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> Find the area of the region bounded by the parabola $y = 4x^2$, the tangent line to this parabola at $(2, 16)$, and the $x$-axis.
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I found the tangent line to be $y=16x-16$ and set up the integral from $0$ to $2$ of $4x^2-16x+16$ with respect to $x$, which is the top function when looking at the graph minus the bottom function. I took the integral and came up with $\frac{4}{3}x^3-8x^2+16x$ evaluated between $0$ and $2$. This came out to be $\frac{32}{3}$ but this was the incorrect answer. Can anyone tell me where I went wrong?