# improper integral

• June 19th 2007, 11:24 AM
viet
improper integral
need a little help starting this problem:
evalute:
$\int_{1}^{\infty} x e^{-x}dx$
• June 19th 2007, 11:37 AM
ThePerfectHacker
Quote:

Originally Posted by viet
need a little help starting this problem:
evalute:
$\int_{1}^{\infty} x e^{-x}dx$

Let $u=x \mbox{ and }v'=e^{-x}$

Then $u' = 1 \mbox{ and }v=-e^{-x}$

Thus,

$-xe^{-x}\big|_1^{\infty} + \int_1^{\infty} e^{-x} dx$

Note $-xe^{-x}\big|_1^{\infty} = \lim_{N\to \infty} - Ne^{-N} + e^{-1} = e^{-1}$

And, $\int_1^{\infty} e^{-x} dx = -e^{-x}\big|_1^{\infty} = \lim_{N\to \infty} -e^{-N} + e^{-1} = e^{-1}$
• June 19th 2007, 12:33 PM
Soroban
Hello, viet!

I got a different answer . . .

Quote:

Evalute: . $\int_{1}^{\infty} x e^{-x}dx$
Integrate by parts . . .

. . $\begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{-x}dx \\
du & = & dv & \quad & v & = & -e^{-x} \end{array}$

We have: . $-xe^{-x} + \int e^{-x}dx \;=\;-xe^{-x} - e^{-x}\;=\;-\frac{x+1}{e^x}$

Evaluate from $1$ to $\infty$

. . $\underbrace{\lim_{x\to\infty}\left(-\frac{x+1}{e^x}\right)} - \underbrace{\left(-\frac{1 + 1}{e}\right)}$

. . . . $= \;\;\;0\qquad\quad-\;\;\left(-\frac{2}{e}\right)$

. . . . . . $= \quad\frac{2}{e}$

• June 19th 2007, 12:41 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Let $u=x \mbox{ and }v'=e^{-x}$

Then $u' = 1 \mbox{ and }v=-e^{-x}$

Thus,

$-xe^{-x}\big|_1^{\infty} + \int_1^{\infty} e^{-x} dx$

Note $-xe^{-x}\big|_1^{\infty} = \lim_{N\to \infty} - Ne^{-N} + e^{-1} = e^{-1}$

And, $\int_1^{\infty} e^{-x} dx = -e^{-x}\big|_1^{\infty} = \lim_{N\to \infty} -e^{-N} + e^{-1} = e^{-1}$

Quote:

Originally Posted by Soroban
Hello, viet!

I got a different answer . . .

Integrate by parts . . .

. . $\begin{array}{ccccccc}u & = & x & \quad & dv & = & e^{-x}dx \\
du & = & dv & \quad & v & = & -e^{-x} \end{array}$

We have: . $-xe^{-x} + \int e^{-x}dx \;=\;-xe^{-x} - e^{-x}\;=\;-\frac{x+1}{e^x}$

Evaluate from $1$ to $\infty$

. . $\underbrace{\lim_{x\to\infty}\left(-\frac{x+1}{e^x}\right)} - \underbrace{\left(-\frac{1 + 1}{e}\right)}$

. . . . $= \;\;\;0\qquad\quad-\;\;\left(-\frac{2}{e}\right)$

. . . . . . $= \quad\frac{2}{e}$

You both have the same answers. TPH simply did not add the two pieces together at the end.

-Dan
• June 20th 2007, 03:43 AM
curvature
Use integration by parts
• June 20th 2007, 11:20 AM
viet
its easier than i expected thanks guys