Find $ f $.
$ f'(t) = t + 1/t^3 $, $ \quad t > 0 $, $ \quad f(1) = 6 $
So for this problem, we are given our function to be f prime of X because X plus one over X cubed. That's what is called a plus next to the negative three and during the anti derivative. What we see is that ffx is going to equal squared over two, and then we'll also get it's, uh, negative, too right to buy. I'm negative, too. So this is our FX, and then we know that that's going to be plus C Well, What we also know is that f of one is six. So if F one is six, then that tells us that since that would ultimately be zero ff of one, it would give us one half plus a negative one half, said B zero. That means that sea must be six, which therefore means this would be our graph. Um, and that's taking the anti derivative, which will ultimately become extremely useful when we talk about integral