Mathematics
OpenStudy (anonymous):

Integrate 1/e^x???

6 years ago
zepdrix (zepdrix):

$\large \int\limits \frac{1}{e^x}dx \quad = \quad \int\limits e^{-x}dx$If you're not sure how to proceed from there, you can always do a u sub. $\large \color{orangered}{u=-x}$$\large \color{royalblue}{-du=dx}$Plugging these in gives us,$\huge \int\limits e^{\color{orangered}{-x}}\color{royalblue}{dx}\qquad \rightarrow \qquad -\int\limits e^{\color{orangered}{u}}\color{royalblue}{du}$

6 years ago
zepdrix (zepdrix):

Eventually you'll get comfortable with integrating exponentials of base e. It's like the chain rule in reverse almost. Like when you would differentiate $$\large e^{3x}$$ you would get an extra factor of 3 right? Well, when you integrate it, you'll divide by that 3 instead.

6 years ago
OpenStudy (anonymous):

its been so long... 12 years since this stuff, its been hard trying to remember. so after integration you would get -e^-x or just -1/e^x?

6 years ago
zepdrix (zepdrix):

Yes, both are correct :)

6 years ago
OpenStudy (anonymous):

and if we were to integrate xe^x, we would do the first times the integral of the second, and the second times the integral of the first?

6 years ago
zepdrix (zepdrix):

No the product rule doesn't quite work the way you might want it to in reverse. We would have to apply what's called Integration by Parts. It's based off of the product rule, but it's a little funky. If you start with the product rule,$\huge (uv)'=u'v+uv'$We'll move some stuff around, giving us,$\huge u'v=(uv)'-uv'$Taking the integral of both sides gives us,$\large \int\limits u'v=uv-\int\limits uv'$The more accurate way to write this would be,$\large \int\limits v \;du=uv-\int\limits u \;dv$ That explanation may have been overkill.... lemme just show you how to do xe^x real quick.

6 years ago
zepdrix (zepdrix):

Crap I wrote that wrong, it doesn't make a big difference, but this is the way you'll actually see it.$\large \int\limits\limits u \;dv=uv-\int\limits\limits v \;du$

6 years ago
OpenStudy (anonymous):

so -1/2x^2+xe^x-e^x

6 years ago
zepdrix (zepdrix):

$\large \int\limits xe^x\;dx$We'll make a substitution, calling one of these pieces u. The one that we pick for u is the one we want to take a derivative of. If we let x be our u it will end up simplifying the integral, because the x will lower in degree each time we take it's derivative. We'll also assign something to be dv and we'll have to integrate that value. $\large u=x, \qquad dv=e^x\; dx$ We've picked our u and dv. Our formula requires 4 pieces, u du v and dv. So we need to find 2 more variables, du and v. Taking the derivative of our u and integrating our dv gives us, $\large du=dx, \qquad \qquad v=e^x$ Now we'll setup the problem according to the integration by part definition. If this is way tedious then I apologize XD I just don't like giving short explanations lol they seem so insufficient.

6 years ago
zepdrix (zepdrix):

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6 years ago
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