单选题 （共 5 题 ），每题只有一个选项正确
设图数 $f(x, y)$ 在 $\mathbf{R}^{\circ}$ 上进续, 交抰祭次积分的顺序 $\int_{-2}^1 \mathrm{~d} x \int_{x^2}^{2-x} f(x, y) \mathrm{d} y=$
$\text{A.}$ $\int_1^4 \mathrm{~d} y \int_{\sqrt{y}}^{2-y} f(x, y) \mathrm{d} x$.
$\text{B.}$ $\int_1^4 \mathrm{~d} y \int_{2-y}^{\sqrt{y}} f(x, y) \mathrm{d} x$.
$\text{C.}$ $\int_0^1 \mathrm{~d} y \int_{\sqrt{y}}^{2-y} f(x, y) \mathrm{d} x+\int_1^4 \mathrm{~d} y \int_{2-y}^{\sqrt{y}} f(x, y) \mathrm{d} x$.
$\text{D.}$ $\int_0^1 \mathrm{~d} y \int_{\sqrt{y}}^{\sqrt{y}} f(x, y) \mathrm{d} x+\int_1^4 \mathrm{~d} y \int_{-\sqrt{y}}^{2-y} f(x, y) \mathrm{d} x$.
设 $0 < a < 1, I_1=\int_0^1 \frac{\mathrm{e}^{a x}-1}{\mathrm{e}^x-1} \mathrm{~d} x, I_2=\int_0^1 \frac{\sqrt{a x}+1}{\sqrt{x}+1} \mathrm{~d} x$, 则
$\text{A.}$ $I_1 < a < I_2$.
$\text{B.}$ $I_2 < a < I_1$.
$\text{C.}$ $a < I_1 < I_2$.
$\text{D.}$ $I_1 < I_2 < a$.
设 $a \neq b$, 若 $\int_0^{+\infty} \frac{\mathrm{e}^{-a x}-\mathrm{e}^{-b x}}{x} \mathrm{~d} x$ 收敛,则 $a, b$ 的取值范围为
$\text{A.}$ $a < 0, b < 0$.
$\text{B.}$ $a < 0, b>0$.
$\text{C.}$ $a>0, b < 0$.
$\text{D.}$ $a>0, b>0$.
已知 $f(x)=\max \left\{1, x^2\right\}$, 则 $\int f(x) d x=$
$\text{A.}$ $\begin{cases}\frac{x^3}{3}-\frac{2}{3}+C, & x < -1 \\ x+C, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+\frac{2}{3}+C, & x>1\end{cases}$
$\text{B.}$ $\left\{\begin{array}{cc}\frac{x^3}{3}+C, & x < -1 \\ x+C, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+C, & x>1\end{array}\right.$
$\text{C.}$ $\left\{\begin{array}{cc}\frac{x^3}{3}+C_1, & x < -1 \\ x+C_2, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+C_3, & x>1\end{array}\right.$
$\text{D.}$ $\begin{cases}\frac{x^3}{3}-\frac{4}{3}+C, & x < -1 \\ x+C, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+\frac{2}{3}+C, & x>1\end{cases}$
若反常积分 $\int_0^1 \frac{\ln x}{x^\alpha\left(\tan \frac{\pi}{2} x\right)^\beta} \mathrm{d} x$ 收敛, 则
$\text{A.}$ $-2 < \beta < 0$ 且 $\alpha+\beta \geqslant 1$.
$\text{B.}$ $0 < \beta < 2$ 且 $\alpha+\beta < 1$.
$\text{C.}$ $\beta < -2$ 且 $\alpha+\beta \geqslant 1$.
$\text{D.}$ $\beta>-2$ 且 $\alpha+\beta < 1$.
填空题 （共 1 题 ），请把答案直接填写在答题纸上
设 $f(x)$ 可表示为 $f(x)=2 x+2 \int_0^1 f(x) \mathrm{d} x$, 则 $f(x)=$