What is the definite integral of $y = x^3$ from $x = 1$ to $x = 4$?

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∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the antiderivative of $ x^3 $. ‖ ‖ step 2 ⋮ The antiderivative of $ x^3 $ is $ \frac{x^4}{4} + C $, where $ C $ is the constant of integration. ‖ ‖ step 3 ⋮ Evaluate the antiderivative from $ x = 1 $ to $ x = 4 $. ‖ ‖ step 4 ⋮ Substitute $ x = 4 $ into the antiderivative to get $ \frac{4^4}{4} = 64 $. ‖ ‖ step 5 ⋮ Substitute $ x = 1 $ into the antiderivative to get $ \frac{1^4}{4} = \frac{1}{4} $. ‖ ‖ step 6 ⋮ Subtract the value at $ x = 1 $ from the value at $ x = 4 $ to find the definite integral: $ 64 - \frac{1}{4} = \frac{255}{4} $. ‖ ∻Answer∻ ⚹ $ \frac{255}{4} $ or $ 63.75 $ ⚹ ∻Key Concept∻ ⚹Definite Integral of a Polynomial Function⚹ ∻Explanation∻ ⚹The definite integral of a polynomial function from $ a $ to $ b $ is found by evaluating the antiderivative at $ b $ and subtracting the antiderivative evaluated at $ a $.⚹◊What is the process for finding the definite integral of a function like $y=x^3$ from $x=1$ to $x=4$?⍭ Generate me a similar question◊

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