Thread: Evaluate the integral

Evaluate:


$\displaystyle \int ^1_0 \frac {d}{dx} (e^{arctan(x)}) dx $


This problem does not make sense to me?

I am reading it as find the derivative $\displaystyle (\frac {d}{dx}) $ of the antiderivative? .... which does not make sense to me at all.

I guess im just unsure how to do this problem in general.

Any help would be great.

Thank you.

Originally Posted by mybrohshi5

Evaluate:


$\displaystyle \int ^1_0 \frac {d}{dx} (e^{arctan(x)}) dx $


This problem does not make sense to me?

I am reading it as find the derivative $\displaystyle (\frac {d}{dx}) $ of the antiderivative? .... which does not make sense to me at all.

I guess im just unsure how to do this problem in general.

Any help would be great.

Thank you.

Surely the antiderivative of a derivative is the original function...

So $\displaystyle \int_0^1{\frac{d}{dx}\left(e^{\arctan{x}}\right)\, dx} = \left[e^{\arctan{x}}\right]_0^1$.

Originally Posted by mybrohshi5

Evaluate:


$\displaystyle \int ^1_0 \frac {d}{dx} (e^{arctan(x)}) dx $


This problem does not make sense to me?

I am reading it as find the derivative $\displaystyle (\frac {d}{dx}) $ of the antiderivative? .... which does not make sense to me at all.

I guess im just unsure how to do this problem in general.

Any help would be great.

Thank you.

The integral

$\displaystyle \int_a^b\frac{d}{dx}f(x)dx=f(b)-f(a)$...

Think about it. You take the derivative of a function, then integrate the function, leaving you with the original function. I think the point to the excersie is to show that differentiation and integration are inverse processes.

Thank you. that does make sense i was not thinking straight at that moment haha.


how would i go about doing

$\displaystyle \frac {d}{dx} \int ^1_0 e^{arctan(x)} dx $


Thank you. This is helping me understand a lot for my final exam coming up =)

Would i evaluate the integral first and then take the derivative of that number?

if so the derivative of a number is just 0 so is that the answer to this second one?

just zero?

Originally Posted by mybrohshi5

Would i evaluate the integral first and then take the derivative of that number?

if so the derivative of a number is just 0 so is that the answer to this second one?

just zero?

Yes. In this case, we are dealing with a definite integral. This implies that you are taking the derivative of a constant.