Show by hand how to find an anti-derivative to
-t E^t^2

- anonymous

Show by hand how to find an anti-derivative to
-t E^t^2

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- schrodinger

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- SolomonZelman

u sub

- SolomonZelman

u=t²

- jim_thompson5910

hint: u-subsitution
u = t^2, so du = 2t*dt

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## More answers

- 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

u =t^2
du = 2t dt
-E^u du = ...

- SolomonZelman

well, not exactly

- anonymous

I suppose (-1) can be pulled out

- nincompoop

show by hand...

- 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

like that - same as you said, BUT you got the 1/2 there (why? I have wrote why in my post)

- anonymous

and that 2 has to be removed I think..
du = 2t dt
du/2 = 2t dt/2
du/2 = t dt
???

- anonymous

thanks solomon...

- SolomonZelman

okay, can you now tell me what your antiderivative would be?

- SolomonZelman

(antiderivative of e^x is just e^x ... (well, +C) )

- anonymous

-1/2 E^t^2

- SolomonZelman

yes, with +C

- anonymous

oh yes.. +C

- 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

thanks, nicely explained.

- anonymous

my professors have been really good with that.

- 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

they had a grumble about how that happened to them.

- 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

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

wait, what does FTC stand for?

- anonymous

fundamental theorem calculus

- SolomonZelman

oh.

- SolomonZelman

With limits of integration, +C ?

- anonymous

I suppose it cancels out ...

- anonymous

unless there's some kind of initial condition

- 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

and yes C cancel's out, but it shouldn't be there in a first place.

- 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

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

gotcha.. makes sense

- SolomonZelman

Alrighty:)

- anonymous

laterzs

- anonymous

thanks again

- SolomonZelman

Anytime

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