解析:选C.(ex+2x)dx=(ex+x2)|=(e+1)-1=e.
sin xdx=________.
解析:sin xdx=-cos x|=(-cos π)-(-cos 0)=2.
答案:2
探究点1 求简单函数的定积分
求下列定积分.
(1)xndx;
(2)(cos x-sin x)dx;
(3)dx.
【解】 (1)xndx
=xn+1|
=×1n+1-×0n+1
=.
(2)(cos x-sin x)dx
=(sin x+cos x)
=(sin 2π+cos 2π)-(sin 0+cos 0)
=0.
(3)dx
=(ex-ln x)|
=(e2-ln 2)-(e1-ln 1)
=e2-e-ln 2.