I have to find the area of the region bounded by $y = 4x^2$, the tangent line at $(3, 36)$ and the $x$-axis.
I found the equation of the tangent line, which is: $y = 24x-36$ Both functions meet at $x = 3$ , so I have enclosed the region I am working with. However, I don't know how to find the area since no function "dominates" the other. So, I tried to find the area under the curve between $\left[0,\frac 32\right]$ and then, I added the area under the curve between $\left[\frac 32, 3\right]$. My result was $\frac {81}{2}$ but the right answer is $9$.
I'd be grateful if you guys can help me realize what I did wrong.
Notice, the equation of the tangent at $(3, 36)$ to the parabola: $y=4x^2$ $$y-36=\left(\frac{dy}{dx}\right)_{(3, 36)}(x-3)$$ $$y-36=24(x-3)\ \ or \ \ \ y=24x-36$$ The above tangent line intersects the x-axis at $\left(\frac 32, 0\right)$
Draw a diagram, drop a perpendicular from the point of tangency $(3, 36)$ to the x-axis at the point $(3, 0)$ & consider a vertical elementary slab of thickness $dx$ & height $y$ in the region bounded between parabola, tangent & the x-axis & move it from $x=0$ to $x=3$ covering bounded region & a region of right triangle, then the required area of bounded region
$$\int_{0}^{3}y\ dx-\text{area of right triangle}$$ $$=\int_{0}^{3}(4x^2)\ dx-\frac{1}{2}\left(3-\frac 32\right)(36)$$ $$=4\left[\frac{x^3}{3}\right]_{0}^{3}-27$$ $$=4\left[\frac{27}{3}-0\right]-27$$ $$=36-27=\color{red}{9}$$