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

- anonymous

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

- chestercat

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- amistre64

anti derive it :)

- amistre64

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

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

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## More answers

- amistre64

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

if im wrong, let me know :)

- anonymous

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

- amistre64

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

- anonymous

isn't the interval [1, 4]?

- amistre64

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

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

- anonymous

Yes, I think.

- amistre64

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

- anonymous

y'=6x+6

- amistre64

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

- anonymous

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

- amistre64

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

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

- anonymous

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

- amistre64

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

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

- amistre64

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

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

- amistre64

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

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

- amistre64

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

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

- anonymous

Thanks anyway.

- amistre64

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

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