यहाँ `a=-1, b=1, f(x)=e^(x), h=(b-a)/(n)=2/n` या
योगफल की सिमा की परीभाषा से,
`underset(a)overset(b)intf(x)dx=underset(h to 0)limh[f(a)+f(a+h)+f(a+2h)+.....+f{a+(n-1)h}]`
`underset(-1)overset(1)intf(x)dx`
`=underset(h to 0)limh [e^(-1)+e^(-1+h)+e^(-1+2h)+......+e^(-1(n-1)h)]`
`=underset(h to 0)limh [e^(-1){1+e^(h)+2^(h)+....+e^((n-1)h)}]`
`=underset(h to 0)limhe^(-1) [1+e^(h)+2^(h)+....+e^((n-1)h)]`
`=underset(h to 0)limhe^(-1) [1+((e^(h))^(n)-1)/(e^(h)-1)] [therefore a+ar+.....+ar^(n-1)=(a(r^(n)-1))/(r-1)]`
`=underset(h to 0)lim [(e^(nh)-1)/(((e^(h)-1)/(h)))]=underset(h to 0)lime^(-1)((e^(2)-1))/(((e^(h)-1)/(h)))`
`=(e^(-1)(e^(2)-1))/(1)" "[therefore underset(h to 0)lim (e^(h)-1)/(h)=1]`
`=e-e^(-1)`
