The area is (1/2)*2x * x^ = x^2
A(1) = 1
A(x) is unbounded for x>0
For the average value, integrate A(x) = x^2 from -1 to +1 and divide by 2.
Looks like 1/3, but check my work.
Let A(x) be the area of the rectangle inscribed under the curve
with vertices at (-x,0) and (x,0), x > 0.
a. Find A(1).
b. What is the greatest value of A(x)? Justify your answer.
c. What is the average value of A(x) on the interval 0 < x < 2?
1 answer