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Numerade Educator

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Problem 33 Easy Difficulty

Find $ f $.

$ f'(t) = 4/(1 + t^2) $, $ \quad f(1) = 0 $

Answer

$$
f(t)=4 \tan ^{-1} t-\pi
$$

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Video Transcript

Okay, so we're being asked to find F and were given a promise to use people there for over one plus two squared and the effort one zero. So let's get right into it. So we're checking the integral are for over one foot two squared. Oh, no, thanks. Oh, one plus two square DT. We can pull out a common factor for injected into your one over one foot. He's squared DJ, and this is going to be equal to four Casian and worse, or tea for some constant C. And that's it. That is our function. And so now we're told that half of one s so it's good unplugging ones of four ten in words of one plus C is equal to zero. So they're mean. Well, tell members of one's private forces before Ty over four or the fourth cancel and it will be prosecuted because zero and C is equal to negative. Hi. So that means our function after Max, it is going to be going to four tangent and worse off key. This is I'm sorry again turned t minus pi, and that is our function