Need Upper Bound of an Integral
Hello everybody,
I need help with a Calculus Problem. Given is a function $f(t)=0,0025t^40,14t^3+1,8t^2$
The function describes the sales of a product in it's lifecycle.
The t is given in months and the y units sold in thousands.
The question is, at when point in time (at which exact value for $t$ ) are 750 thousand units sold ($f(t)=750$ ).
My approach was building the antiderivative and setting it equal to 750 ($\int_{0}^{x}f(t)\;\mathrm{d}t = 750$, where x is the boundary (the second xaxis section which is needed to solve the problem))
I tried to solve this but couldn't get a correct result. What would be the proper way to solve it?
(I would want not just the result but how you got there as well)
((I also haven't got much money to offer for this but I think it shouldn't be that demanding))

I should also add, that the answer should be in the Interval [0; 19]. This probably isn't necessary though, since I did a rough graphical approximation and I believe it was somewhere around the 17 months Mark.

(Please replace the commas with dots when doing this, I'm german and we note decimal places with a comma rather than a dot)

I think that the amount is quite low for this question, and also that it's poorly stated, since you're asking two separate things. I suppose the original text was in German, but right now it's not clear if f(t) is the amount of goods sold "in the tth month" or "until the tth month", and of course the problem is different in the two cases.

So f=0.0025t ^4−0.14t^3 +1.8t+8t^2 and it represents the number of units (in thousands) sold between the time 0 and t?

Alessandro is right. Please clarify what f(t) represents.

Also note that setting the antiderivative of that equal to 750 requires finding a zero of a 5th degree polynomial, which is impossible to do exactly except in some very special cases. Not sure, but I think it's unlikely that it's one of these cases.

f(t) represents the units sold in tth month. So if finding the zero to this 5th degree polynomial is likely to be impossible, what is the correct way to do it instead?

It would take about 20 minutes to answer this question. Your offered bounty is too low.

What would be a proper bounty then? I'm new to this platform and if course don't know such things... As I stayed already though, I don't really habe much more to offer. It's fine though. I'm sorry that I can't offer more and will accept the fact that my bounty is too low.
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As pointed out by other users you offered bounty was low, but I decided to answer as this seems to be the first time you are asking a question here. If possible, offer higher a bounty for your future posts to support the Matchmaticians community.
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