anonymous
  • anonymous
Show by hand how to find an anti-derivative to -t E^t^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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SolomonZelman
  • SolomonZelman
u sub
SolomonZelman
  • SolomonZelman
u=t²
jim_thompson5910
  • jim_thompson5910
hint: u-subsitution u = t^2, so du = 2t*dt

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SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}f'(x)\times e^{f(x)}~dx}\) SUBSTITUTION: `u=f(x) du=f'(x) dx` \(\large\color{black}{\displaystyle\int\limits_{~}^{~}e^{u}~du=e^u+C=e^{f(x)}+C}\) (this is an abstract case for some differentiable f(x) )
anonymous
  • anonymous
u =t^2 du = 2t dt -E^u du = ...
SolomonZelman
  • SolomonZelman
well, not exactly
anonymous
  • anonymous
I suppose (-1) can be pulled out
nincompoop
  • nincompoop
show by hand...
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\int\limits_{~}^{~}-t\times e^{t^2}~dt}\) \(\large\color{black}{\displaystyle-\int\limits_{~}^{~}t\times e^{t^2}~dt}\) \(\large\color{black}{\displaystyle u=t^2}\) \(\large\color{black}{\displaystyle du=2t~du~~~~\rightarrow~~~~\frac{1}{2}du=t~dt}\) \(\large\color{black}{\displaystyle-\frac{1}{2}\int\limits_{~}^{~} e^{u}~du}\)
SolomonZelman
  • SolomonZelman
like that - same as you said, BUT you got the 1/2 there (why? I have wrote why in my post)
anonymous
  • anonymous
and that 2 has to be removed I think.. du = 2t dt du/2 = 2t dt/2 du/2 = t dt ???
anonymous
  • anonymous
thanks solomon...
SolomonZelman
  • SolomonZelman
okay, can you now tell me what your antiderivative would be?
SolomonZelman
  • SolomonZelman
(antiderivative of e^x is just e^x ... (well, +C) )
anonymous
  • anonymous
-1/2 E^t^2
SolomonZelman
  • SolomonZelman
yes, with +C
anonymous
  • anonymous
oh yes.. +C
SolomonZelman
  • SolomonZelman
One of my past professors took off most of the partial credit for +C. He explained that by saying that you gave only 1/∞ of all possible answers.
anonymous
  • anonymous
thanks, nicely explained.
anonymous
  • anonymous
my professors have been really good with that.
SolomonZelman
  • SolomonZelman
Enjoy it any time I'm on. Although sometimes I will be gone. I wish good luck to you in all, Yes, that's the way! lets make it roll 0~0
anonymous
  • anonymous
they had a grumble about how that happened to them.
SolomonZelman
  • SolomonZelman
If you have further questions, then u know - whenever I'm.... (or, there are many other math peeps here) Just always watch out how you solve for du in your substitution ....
anonymous
  • anonymous
actually if I had to use FTC with something like this equation.. for limits of [a,b] would they expect me to put +C on that too? say... -1/2 E^b^2 - (-1/2 E^a^2)
SolomonZelman
  • SolomonZelman
wait, what does FTC stand for?
anonymous
  • anonymous
fundamental theorem calculus
SolomonZelman
  • SolomonZelman
oh.
SolomonZelman
  • SolomonZelman
With limits of integration, +C ?
anonymous
  • anonymous
I suppose it cancels out ...
anonymous
  • anonymous
unless there's some kind of initial condition
SolomonZelman
  • SolomonZelman
If you are integrating a definite integral, you don't have the +C, because when you are integrating a definite integral you are finding the (numerical) area under the curve over a specific interval. And you are not finding the family of functions. So they say I guess: \(\large\color{black}{\displaystyle\int\limits_{a}^{b}f(x)~dx=\left( f(x)+C\right){\LARGE |}^{x=b}_{x=a}}\) (where F'=f)
SolomonZelman
  • SolomonZelman
and yes C cancel's out, but it shouldn't be there in a first place.
SolomonZelman
  • SolomonZelman
+C is a family of antiderivative functions F(x) (and this family of F(x)+C is a set of functions that has a derivative of f) when finding area under the corve over some interval I , that is not a family of functions you are finding - that is an area calculation.
SolomonZelman
  • SolomonZelman
Not that the result would differ if you have +C there, since it will cancel, but as far as the concept goes +C, again, should not be there)
anonymous
  • anonymous
gotcha.. makes sense
SolomonZelman
  • SolomonZelman
Alrighty:)
anonymous
  • anonymous
laterzs
anonymous
  • anonymous
thanks again
SolomonZelman
  • SolomonZelman
Anytime

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