Trouble with complex definite integral

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

1 Attachment(s)

Re: Trouble with complex definite integral

ooops sorry it should read Attachment 26084

Re: Trouble with complex definite integral

Quote:

Originally Posted by chizmin10

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

I suspect that you are to find the derivative of that integral with respect to $x$ .

There is no simple integral. See here..

Re: 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: $F(0)- F(\sqrt{x})$
If that really is $\sqrt{1+ t^4}$ dx, then F is just $x\sqrt{1+ t^4}$ and so the integral is $-\sqrt{x}\sqrt{1+ t^4}= -\sqrt{x(1+ t^4)}$ .

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:
$f(x)= \int_{\sqrt{x}}^0 \sqrt{1+ t^4}dt= -\int_0^{\sqrt{x}}\sqrt{1+ t^4}dt= -\int_0^u\sqrt{1+ t^4}dt$ with $u= \sqrt{x}$
Now use the "fundamental theorem of Calculus" to find $\frac{df}{du}$ and then the chain rule to find $\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}$ .