Calculate the area under the curve $y = x^3$ from $x = 1$ to $x = 4$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the area under the curve $y = x^3$ from $x = 1$ to $x = 4$, we need to evaluate the definite integral $\int_{1}^{4} x^3 \, dx$. ‖ ‖ step 2 ⋮ The integral of $x^3$ is given by $\int x^3 \, dx = \frac{x^4}{4} + C$. ‖ ‖ step 3 ⋮ We now evaluate this antiderivative at the bounds $x = 4$ and $x = 1$: $$ \left. \frac{x^4}{4} \right|_{1}^{4} = \frac{4^4}{4} - \frac{1^4}{4}. $$ ‖ ‖ step 4 ⋮ Simplifying the expression: $$ \frac{4^4}{4} = \frac{256}{4} = 64, $$ $$ \frac{1^4}{4} = \frac{1}{4}. $$ Thus, the area is: $$ 64 - \frac{1}{4} = \frac{256}{4} - \frac{1}{4} = \frac{255}{4} = 63.75. $$ ‖ ∻Answer∻ ⚹ The area under the curve $y = x^3$ from $x = 1$ to $x = 4$ is $63.75$. ⚹ ∻Key Concept∻ ⚹Definite Integral⚹ ∻Explanation∻ ⚹The definite integral of a function over an interval gives the area under the curve of that function within the specified bounds.⚹◊What is the definite integral of $y = x^3$ from $x = 1$ to $x = 4$?⍭ Generate me a similar question◊

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