Just need help with the area portion

Question

Find the equation of the tangent line to the curve  

                              y³ + y – 5x = 0    at  (2, 2).

         Find the area of the region bounded by the curve, the x-axis and the line x = 2.

Answer 1

x = (1/5)(y³ - y)

Now integrate: from 0 to 2: (1/5)∫(y³ + y) dy = 6/5

The area you need is 4-(6/5) = 2 4/5 since the area of the square with vertex at (2,2) is 4

Answer 2

  Find the area bounded by:

 y³ + y – 5x = 0, x = 2, and the x-axis, which is y = 0

To find the area between 2 functions, write the difference between them and then integrate over the appropriate interval.

This one is best done by writing x as a function of y:

boundary 1: x = f(y) = 1/5 (y³ + y)

boundary 2: x = 2

The third boundary is a y value, so that will be one limit of the definite integral. So---where is the other limit? The function is symmetrical (with a sign change), and so we can integrate from either + or - infinity to 0.

By inspection, the area in question is in fact infinite---AKA "undefined".

So, maybe there is an error in the problem statement??