If f is the antiderivative of x^2/(1+x^5) such that f(1)=0, then what would f(4) be?

Question

amistre64 · Answer

anti derive it :)

amistre64 · Answer

it gives you the equation to antiderive and the initial condition of (1,0).... the trick is figuring out a way to get it to antiderive

anonymous · Answer

The limits of integrals are from 1 to 4. I tried approximating it but f'(1) is not 0.

amistre64 · Answer

i dont think this its asking for the interval of integration when it says f(1) and f(4); its asking you to find the equation that this was gotten from, pinning it down to the point (1,0)  and getting the answer for (4,y)

amistre64 · Answer

if im wrong, let me know :)

anonymous · Answer

Will it be right to say? $$f = \int\limits_{a}^{b}x^2/(1+ x^5) dx$$
$$f' = x^2/(1+x^5)$$

amistre64 · Answer

that looks proper, except for the a and b part... it gave no interval to find an "area" for right?

anonymous · Answer

isn't the interval [1, 4]?

amistre64 · Answer

nope, there is no interval.  they want to know a specific value of f at x=4.  they give you f(1) = 0 so that you have something to anchor this f(x) with.  otherwise it just floats up and down the y axis like a roaming gnome.

amistre64 · Answer

tell me, can a derivative have more than 1 antiderivative?

anonymous · Answer

Yes, I think.

amistre64 · Answer

lets verify that :)

whats the derivative of these 2 equations:

y = 3x^2 +6x +10

y = 3x^2 +6x -3

anonymous · Answer

y'=6x+6

amistre64 · Answer

when you suit it back up it begins to float up and down the y axis right?

anonymous · Answer

I see how integrating a derivative can produce a family of functions with a different vertical translations.

amistre64 · Answer

good, then when we find a suitable antiderivative, we add a constant to it, a generic "+C" as a place holder; apply the "initial condition" that f(1)=0 and sove for "+C" then we have a valid function with which to determine the value of f(4)

amistre64 · Answer

the real issue becomes, what is the integral of that function :)  I have not seen an easy way to get it....

anonymous · Answer

Since my teacher rushed through approximation methods today, I guess Ill have to use that. Trapezoidal approximation maybe?

amistre64 · Answer

thats still looking to find the area under the curve, but that is not the question you posted above.  do you have the question right?

anonymous · Answer

Im pretty sure that was the question. But is the area under x^2/(1+x^5) f?

amistre64 · Answer

No, the area underneath x^2/(1+x^5)  is not f.

"f"  is the function that will originally be derived down to:

 x^2/(1+x^5)

amistre64 · Answer

its like you found someones wallet and are trying to find the owner by the limited clues available to you.

amistre64 · Answer

we know that it is some type of cubic rational function; that it has a critical point at x=0.... and if we take another derivative we might be able to see some other clues to it, but figureing out the actual function it came from will be tricky nonetheless

anonymous · Answer

I got an email from my teacher. She said to estimate it and gave three choices: 0.016, 0.376, 0.629.

amistre64 · Answer

then you try that trap rule and see if that gets you an answer near one of these, if so, then go for it :)  but i think I am right about it not being an "area" question.... but Ive been known to be wrong :)

anonymous · Answer

Haha... I'm so clueless in calc.

anonymous · Answer

Thanks anyway.

amistre64 · Answer

wish I coulda been more help :)  good luck!!