Integration question (1 Viewer)

HeroicPandas · Jan 21, 2013

$\int xe^{-x}.dx$

Me and my friend found 2 methods, but end up with 2 different answers...

Drongoski · Jan 21, 2013

HeroicPandas · Jan 21, 2013

Drongoski said:
$=e^{-x}(1+x) + C$

Can u briefly explain the method u used please?

barbernator · Jan 21, 2013

HeroicPandas · Jan 21, 2013

barbernator said:
IBP

That's a 4U technique that i haven't learnt

barbernator · Jan 21, 2013

HeroicPandas said:
That's a 4U technique that i haven't learnt

well thats how u have to do it

Sy123 · Jan 21, 2013

HeroicPandas said:
That's a 4U technique that i haven't learnt

I can''t find a really good substitution that will fit it,this is the closest I got:

$\int x e^{-x} \ dx$

$x= -\ln u \ \ dx = \frac{-du}{u}$

$\int -\ln u \cdot u \cdot -1/u \ du$

$=\int \ln u \ du$

We don't need 4U to do this integral, this is a question very commonly asked in textbooks for 3U

$$By considering$ \ \ d/dx x \sin^{-1} x \ \ \ $find$ \ \ \int \sin^{-1}x \ dx$

We will do the same thing:

$d/du u \ln u = \ln u + u \cdot 1/u = \ln u + 1$

$\therefore u\ln u + C = \int \ln u \ du + \int 1 \ du$

$\therefore \int \ln u \du = u\ln u - u$

Sub in x again

$\int xe^{-x} \ dx = -e^{-x}x - e^{-x} = -e^{-x}(1+x)$

HeroicPandas · Jan 21, 2013

Method 1:
$\frac{d[xe^{-x}]}{dx} = e^{-x} - xe^{-x} \\ \\ \int \frac{d[xe^{-x}]}{dx} = \int e^{-x}.dx - \int xe^{-x}.dx\\ \\ \int xe^{-x}.dx = \int e^{-x}.dx - \int \frac{d[xe^{-x}]}{dx} \\ \\ \int xe^{-x}.dx = -e^{-x} -xe^{-x} + C \\ \\ \therefore \int xe^{-x}.dx = -e^{-x}(x+1) + C$

Method 2:
$\int xe^{-x}.dx = \int e^{log_e (xe^{-x})}.dx \\ \\ \int e^{log_e x + log_e e^{-x}}.dx = \int e^{log_e x -x}.dx = \frac{xe^{-x}}{\frac{1}{x}-1} + C = \frac{x^2 e^{-x}}{1-x} + C$

Why are they different?

EDIT: added more steps

HeroicPandas · Jan 21, 2013

Sy123 said:
I can''t find a really good substitution that will fit it,this is the closest I got:

$\int x e^{-x} \ dx$

$x= -\ln u \ \ dx = \frac{-du}{u}$

$\int -\ln u \cdot u \cdot -1/u \ du$

$=\int \ln u \ du$

We don't need 4U to do this integral, this is a question very commonly asked in textbooks for 3U

$$By considering$ \ \ d/dx x \sin^{-1} x \ \ \ $find$ \ \ \int \sin^{-1}x \ dx$

We will do the same thing:

$d/du u \ln u = \ln u + u \cdot 1/u = \ln u + 1$

$\therefore u\ln u + C = \int \ln u \ du + \int 1 \ du$

$\therefore \int \ln u \du = u\ln u - u$

Sub in x again

$\int xe^{-x} \ dx = -e^{-x}x - e^{-x} = -e^{-x}(1+x)$

Wow, nice!!!

HeroicPandas · Jan 21, 2013

Sy123 said:
I can''t find a really good substitution that will fit it,this is the closest I got:

$\int x e^{-x} \ dx$

$x= -\ln u \ \ dx = \frac{-du}{u}$

$\int -\ln u \cdot u \cdot -1/u \ du$

$=\int \ln u \ du$

We don't need 4U to do this integral, this is a question very commonly asked in textbooks for 3U

$$By considering$ \ \ d/dx x \sin^{-1} x \ \ \ $find$ \ \ \int \sin^{-1}x \ dx$

We will do the same thing:

$d/du u \ln u = \ln u + u \cdot 1/u = \ln u + 1$

$\therefore u\ln u + C = \int \ln u \ du + \int 1 \ du$

$\therefore \int \ln u \du = u\ln u - u$

Sub in x again

$\int xe^{-x} \ dx = -e^{-x}x - e^{-x} = -e^{-x}(1+x)$

Wow, nice!!!

nightweaver066 · Jan 21, 2013

Your method is just using a formula that shouldn't even be a formula because of the mistakes it causes people which led you to a wrong answer.

More correctly, the formula should be $\int f'(x) e^{f(x)} dx = e^{f(x)}+c$ .

HeroicPandas · Jan 21, 2013

nightweaver066 said:
Your method is just using a formula that shouldn't even be a formula because of the mistakes it causes people which led you to a wrong answer.

More correctly, the formula should be $\int f'(x) e^{f(x)} dx = e^{f(x)}+c$ .

i don't understand...what was wrong with method 2?

nightweaver066 · Jan 21, 2013

HeroicPandas said:
i don't understand...what was wrong with method 2?

The formula $\int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} + c$ only applies when f(x) is a linear function.

HeroicPandas · Jan 21, 2013

nightweaver066 said:
The formula $\int e^{f(x)} dx = \frac{e^{f(x)}}{f'(x)} + c$ only applies when f(x) is a linear function.

Omg, oh yeh, totally forgot about that

Integration question (1 Viewer)

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