The graph of is a horizontal line. So the area under
on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.
The graph of is the upper half of a circle with radius
. It's symmetric about
, so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is
.
Then use the fact that the definite integral is distributive over sums, meaning
We are given coordinates of the triangle: A(2,2), B(7,1) and C(8,-4).
We need to rotated 90° counterclockwise about the origin.
In order to find the new coordinates of rotatation 90°counterclockwise about the origin, we can apply rule (h, k) ---> (-k,h).
Where (h,k) are the coordinates of original image on axes and (-k,h) are the coordinates of rotated image.
In resulting coordinates of the image first swap the x and y coordinates of the original image and then make the sign opposite of each x-coordinate.
On applying rule (h, k) ---> (-k,h), we get
A(2,2) --> A'(-2,2)
B(7,1) --> B'(-1,7)
C(8,-4) --> C'(4,8)
