Question

利用定义求定积分

定积分号（积分下限0积分上限1）e^x dx

Answer 1

解：原式=∫(0,1)e^xdx
=lim(n->∞)[e^(1/n)/n+e^(2/n)/n+e^(3/n)/n+....+e^(n/n)/n] (由定积分定义得)
=lim(n->∞){(1/n)[e^(1/n)+e^(2/n)+e^(3/n)+....+e^(n/n)]}
=lim(n->∞){(e^(1/n)/n)[1+e^(1/n)+e^(2/n)+....+e^((n-1)/n)]}
=lim(n->∞){(e^(1/n)/n)[(1-e^(n/n))/(1-e^(1/n))]}
=lim(n->∞){[(1-e)e^(1/n)]*[(1/n)/(1-e^(1/n))]}
=lim(n->∞)[(1-e)e^(1/n)]*lim(n->∞)[(1/n)/(1-e^(1/n))]
=(1-e)*(-1) (∵lim(n->∞)[(1/n)/(1-e^(1/n))]=-1)
=e-1

Answer 2

把X写成1/n。然后让n趋于无穷大。逐项求和。好像用到泰勒公式展开形式。

Answer 3

e-1

指数积分是它本身啊，带入上下限就好

Answer 4

1 x x 1 1 0
∫ e dx=e / =e - e =e-1 仔细看下！
o 0