Evaluate the following definite integrals as limit of sums. `int_-1^1e^x dx`

Explanation

It is known that,
`int_{-1}^{1} f(x) d x=lim _{h rightarrow 0} h[f(a)+f(a+h)+f(a+2 h)+ldots .+f{a+(n-1) h}]`
where, `n h=b-a`
Given, `int_{-1}^{1} e^{x} d x`

Here, `a=-1, b=1 and n h=b-a=2`

and
`f(x)=e^{x}`
, then`f (a)=f(1)=e^{-1}`, `f(-1) f(-1+h)=e^{(-1+h)} ldots f[(-1+(n-1) h)]=e^{(-1+(n-1) h)}`

therefore, `int_{-1}^{1} {e}^{{x}} {dx}=lim _{{h} rightarrow 0} {h}[{e}^{-1}+{e}^{(-1+{h})}+{e}^{-1+2 {~h}}+ldots+{e}^{(-1+({n}-1) {h})}]`


`=lim _{h rightarrow 0} h e^{-1}[1+e^{h}+e^{2 h}+ldots+e^{(n-1) h}]`

[because `a+a r+a r^{2}+ldots .+a r^{n}=frac{a(r^{n}-1)}{(r-1)}]`

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