Find the area of the region bounded by the parabola y = 3x^2, the tangent line to this parabola at (1, 3), and the x-axis.

Answer 1

Well, you know the tangent line at (1,3) is

y-3 = 6(x-1)
y = 6x-3
x = (y+3)/6

See the graph at

http://www.wolframalpha.com/input/?i=plot+y%3D3x^2%2C+y%3D6x-3

Now, if you integrate along x, you have to split the region at x=1/2, where the lower boundary changes from the x-axis to the tangent line.

So, let's integrate along y, so that the tangent line is the right edge all the way up.

a = ∫[0,3] (y+3)/6 - √(y/3) dy = 1/4

As a check, you might want to integrate along x, so that

a = ∫[0,.5] 3x^2 dx + ∫[.5,1] (3x^2-(6x-3)) dx
= 1/8 + 1/8 = 1/4