Evaluate : `int _(-1) ^(1) e^(x) dx` .

Explanation

we know that
`overset(a)underset(b)(int)f(x)dx=underset(h to0)("lim") h[f(a)+f(a+h)`
`+f(a+2h)+....+f{a+(n-1)}h]`
where nh=b-a
`" Given " overset(1)underset(-1)(int) e^(x) dx`
`" Here "a=-1 ,b=1 " and " nh=b-a=2`
`" and " f(x) =e^(x)`
`:. overset(1)underset(-1)(int) e^(x) dx= underset(h to 0)(" lim") [e^(-1)+e^((-1+h))+e^((-1+2h))`
`+.....+e^({-1+(n-1)h})]`
`=underset(h to0)("lim") he^(-1)[1+e^(h)+e^(2h)+......+e^((n-1)h)]`
`=underset(h to0)("lim") (h)/(e) ({(e^(h))^(n)-1})/(e^(n)-1)=(1)/(e)underset(h to0)("lim") (e^(hn)-1)/((e^(h)-1)/(h))`
`(:. |e^(h)|gt 1)`
`=(1)/(e)underset(h to0)("lim") (e(2)-1)/((e^(h)-1)/(h))=(1)/(e)((e^(2)-1)/(1))`
`"(":' underset(h to0)("lim") (e^(x)-1)/(x)=1 " and nh"=2")"`
`=(e^(2)-1)/(e)=e-(1)/(e)`