Show-that-the-function-f-x-e-x-is-of-Reimann-within-x-1-5-hence-calculate-1-5-e-x-dx-

Question Number 93941 by Ar Brandon last updated on 16/May/20

Show that the function f (x) = e^{x} is of Reimann within

x \in [1; 5] hence calculate \int_{1}^{5} e^{x} dx

Commented by mathmax by abdo last updated on 16/May/20

f c o n t i n u e w e h a v e \int_{1}^{5} e^{x} d x = {l i m}_{n \to + \infty} \frac{5 - 1}{n} \sum_{k = 1}^{n} e^{1 + \frac{4 k}{n}}

= {l i m}_{n \to + \infty} \frac{4 e}{n} \sum_{k = 0}^{n - 1} {(e^{\frac{4}{n}})}^{k} = {l i m}_{n \to + \infty} \frac{4 e}{n} \times \frac{1 - {(e^{\frac{4}{n}})}^{n}}{1 - e^{\frac{4}{n}}}

= {l i m}_{n \to + \infty} \frac{4 e}{n} \times \frac{1 - e^{4}}{1 - e^{\frac{4}{n}}} = {l i m}_{n \to + \infty} 4 e (1 - e^{4}) \times \frac{1}{n (1 - e^{\frac{4}{n}})}

w e h a v e e^{\frac{4}{n}} \sim 1 + \frac{4}{n} \Rightarrow 1 - e^{\frac{4}{n}} \sim - \frac{4}{n} \Rightarrow n (1 - e^{\frac{4}{n}}) \sim - 4 \Rightarrow

{l i m}_{n \to + \infty} \frac{1}{n (1 - e^{\frac{4}{n}})} = - 4 \Rightarrow \int_{1}^{5} f (x) d x = - \frac{1}{4} \times 4 e (1 - e^{4})

= e^{5} - e .

Commented by Ar Brandon last updated on 16/May/20

Thanks ��