Question #c62e7

The area is bounded by the maroon lines and the blue dots.
So we can break down the region into 2 parts: #[-2, 0) uu [0,4]#
Note we are interested in the upper region so the area in each case is: #" Area of rectangle - Area under the curve "#

#[A_square - A_(y=x^2)] + [A_square- A_(y=x)]#
#[A_square - A_(y=x^2)] = 8 - int_-2^0 x^2dx = 16/3#
#[A_square- A_(y=x)]= 16/2 #
#A_T= 16/3 + 16/2 = 40/3#

Referencing the picture in the other answer, we can split this into two integrals. The first is the region bounded on the interval #-2<=x<=0# by #y=4# and #y=x^2#. The area of this region can be found by integrating the upper function, #y=4#, minus the lower function, #y=x^2#.

#int_-2^0(4-x^2)dx=[4x-x^3/3]_-2^0#

#=[4(0)-0^3/3]-[4(-2)-(-2)^3/3]#

#=0-[-8+8/3]#

#=16/3#

We have to add this value to the amount bounded by #y=x# and #y=4#, which is just a triangle.

The triangle has a height of #4# and a length of #4#, which creates a triangle of area #8#.

The area of #8# added to the other area of #16/3# gives a total area of #40/3#.

The region is shown below:

We can think of the region as the collection of points whose #y# coordinates vary from #0# to #4# while the #x# coordinates go from a minimum of #-sqrty# to a maximum of #y#. (The black horizontal lines in the image above.)

Note that the left branch of #y=x^2# gives us #x=-sqrty#.

On the right it is easy to see that #y=x# gives us #x=y#.

Our method is analogous to finding an area between #f(x)# and #g(x)# by integrating, with respect to #x#, the greater minus the lesser #y# values,

we can integrate with respect to #y#:
Greater #x# - lesser #x# or
"#x# on the right - #x# on the left#

#int_0^4(y-(-sqrty)) dy = int_0^4 (y+sqrty) dy#

# = y^2/2+2/3sqrty^3]_0^4#

# = 8 + 16/3 = 40/3#