Question

Evaluate the following improper integral. If the integral diverges, enter "DIV".

[₁
IH
Te dx=

Answer 1

`2e^(-1)}`

Explanation:

Evaluate the given improper integral.

`\int\limits^( \infty)_(1) {xe^(-x)} \, dx\\\\\\\hrule`

Using integration by parts.

`\boxed{\left\begin{array}{ccc}\text{\underline{Integration by Parts}}\\\\uv-\int vdu\end{array}\right}`

Letting...

u = x => du = dx

dv = e^{-x} dx => v = -e^{-x}

`\int\limits^( \infty)_(1) {xe^(-x)} \, dx\\\\\\\\\Longrightarrow (x)(-e^(-x))-\int\limits^( \infty)_(1) {-e^(-x)} \, dx\\\\\\\\\Longrightarrow -xe^(-x)\Big|\limits^( \infty)_(1)+\int\limits^( \infty)_(1) {e^(-x)} \, dx\\\\\\\\\Longrightarrow \Big[ -xe^(-x)-e^(-x)\Big]\limits^( \infty)_(1)\\\\ \\\\\Longrightarrow \Big[ -(\infty)e^(-\infty)-e^(-\infty)\Big]-\Big[ -(1)e^(-(1))-e^(-(1))\Big]\\\\\\ \\\Longrightarrow \Big[-0-0\Big]-\Big[ -e^(-1)-e^(-1)\Big]`

`\Longrightarrow e^(-1)+e^(-1)\\\\\\\\\therefore \int\limits^( \infty)_(1) {xe^(-x)} \, dx =\boxed{2e^(-1)}`

Thus, the problem is solved.