Question #c62e7

3 Answers
May 8, 2016

The area is bounded by the maroon lines and the blue dots.
A_T= 40/3
enter image source here

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

May 9, 2016

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.

May 10, 2016

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

Explanation:

The region is shown below:

enter image source here

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