# increasing or decreasing

• Jul 14th 2009, 06:04 PM
acosta0809
increasing or decreasing
the question is

Let F(a) be the function which gives the area under the graph of y=x*e^-x between x=0 and x=a for a>0

i found F(a) to be (-x*e^-x)-(e^-x)
then evaluated with the given endpoints i got

(-a*e^-a)-e^a+1
(not sure if this is correct)

the second question is

is F increasing or a decreasing function?
is it the same that if the second derivative tells the concavity? pointers anyone?
• Jul 14th 2009, 06:30 PM
VonNemo19
Quote:

Originally Posted by acosta0809
is F increasing or a decreasing function?
is it the same that if the second derivative tells the concavity? pointers anyone?

Look at it like any other function.

$\frac{dF}{dx}=x*e^{-x}\Rightarrow\frac{d}{dx}(x*e^{-x})=e^{-x}-xe^{-x}=0$

$\frac{1}{e^x}(1-x)=0\Rightarrow\text{ critical values at }x=1$
• Jul 14th 2009, 07:02 PM
Soroban
Hello, acosta0809!

Quote:

Let $F(a)$ be the function which gives the area under the graph of $y\,=\,xe^{-x}$
between $x=0$ and $x=a$ for $a>0.$

i found $F(a)$ to be: . $-xe^{-x} -e^{-x}$

then evaluated with the given endpoints i got: . $-ae^{-a} -e^{-a}+1$
(not sure if this is correct)

It's correct!

Quote:

The second question is:

Is $F(a)$ an increasing or a decreasing function?

Find the first derivative (slope).

We have: . $F(a) \:=\:ae^{-a} - e^{-a} + 1$

Then: . $F'(a) \;=\;ae^{-a} - e^{-} + e^{-a} + 0 \;=\;\frac{a}{e^a}$

Since $a > 0$, then: . $F'(a) \:=\:\frac{a}{e^a}$ is always positive.

Therefore, $F(a)$ is an increasing function.

If we look at the graph, it's obvious . . .
Code:

. . . - | . . . - |    ..*. . . . - |  *::::::*. . . . - | *::::::::::*.. . . . - |*::::::::::::::::* . . . - |:::::::::::::::::|        * . . . - * - - - - - - - - + - - - - - . . . - |                a
As $a$ increases, the area under the curve increases.