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Find $ f $.

$ f"(t) = \sqrt[3]{t} - \cos t $, $ \quad f(0) = 2 $, $ \quad f(1) = 2 $

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$$F(x)=9 / 28 t(7 / 3)+\cos (t)-0.32 t+1$$

03:17

Wen Zheng

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 9

Antiderivatives

Derivatives

Differentiation

Volume

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So for this problem were given a second derivative function which is the cubed root. So I have the cube root here of T. Broke our X minus cosine X. And what we see here is that when we take the anti derivative, what we're going to end up getting is, and this is X to the 1/3. We'd end up getting, we have to add the exponents, we get to the 4/3 and then we divided by 3/4 and then we have to apply the initial condition so then we repeat this process. Given the initial conditions we have and our final answer is going to end up being 9/28. Mhm. As a result of I'm taking the integral twice Times T to the 7/3, we add the exponents twice. And then that's going to be a code sine of X. So call it X minus cosine, X plus cosine X. In this case because we picked the integral twice and then it'll be minus. This is where the initial condition come in .32 T if you do X plus one side what the initial condition cost. And based on that, we see that if we were to call this F of X that are F prime of X or F double prime of X is going to end up giving us the exact same function that we were expecting, which was the cubed root of X plus cosign X.

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