The integral of $3x^2$ with respect to $y$ is $3x^2y$ because $3x^2$ is considered a constant when integrating with respect to $y$. The integral of $4y^3$ with respect to $y$ is $y^4$. So, the inner integral becomes:
$$\int_{0}^{1}\left(3 x^{2}+4 y^{3}\right) d y
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