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## Homework Statement

My book states that "The only way that we can show that the function [tex] f(x) = \sqrt {1 + x^{4}} [/tex] has an anti-derivative is to take a definite integral,

[tex]\int^{x}_{0} \sqrt {1 + t^{4}} dt[/tex]

This is a new function that cannot be expressed in terms of algebraic, trigonometric or exponential functions without calculus."

What are they really saying here? I don't understand the point. Why is the anti-derivative of

[tex] f(x) = \sqrt {1 + x^{4}} [/tex] an integral? And why do we switch the variables from x to t with respect to t?

**Second Part Of My Question**

From my book:

"The Fundamental Theorem can also be used to find the derivative of a function which is defined as a definite integral with a variable limit of integration. This can be done without actually evaluating the integral."

Examples:

Let

[tex] y = \int^{2}_{x} \sqrt{1+t^{2}} dt [/tex]

then

[tex] y = - \int^{x}_{2} \sqrt{1+t^{2}} dt [/tex]

and

[tex] dy = -d( \int^{x}_{2} \sqrt{1+t^{2}}dt) = -\sqrt{1+x^{2}} dx [/tex]

My question here arises from the same thing, why do they take a negative integral before taking a derivative and why does the derivative switch from t dt to x dx? The only reason I can see from the example is that x is the upper limit of the integral.. is that why?

A second example given:

Let

[tex] y = \int^{x^{2}+x}_{3} \frac {1} {t^{3}+1} dt [/tex]

Let

[tex] u = x^{2}+x [/tex]

Then

[tex] \frac {du}{dx} = (2x+1) [/tex] ....... [tex] y = \int^{u}_{3} \frac {1}{t^{3}+1} dt [/tex]

[tex] \frac {dy}{du} = \frac {1}{u^{3}+1} [/tex]

then by the chain rule

[tex] \frac {dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \frac{1}{u^{3}+1} (2x+1) = \frac {2x +1} {(x^{2}+x)^{3} +1} [/tex]

I have basically the same questions as before here, so any advice would be great.

The only thing that is after this in my book is questions for the chapter so I am kind of at a loss right now.

Thanks!