Integrate 1/e^x???

6 years ago\[\large \int\limits \frac{1}{e^x}dx \quad = \quad \int\limits e^{-x}dx\]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 agoEventually 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 agoits 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 agoYes, both are correct :)

6 years agoand 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 agoNo 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 agoCrap 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 agoso -1/2x^2+xe^x-e^x

6 years ago\[\large \int\limits xe^x\;dx\]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|dw:1359436512411:dw|

6 years ago