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Câu hỏi: Tìm nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)={{e}^{2x}}$, biết $F\left( 0 \right)=1$.
A. $F\left( x \right)={{e}^{2x}}$.
B. $F\left( x \right)=\dfrac{{{e}^{2x}}}{2}+\dfrac{1}{2}$.
C. $F\left( x \right)=2{{e}^{2x}}-1$.
D. $F\left( x \right)={{e}^{x}}$.
A. $F\left( x \right)={{e}^{2x}}$.
B. $F\left( x \right)=\dfrac{{{e}^{2x}}}{2}+\dfrac{1}{2}$.
C. $F\left( x \right)=2{{e}^{2x}}-1$.
D. $F\left( x \right)={{e}^{x}}$.
Ta có $\left\{ \begin{aligned}
& F\left( x \right)=\int{{{e}^{2x}}dx}=\dfrac{1}{2}{{e}^{2x}}+C \\
& F\left( 0 \right)=1\Rightarrow C=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow F\left( x \right)=\dfrac{1}{2}{{e}^{2x}}+\dfrac{1}{2}$.
& F\left( x \right)=\int{{{e}^{2x}}dx}=\dfrac{1}{2}{{e}^{2x}}+C \\
& F\left( 0 \right)=1\Rightarrow C=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow F\left( x \right)=\dfrac{1}{2}{{e}^{2x}}+\dfrac{1}{2}$.
Đáp án B.