∫(-1->2) x^2dx
=(1/3)[x^3](-1->2)
=(1/3)(8+1)
=3是定义法。。。就是f（ξi）△ Xi的那种divide (-1,2) into n equal interval (3/n)-1+3/n , -1+6/n,...., 2 ∫(-1->2) x^2dx =lim(n->无穷) summation(i:1->n)f（ξi）△ xi=lim(n->无穷) summation(i:1->n) (3/n) (-1+ 3i/n)^2= lim(n->无穷) summation(i:1->n) (3/n) ( 1 - 6i/n + 9i^2/n^2 )=lim(n->无穷) 3 (1+ 3(n+1)/n + (3/2) (n+1)(2n+1)/n^2 )=3