Đề kiểm tra 15 phút Toán 12 chương 3: Nguyên hàm - Đề số 3

\(\begin{align}  & f'\left( x \right)=-3.\frac{a}{{{\left( x+1 \right)}^{4}}}+b{{e}^{x}}+bx{{e}^{x}} \\  & \Rightarrow f'\left( 0 \right)=-3a+b=-22\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right) \\  & \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \frac{a}{{{\left( x+1 \right)}^{3}}}+bx{{e}^{x}} \right)dx}=a\int\limits_{0}^{1}{{{\left( x+1 \right)}^{-3}}dx}+b\int\limits_{0}^{1}{x{{e}^{x}}dx}=a{{I}_{1}}+b{{I}_{2}} \\  & {{I}_{1}}=\int\limits_{0}^{1}{{{\left( x+1 \right)}^{-3}}dx}=\left. \frac{{{\left( x+1 \right)}^{-2}}}{-2} \right|_{0}^{1}=\frac{-1}{2}\left( \frac{1}{4}-1 \right)=\frac{3}{8} \\ \end{align}\)

Đặt\(\left\{ \begin{array}{l}
u = x\\
dv = {e^x}dx
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
du = dx\\
v = {e^x}
\end{array} \right. \Rightarrow {I_2} = \left. {x{e^x}} \right|_0^1 - \int\limits_0^1 {{e^x}dx} = e - \left. {{e^x}} \right|_0^1 = e - \left( {e - 1} \right) = 1\)

\(\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\frac{3}{8}a+b=5\,\,\,\,\,\,\left( 2 \right)\) 

Từ (1) và (2) \(\Rightarrow \left\{ \begin{align}  & a=8 \\  & b=2 \\ \end{align} \right.\)