Giải thích các bước giải:
Câu 29:
f(x) là hàm số chẵn nên \(f\left( x \right) = f\left( { - x} \right)\)
Ta có:
\(\begin{array}{*{20}{l}}
{I = \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{f\left( x \right)}}{{{{2018}^x} + 1}}dx}\\
{t = {\rm{}} - x \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{dt = {\rm{}} - dx}\\
{x = {\rm{}} - \pi {\rm{}} \Rightarrow t = \pi }\\
{x = \pi {\rm{}} \Rightarrow t = {\rm{}} - \pi {\rm{}}}
\end{array}} \right.}\\
{ \Rightarrow I = \int\limits_\pi ^{ - \pi } {\dfrac{{f\left( { - t} \right)}}{{{{2018}^{ - t}} + 1}}\left( { - dt} \right)} }\\
{ = \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{f\left( { - t} \right)}}{{{{2018}^{ - t}} + 1}}dt}\\
{ = \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{f\left( { - x} \right)}}{{{{2018}^{ - x}} + 1}}dx}\\
{ = \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{f\left( x \right)}}{{\dfrac{1}{{{{2018}^x}}} + 1}}dx}\\
{ = \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{{{2018}^x}f\left( x \right)}}{{1 + {{2018}^x}}}dx}\\
{{\rm{}} \Rightarrow 2I = \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{f\left( x \right)}}{{{{2018}^x} + 1}}dx{\rm{}} + \int\limits_{ - \pi }^\pi {\rm{}} \dfrac{{{{2018}^x}f\left( x \right)}}{{{{2018}^x} + 1}}dx{\rm{}} = \int\limits_{ - \pi }^\pi {\rm{}} f\left( x \right)dx{\rm{}}}\\
{\int\limits_{ - \pi }^0 {f\left( x \right)dx} = \int\limits_\pi ^0 {f\left( { - x} \right).\left( { - dx} \right)} = - \int\limits_\pi ^0 {f\left( x \right)dx} = \int\limits_0^\pi {f\left( x \right)dx} }
\end{array}\)
\(\begin{array}{l}
\Rightarrow \int\limits_{ - \pi }^\pi {f\left( x \right)dx} = \int\limits_{ - \pi }^0 {f\left( x \right)dx} + \int\limits_0^\pi {f\left( x \right)dx} = 2\int\limits_0^\pi {f\left( x \right)dx} \\
\Rightarrow I = \int\limits_0^\pi {f\left( x \right)dx} = 2018
\end{array}\)
Câu 28:
f(x) là hàm số chẵn nên \(f\left( x \right) = f\left( { - x} \right)\)
Ta có:
\(\begin{array}{l}
t = - x \Rightarrow \left\{ \begin{array}{l}
dt = \left( { - x} \right)'.dx = - dx\\
x = - 1 \Rightarrow t = 1\\
x = 1 \Rightarrow t = - 1
\end{array} \right.\\
I = \int\limits_{ - 1}^1 {\dfrac{{f\left( x \right)}}{{1 + {e^x}}}dx} = \int\limits_1^{ - 1} {\dfrac{{f\left( { - t} \right)}}{{1 + {e^{ - t}}}}.\left( { - dt} \right)} = \int\limits_{ - 1}^1 {\dfrac{{f\left( { - x} \right)}}{{1 + {e^{ - x}}}}dx} = \int\limits_{ - 1}^1 {\dfrac{{f\left( x \right)}}{{1 + {e^{ - x}}}}dx} \\
2I = \int\limits_{ - 1}^1 {\dfrac{{f\left( x \right)}}{{1 + {e^x}}}dx} + \int\limits_{ - 1}^1 {\dfrac{{f\left( x \right)}}{{1 + {e^{ - x}}}}dx} \\
= \int\limits_{ - 1}^1 {f\left( x \right).\left( {\dfrac{1}{{1 + {e^x}}} + \dfrac{1}{{1 + {e^{ - x}}}}} \right)dx} \\
= \int\limits_{ - 1}^1 {f\left( x \right).\left( {\dfrac{1}{{1 + {e^x}}} + \dfrac{{{e^x}}}{{{e^x} + 1}}} \right)dx} \\
= \int\limits_{ - 1}^1 {f\left( x \right).1.dx} = \int\limits_{ - 1}^1 {f\left( x \right)dx} = 4\\
\Rightarrow I = 2
\end{array}\)
