evaluate Attachment 26083 i know how to definite integrals but i don't understand having two variables?

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- Dec 5th 2012, 05:04 AMchizmin10Trouble with complex definite integral
evaluate Attachment 26083 i know how to definite integrals but i don't understand having two variables?

- Dec 5th 2012, 05:18 AMchizmin10Re: Trouble with complex definite integral
ooops sorry it should read Attachment 26084

- Dec 5th 2012, 06:08 AMPlatoRe: Trouble with complex definite integral
I suspect that you are to find the derivative of that integral with respect to $\displaystyle x$.

There is no simple integral. See here.. - Dec 5th 2012, 06:23 AMHallsofIvyRe: Trouble with complex definite integral
First, please don't use the word "complex" in mathematics unless you are referring to complex numbers! Use "complicated" instead.

Second, what you have is really the same as a definite integral- find an "anti-derivative", F, and evaluate: $\displaystyle F(0)- F(\sqrt{x})$

**If**that really is $\displaystyle \sqrt{1+ t^4}$ d**x**, then F is just $\displaystyle x\sqrt{1+ t^4}$ and so the integral is $\displaystyle -\sqrt{x}\sqrt{1+ t^4}= -\sqrt{x(1+ t^4)}$.

However, I suspect that the problem**really**says (or meant to say) "dt" and, as Plato said, the anti-derivative of that function cannot be written in terms of "elementary functions". But you are*not*asked to find the integral, you are asked to find the derivative of that integral

To find the**derivative**, use the "fundamental theorem of Calculus" and the chain rule:

$\displaystyle f(x)= \int_{\sqrt{x}}^0 \sqrt{1+ t^4}dt= -\int_0^{\sqrt{x}}\sqrt{1+ t^4}dt= -\int_0^u\sqrt{1+ t^4}dt$ with $\displaystyle u= \sqrt{x}$

Now use the "fundamental theorem of Calculus" to find $\displaystyle \frac{df}{du}$ and then the chain rule to find $\displaystyle \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}$.