# Finding the area bounded by two curves

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.

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?

Hint: After drawing it, note that you have to calculate $\int_0^1 4x^2\;dx + \int_1^2 4x^2-16x+16\;dx$.
• I got $\frac{8}{3}$. I'm sorry but did you do it right? – Rodrigo Dias Nov 22 '16 at 0:22
• Any time! ${}{}$ – Rodrigo Dias Nov 22 '16 at 0:28
The tangent crosses the $x$ axis at $x=1$, so your integral is including (with the plus sign) also the triangle made by the tangent below the $x$ axis.
The correct way is to integrate only the parabola for $x=0 \cdots 2$ (which is $32/3$ and then subtract the area of the triangle$(1,0),(2,16),(2,0)$, which is $8$, so the net area is $8/3$ .