# How to solve this using limit of sum? int_0^1xe^xdx

Jun 27, 2018

${\int}_{0}^{1} x {e}^{x} \mathrm{dx} = 1$

#### Explanation:

Integrating by parts:

${\int}_{0}^{1} x {e}^{x} \mathrm{dx} = {\int}_{0}^{1} x d \left({e}^{x}\right)$

${\int}_{0}^{1} x {e}^{x} \mathrm{dx} = {\left[x {e}^{x}\right]}_{0}^{1} - {\int}_{0}^{1} {e}^{x} \mathrm{dx}$

${\int}_{0}^{1} x {e}^{x} \mathrm{dx} = e - {\left[{e}^{x}\right]}_{0}^{1}$

${\int}_{0}^{1} x {e}^{x} \mathrm{dx} = e - e + 1 = 1$

If you want to demonstrate as limit of the Riemann sum, consider the partitions of the interval $\left[0 , 1\right]$ in $n$ intervals of length $\frac{1}{n}$ limited by the points:

${x}_{k , n} = \frac{k}{n}$ for $k = 0 , 1 , \ldots , n$

for every $n \in \mathbb{N}$.

The left Riemann sum is then:

${s}_{n} = {\sum}_{k = 0}^{n - 1} \frac{{x}_{k} {e}^{{x}^{k}}}{n} = {\sum}_{k = 1}^{n - 1} \frac{k {e}^{\frac{k}{n}}}{n} ^ 2 = \frac{1}{n} ^ 2 {\sum}_{k = 1}^{n - 1} k {e}^{\frac{k}{n}}$

While the right Riemann sum is:

${S}_{n} = {\sum}_{k = 0}^{n - 1} \frac{{x}_{k + 1} {e}^{{x}^{k + 1}}}{n} = {\sum}_{k = 1}^{n} \frac{k {e}^{\frac{k}{n}}}{n} ^ 2$

so that:

$\left(1\right) \text{ } {\int}_{0}^{1} x {e}^{x} \mathrm{dx} = {\lim}_{n \to \infty} \frac{1}{n} ^ 2 {\sum}_{k = 1}^{n} k {e}^{\frac{k}{n}}$

Now consider the geometric series:

$\frac{1}{1 - x} = {\sum}_{k = 0}^{\infty} {x}^{k}$

converging for $x \in \left(- 1 , 1\right)$ and differentiate it term by term:

$\frac{1}{1 - x} ^ 2 = {\sum}_{k = 1}^{\infty} k {x}^{k - 1} = \frac{1}{x} {\sum}_{k = 1}^{\infty} k {x}^{k}$

For $x = {e}^{\frac{1}{n}}$:

$\frac{1}{1 - {e}^{\frac{1}{n}}} ^ 2 = \frac{1}{e} ^ \left(\frac{1}{n}\right) {\sum}_{k = 1}^{\infty} k {\left({e}^{\frac{1}{n}}\right)}^{k}$

${e}^{\frac{1}{n}} / {\left(1 - {e}^{\frac{1}{n}}\right)}^{2} = {\sum}_{k = 1}^{\infty} k {e}^{\frac{k}{n}}$

Dividing both sides by ${n}^{2}$ we have:

${e}^{\frac{1}{n}} / \left({n}^{2} {\left(1 - {e}^{\frac{1}{n}}\right)}^{2}\right) = \frac{1}{n} ^ 2 {\sum}_{k = 1}^{\infty} k {e}^{\frac{k}{n}}$

and comparing this to equation $\left(1\right)$ we get:

$\left(2\right) \text{ } {\int}_{0}^{1} x {e}^{x} \mathrm{dx} = {\lim}_{n \to \infty} {e}^{\frac{1}{n}} / \left({n}^{2} {\left(1 - {e}^{\frac{1}{n}}\right)}^{2}\right)$

Using the well known limit:

${\lim}_{x \to 0} \frac{{e}^{x} - 1}{x} = 1$

we can see that:

${\lim}_{n \to \infty} \frac{{e}^{\frac{1}{n}} - 1}{\frac{1}{n}} = {\lim}_{n \to \infty} n \left({e}^{\frac{1}{n}} - 1\right) = 1$

and then:

${\lim}_{n \to \infty} {e}^{\frac{1}{n}} / \left({n}^{2} {\left(1 - {e}^{\frac{1}{n}}\right)}^{2}\right) = {\lim}_{n \to \infty} {e}^{\frac{1}{n}} \cdot \frac{1}{{\lim}_{n \to \infty} n \left({e}^{\frac{1}{n}} - 1\right)} ^ 2 = 1$

and we can conclude that:

${\int}_{0}^{1} x {e}^{x} \mathrm{dx} = 1$