Lời giải
Bấm: Math error !
Đặt \( t = -x \Rightarrow dx = -dt \)
\( \begin{cases} x = 2 & \Rightarrow t = -2 \\ x = -2 & \Rightarrow t = 2 \end{cases} \)
\(I = \int_{2}^{-2} \frac{t^{2018}}{1 + e^{-t}} (-dt) = \int_{-2}^{2} \frac{e^t . t^{2018}}{1 + e^t} \, dt\)
\( I = \int_{-2}^2 \frac{(e^t + 1 - 1).t^{2018}}{1 + e^t} \, dt = \int_{-2}^2 t^{2018} dt - I\)
\( \Rightarrow 2I = \frac{t^{2019}}{2019} \bigg|_{-2}^2 \Rightarrow I = \frac{2^{2019}}{2019} \)
\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{C}} \)
*Tổng quát: Nếu \( y = f(x) \) là hàm số chẵn liên tục trên đoạn\( [-a, a] \) thì:
\( \int_{-a}^a \frac{f(x)}{1 + a^x} \, dx = \frac{1}{2} \int_{-\alpha}^\alpha f(x) \, dx \)
page 82
Lời giải
• \( y = f(x) \), \( dy = \left(ax + \frac{b}{x^2}\right) dx \) \( \Rightarrow f'(x) = ax + \frac{b}{x^2} \)
\( f'(1) = 0 \Rightarrow a + b = 0 \)
• \( dy = \left(ax + \frac{b}{x^2}\right) dx \Rightarrow \int dy = \int \left(ax + \frac{b}{x^2}\right) dx \) \( \Rightarrow y = \frac{a x^2}{2} - \frac{b}{x} + c = f(x) \)
• \( f(1) = 4 \Leftrightarrow \frac{a}{2} - b + c = 4 \)
• \( f(-1) = 2 \Leftrightarrow \frac{a}{2} + b + c = 2 \)
Vậy \( \begin{cases}
a + b = 0 \\
\frac{a}{2} - b + c = 4 \\
\frac{a}{2} + b + c = 2
\end{cases} \Rightarrow
\begin{cases}
a = 1\\
b = -1\\
c = \frac{5}{2}
\end{cases}\)
\( f(x) = \frac{ x^2}{2} + \frac{1}{x} + \frac{5}{2} \Rightarrow f(-2) = 2 - \frac{1}{2} + \frac{5}{2} = 4 \)
\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)
page 83
Đặt \( t = \frac{\pi}{2} - \Rightarrow I = \frac{\pi}{4} \)
Đặt \( t = \frac{\pi}{2} - x \)
page 84
Đặt \( t = \frac{\pi}{4} - x \Rightarrow x = \frac{\pi}{4} - t \), \( dx = -dt \)
\( I = - \int_{\frac{\pi}{4}}^{0} \ln \left(1 + \tan \left(\frac{\pi}{4} - t\right)\right) dt \)
\( = \int_{0}^{\frac{\pi}{4}} \ln \left(1 + \frac{1 - \tan t}{1 + \tan t}\right) \, dt \)
\( = \int_{0}^{\frac{\pi}{4}} \ln \left(\frac{2}{1 + \tan t}\right) \, dt \)
\( = \int_{0}^{\frac{\pi}{4}} \ln 2 - I \)
\( \Rightarrow I = \frac{1}{2} \ln 2 \bigg|_0^{\frac{0}{\pi}} = \frac{\pi}{8} \ln 2 \)
page 85
Đặt \( t = \pi - x \)
\( I = \int_{0}^{\pi} x f(\sin x) \, dx = - \int_{\pi}^{0} (\pi - t) f(\sin t)dt \)
\( \Rightarrow \frac{\pi}{2} \int_{0}^{\pi} f(\sin t) \, dt \)
\( I = \frac{\pi^2}{4} \)
page 86