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

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- SolomonZelman

u sub

- SolomonZelman

u=t²

- jim_thompson5910

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.