Evaluate the integral

**mybrohshi5** · Dec 8th 2009, 08:14 PM

Evaluate:

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

This problem does not make sense to me?

I am reading it as find the derivative $(\frac {d}{dx})$ 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.

**Prove It** · Dec 8th 2009, 08:25 PM

Originally Posted by mybrohshi5

Evaluate:

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

This problem does not make sense to me?

I am reading it as find the derivative $(\frac {d}{dx})$ 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 $\int_0^1{\frac{d}{dx}\left(e^{\arctan{x}}\right)\, dx} = \left[e^{\arctan{x}}\right]_0^1$ .

**VonNemo19** · Dec 8th 2009, 08:26 PM

Originally Posted by mybrohshi5

Evaluate:

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

This problem does not make sense to me?

I am reading it as find the derivative $(\frac {d}{dx})$ 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

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

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.

**mybrohshi5** · Dec 8th 2009, 08:30 PM

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

how would i go about doing

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

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

**mybrohshi5** · Dec 8th 2009, 08:35 PM

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?

**VonNemo19** · Dec 8th 2009, 08:37 PM

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.

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