题目
3.已知函数f(x )= {x)^2,xgt 1f(x+1)dx.
题目解答
答案
解析
步骤 1：求F(x)的表达式
当 $0\lt x\leqslant 1$ 时, $F(x)={\int }_{0}^{x}(t+1)dt=\dfrac {{x}^{2}}{2}+x$
当 $x\gt 1$ 时, $F(x)={\int }_{0}^{1}(t+1)dt+\int \dfrac {1}{2}{t}^{2}dt=\dfrac {3}{2}+\dfrac {{x}^{3}}{6}-\dfrac {1}{6}=\dfrac {{x}^{3}}{6}+\dfrac {4}{3}$
步骤 2：计算 ${\int }_{-2}^{1}f(x+1)dx$
$f(x+1)= \left \{ \begin{matrix} x+2,\quad x\leqslant 0\\ \dfrac {1}{2}{(x+1)}^{2},\quad x\gt 0\end{matrix} \right.$
${\int }_{-2}^{1}f(x+1)dx={\int }_{-2}^{0}(x+2)dx+{\int }_{0}^{1}\dfrac {1}{2}{(x+1)}^{2}dx=\dfrac {19}{6}$
当 $0\lt x\leqslant 1$ 时, $F(x)={\int }_{0}^{x}(t+1)dt=\dfrac {{x}^{2}}{2}+x$
当 $x\gt 1$ 时, $F(x)={\int }_{0}^{1}(t+1)dt+\int \dfrac {1}{2}{t}^{2}dt=\dfrac {3}{2}+\dfrac {{x}^{3}}{6}-\dfrac {1}{6}=\dfrac {{x}^{3}}{6}+\dfrac {4}{3}$
步骤 2：计算 ${\int }_{-2}^{1}f(x+1)dx$
$f(x+1)= \left \{ \begin{matrix} x+2,\quad x\leqslant 0\\ \dfrac {1}{2}{(x+1)}^{2},\quad x\gt 0\end{matrix} \right.$
${\int }_{-2}^{1}f(x+1)dx={\int }_{-2}^{0}(x+2)dx+{\int }_{0}^{1}\dfrac {1}{2}{(x+1)}^{2}dx=\dfrac {19}{6}$