
Recall that a circle of radius 2 centered at the origin has equation

where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius

is

, it follows that the area of a quarter-circle would be

.
You have

, so the definite integral is equal to

.
Another way to verify this is to actually compute the integral. Let

, so that

. Now

Recall the half-angle identity for cosine:

This means the integral is equivalent to
Answer:
Translation
Step-by-step explanation:
To keep a shape congruent or the same then you can Translate or to slide the shape not to change any sides.
(5,0) and (8,4)
it’s the only one when graphed that the line (from left to right) goes up,
Answer:
Your selection is correct
Step-by-step explanation:
The graph of cos(3x) will have an amplitude of 1 and will have 3 periods in the interval 0 to 2π. It will cross the y-axis at (0, 1). That is the graph that is shown highlighted in your problem statement.
Your selection is correct.