设圆x^2+y^2=a^2的面积为S，则int_(-a)^asqrt(a^2)-x^(2)dx=（）A. $S$B. $S/2$C. $S/4$D. $S/8$
int_(0)^pi/2(cos x+2)dx=( ).A. $\pi+1$B. $\pi$C. 2D. $\pi/2+2$
【例4】已知函数f(x)在[-1,2]上连续，且int_(-1)^0f(x)dx=2，int_(0)^1f(2x)dx=1，则int_(-1)^2f(x)dx=
积分 int_(0)^1 xe^2x , dx = ____.A. $\frac{e^2+1}{4}$B. $1$C. $\frac{e^2+1}{2}$D.
2.(2020山东高数Ⅲ)已知函数f(x)在[-1,2]上连续，且int_(-1)^0f(x)dx=2,int_(0)^1f(2x)dx=1,则int_(-1)
int_(0)^a (x^2sqrt(a^2-x^2)),(rm dx)(a>0)$\int_{0}^{a} {x^2\sqrt{a^2-x^2}}\,{\rm
int_(0)^2(1)/(1+sqrt[3](x))dx;$\int_{0}^{2}\frac{1}{1+\sqrt[3]{x}}dx;$
(2)int_(-infty)^+infty(dx)/(x^2)+2x+2;(2)$\int_{-\infty}^{+\infty}\frac{dx}{x^{2
(d)/(dx) int_(1)^x x ln (x^2 + 1) , dx = $\frac{d}{dx} \int_{1}^{x} x \ln (x^2 +
9.计算下列定积分:-|||-(1) (int )_(0)^asqrt ({a)^2-(x)^2}dx(agt 0);