liate rule i think

**cummings15** · May 6th 2010, 04:31 AM

integral from 0 to 8 of (5xe^-x)(dx)

so far i think

5(e^-x)

so then do i plug in 8

**tom@ballooncalculus** · May 6th 2010, 04:46 AM

Just in case a picture helps...

Trial and error will quickly establish (and liate confirm) that you need the legs-crossed version of...

... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x.

The general drift is...

In this case...

And the rest...

Spoiler:

_________________________________________
Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

**cummings15** · May 6th 2010, 04:57 AM

so is it

(5x)(e^-x) + (5) (e^-x) + C

**tom@ballooncalculus** · May 6th 2010, 05:02 AM

Originally Posted by cummings15

so is it

(5x)(e^-x) + (5) (e^-x) + C

The negative of that, as you can maybe see above now that I've fixed the second pic (in the spoiler).

I think you missed that we need the chain rule for e^(-x). I didn't bother zooming in on that, but (gimme a minute)...

... where

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

I uncrossed the legs for clarity - i.e. switched the 5x and e^-x.

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

**Soroban** · May 6th 2010, 08:33 AM

Hello, cummings15!

$5\int^8_0 xe^{-x}\,dx$

Integrate by parts . . .

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

We have: . $5\bigg[-xe^{-x} + \int e^{-x}\,dx\bigg] \;=\;5\bigg[-xe^{-x} - e^{-x}\bigg] + C \;=\;-5e^{-x}(x+1) + C$

Evaluate: . $-\frac{5}{e^x}(x+1)\,\bigg]^8_0 \;=\;\bigg[-\frac{5}{e^8}(8+1) \bigg] - \bigg[-\frac{5}{e^0}(0+1)\bigg] \;=\;5 - \frac{45}{e^8}$

**skeeter** · May 6th 2010, 08:55 AM

Originally Posted by cummings15

integral from 0 to 8 of (5xe^-x)(dx)

so far i think

5(e^-x)

so then do i plug in 8

tabular integration ...

sign ........... u ........... dv

+............... 5x .......... $e^{-x}$

- ................ 5 .......... $-e^{-x}$

................... 0 .......... $e^{-x}$

$ \left[-5xe^{-x} - 5e^{-x}\right]_0^8 $

$ \left[-5e^{-x}(x+1) \right]_0^8 $

$ 5 -\frac{45}{e^8} $

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