Find the area of the region bounded by the parabola y = 4x^2, the tangent line

Similar question:

Find the area of the region bounded by the parabola y=5x2, the tangent line to this parabola at (3,45) and the x axis.

The answer:

The parabola is y = 5x².

Its gradient = 10x so at x=3 the parabola, and thereby also the tangent, have gradients of 10×3 = 30. The tangent passes through (3, 45) and therefore also – due to its gradient of 30 - through (0, -45) This makes the equation of the tangent:

y = 30x-45

Here is a sketch of the situation – just click on it:
http://www.wolframalpha.com/input/?i=y%3D5x%5E2+and+y%3D30x-45+

The tangent crosses the x axis at x = 1½ and this makes a change in the calculation at that point. We have to integrate from 0 to 1½ and then (with a different formula from 1½ to 3. I must assume you only want area in the region x≥0.
 
The calculation is then :

Area A = ∫ [5x²].dx from 0 to 1½
+ ∫ [5x²-30x+45].dx from 1½ to 3
A = [5x³/3] (from 0 to 1½) +[5x³/3-15x²+45x] (from x=1½ to 3)
5.625 + 45 – 135 + 135 – 5.625 + 33.75 – 67.5 = 11.25 square units.
which, by visual inspection, looks to be about right!