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Problem 12 Easy Difficulty

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

$ \displaystyle R(y) = \int^2_y t^3 \sin t \,dt $

Answer

$R^{\prime}(y)=-y^{3} \sin y$

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Video Transcript

before we do the fundamental theme of calculus. Let's flip the original function to make it easier to the fundamentals of calculus. Because now this is our Why so now we can do our prime of wire derivative. There's negative G over. Do you? Why? From to toe Why two cubed Santi DT Take away the tea's replaced them with wise, that's are variable and we end up with us is equivalent to our prime of why, yeah.