Cho F(x) là một nguyên hàm của f(x) = 2^x + x + 1. Biết F(0) = 1. Tính F(-1) kết quả là.

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Cho \(F\left( x \right)\) là một nguyên hàm của \[f\left( x \right) = {2^x} + x + 1\]. Biết \(F\left( 0 \right) = 1\). Tính \(F\left( { - 1} \right)\) kết quả là.

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

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

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

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

Giải thích

Chọn B

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

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

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

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

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