利用定义计算定积分 例1利用定义计算定积分[x2ax 解把区间[0,1分成n等份,分点为和小区间长度为 x=(i=1,2,…,n-1) 取5=(=1,2,…,n),作积分和 ∑1(5)x=Ax=÷a110+2× 因为2=1,当>0时,m>,所以 x2x=lmn/5)x=imn2(+12+1)=1 n→>0 页返回 页结束铃首页 上页 返回 下页 结束 铃 •利用定义计算定积分 解 把区间[0, 1]分成n等份, 分点为和小区间长度为 n i xi = (i=1, 2,  , n−1), n xi 1  = (i=1, 2,  , n) . 取 n i i = (i=1, 2,  , n), 作积分和    = = =  =  =  n i i n i i i n i i n n i f x x 1 2 1 2 1 1 ( )  ( ) ) 1 )(2 1 (1 6 1 n n = + + . 3 1 ) 1 )(2 1 (1 6 1 lim ( ) lim 1 0 2 1 0 =  = + + = → → =   n n x dx f x n n i i i  . 因为 n 1 = , 当→0 时, n→, 所以 例 例 1 1 利用定义计算定积分 1 0 2 x dx . n i xi = (i=1, 2,  , n−1), n xi 1  = (i=1, 2,  , n) .    = = =  =  =  n i i n i i i n i i n n i f x x 1 2 1 2 1 1 ( )  ( ) ) 1 )(2 1 (1 6 1 n n    = + + . = = =  =  =  n i i n i i i n i i n n i f x x 1 2 1 2 1 1 ( )  ( ) ) 1 )(2 1 (1 6 1 n n    = + + . = = =  =  =  n i i n i i i n i i n n i f x x 1 2 1 2 1 1 ( )  ( ) ) 1 )(2 1 (1 6 1 n n = + + . 因为 n 1 = , 当→0 时, n→, 所以 3 1 ) 1 )(2 1 (1 6 1 lim ( ) lim 1 0 2 1 0 =  = + + = → → =   n n x dx f x n n i i i  . 3 1 ) 1 )(2 1 (1 6 1 lim ( ) lim 1 0 2 1 0 =  = + + = → → =   n n x dx f x n n i i i  . 下页