find indefinite integral of: (6 - 3e^x) / (e^(2x))?
with working please, not sure how to do this integral, if possible solve with substitution because I understand that.
- 8 years agoFavorite Answer
First thing to note is that we can break this guy into separate integrals:
(6 - 3e^x)/e^(2x) = 6/e^(2x) - 3e^(x-2x) = 6e^(-2x) - 3e^(-x), and now you can solve as two separate integrals:
(1) 6e^(-2x); and
To integrate (1), use a "u" substitution: u = -2x with du = -2dx. This transforms (1) into:
Integral(6e(-2x),x) = Integral(-3e^u,u) = -3e^u + c = -3e^(-2x) + c.
Likewsie in (2), let u = -x with du = -dx:
Integral(3e^(-x),x) = Integral(-3e^u,u) = -3e^u + c = -3e^(-x) + c.
Subtracting (1) - (2), means our answer is:
-3e^(-2x) + 3e^(-x) + C.
To check, differentiate:
Dx[-3e^(-2x) + 3e^(-x) + C] = 6e(-2x) - 3e(-x), which is what we started with. We can also check with the Wolfram online integrator, which is referenced below.
To help reinforce your understanding of the solving indefinite integrals with exponential functions, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.
As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.
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- YirmiyahuLv 78 years ago
No need for substitution. It's just algebra:
(6 - 3e^x) / e^(2x) = 6/e^(2x) - 3e^x/e^(2x) = 6e^(-2x) - 3e^(-x)
Integrating these terms should be trivial.
- seguinLv 43 years ago
enable u=6-x^3 then you quite get u'=-3x^2 you have already got x^2 interior the equation so which you purely desire neg. 3, so multiply neg.3 interior the front of the crucial sign and multiply the finished element with the help of -a million/3 so as that they make one (so as that they cancel out) then you quite've crucial of u^a million/2= 2/3u^(3/2)=2/3(6-x^3)^(3/2) and that's the respond.