Cho F ( x ) là một nguyên hàm của f ( x ) = 3^x + x + 1 . Biết F ( 0 ) = 1 . Khi đó, F ( − 1 ) bằng:

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Cho \(F\left( x \right)\) là một nguyên hàm của \[f\left( x \right) = {3^x} + x + 1\]. Biết \(F\left( 0 \right) = 1\). Khi đó,\(F\left( { - 1} \right)\)bằng:

\(F\left( { - 1} \right) = \frac{1}{{2\ln 3}}\).

\(F\left( { - 1} \right) = \frac{1}{2} - \frac{2}{{3\ln 3}}\).

\(F\left( { - 1} \right) = 1 + \frac{1}{{3\ln 3}}\).

\(F\left( { - 1} \right) = \frac{1}{2} - \frac{1}{{\ln 3}}\).

Giải thích

\[F\left( x \right) = \int {\left( {{3^x} + x + 1} \right){\rm{d}}x} = \frac{1}{{\ln 3}} \cdot {3^x} + \frac{1}{2}{x^2} + x + C.\]

\[ \Rightarrow F\left( x \right) = \frac{1}{{\ln 3}} \cdot {3^x} + \frac{1}{2}{x^2} + x + C.\]

\(F\left( 0 \right) = 1 \Rightarrow 1 = \frac{1}{{\ln 3}}{.3^0} + \frac{1}{2}{.0^2} + 0 + C \Rightarrow C = 1 - \frac{1}{{\ln 3}}\)

\[ \Rightarrow F\left( x \right) = \frac{1}{{\ln 3}} \cdot {3^x} + \frac{1}{2}{x^2} + x + 1 - \frac{1}{{\ln 3}}\].

\( \Rightarrow F\left( { - 1} \right) = \frac{1}{{\ln 3}} \cdot {3^{ - 1}} + \frac{1}{2} - 1 + 1 - \frac{1}{{\ln 3}} \Rightarrow F\left( { - 1} \right) = \frac{1}{2} - \frac{2}{{3\ln 3}}\). Chọn B.