\text{e}}^{4038}}-1$.
D. $I={{\text{e}}^{4038}}$.
A. $I={{\text{e}}^{4038}}-1$.
B. $I=\dfrac{1}{2}\left( {{\text{e}}^{4038}}-1 \right)$.
C. $I=\dfrac{1}{2}{{\text{e}}^{4038}}-1$.
D. $I={{\text{e}}^{4038}}$.
$I=\int\limits_{0}^{2019}{{{\text{e}}^{2x}}\text{d}x}$ $=\dfrac{1}{2}\int\limits_{0}^{2019}{{{\text{e}}^{2x}}\text{d}\left( \text{2}x \right)}$ $=\dfrac{1}{2}\left. {{\text{e}}^{2x}} \right|_{0}^{2019}$ $=\dfrac{1}{2}\left( {{\text{e}}^{4038}}-1 \right)$.
Đáp án B.