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  • Evaluate the following definite integrals as limit of sums. integral -1 to 1 e^x dx

Evaluate the following definite integrals as limit of sums.

    Q5.    \int_{-1}^1 e^xdx

Answers (1)

best_answer

let I = \int_{-1}^{1}e^xdx
We know that
\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}
Here a =-1,  b = 1 and f(x) = e^x
therefore h = 2/n
I = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[f(-1)+f(-1+\frac{2}{n})+.....+f(-1+(n-1).\frac{2}{n})]
    \\ =2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}+e^{-1+\frac{2}{n}}+e^{-1+2.\frac{2}{n}}+...+e^{-1+(n-1).\frac{2}{n}}]\\ = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}(1+e^{2/n}+e^{4/n}+...+e^{(n-1).\frac{2}{n}})]\\ =
By using sum of n terms of GP S =\frac{a(r^n-1)}{r-1}....where a = 1st term and r = ratio

\\=2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}[\frac{1.(e^{\frac{2}{n}.n}-1)}{e^\frac{2}{n}-1}]\\ =2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}(\frac{e^2-1}{e^{2/n}-1})\\ =\frac{e^{-1}(e^2-1)}{\lim_{\frac{2}{n}\rightarrow \infty }\frac{e^{2/n}-1}{2/n}}\\ =\frac{e^2-1}{e}.........using [\lim_{x\rightarrow \infty }(\frac{e^x-1}{x})=1]
 

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manish

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