Question

Question #c62e7

Answer 1

The area is bounded by the maroon lines and the blue dots.
`A_T= 40/3`

Explanation:

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`

Answer 2

`40/3`

Explanation:

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`.

Answer 3

Here is a third way to find the answer to this question. Rotate your thinking `90^@`

Explanation:

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`