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?
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.
a. To find A(1), we need to find the area of the rectangle when x = 1.
The rectangle is inscribed under the curve with vertices at (-x,0) and (x,0).
So when x = 1, the vertices of the rectangle are (-1,0) and (1,0).
The formula for the area of a rectangle is A = length * width.
In this case, the length is the distance between the x-coordinates of the vertices, which is 2 (1 - (-1) = 2).
The width is the y-coordinate of the vertices, which is 0.
Therefore, A(1) = length * width = 2 * 0 = 0.
b. To find the greatest value of A(x), we need to maximize the area of the rectangle.
Since the rectangle is inscribed under the curve, we can see that the height of the rectangle (y-coordinate of the vertices) can be any non-negative value. However, the length of the rectangle (distance between the x-coordinates of the vertices) is fixed at 2x.
To maximize the area, we need to maximize the product of the length and the width (height).
Since the length (2x) is fixed, the area will be maximized when the width (height) is maximized.
The width (height) is maximized when it is equal to the maximum y-coordinate of the curve.
Since the curve is not specified, we cannot determine the exact maximum value of A(x) without additional information. However, we can say that the greatest value of A(x) occurs when the width (height) is equal to the maximum y-coordinate of the curve.
c. To find the average value of A(x) on the interval 0 < x < 2, we need to calculate the average area of the rectangles for all values of x in that interval.
We can do this by integrating the function A(x) over the interval (0, 2) and dividing by the width of the interval (2 - 0 = 2).
A(x) = length * width = (2x) * 0 = 0
∫ A(x) dx = ∫ 0 dx = C
where C is the constant of integration.
To find the average value, we divide the integral ∫ A(x) dx by the width of the interval:
Average value of A(x) = (∫ A(x) dx) / (2 - 0) = C / 2
Therefore, the average value of A(x) on the interval 0 < x < 2 is C/2, where C is the constant of integration.
To find the area of the rectangle inscribed under the curve, follow these steps:
Step 1: Determine the equation of the curve.
Since the curve is described as having vertices at (-x, 0) and (x, 0), it is a straight line passing through the x-axis.
Step 2: Find the equation of the line.
To find the equation of the line, we need two points: (-x, 0) and (x, 0).
The slope of the line can be found using the formula: slope = (y2 - y1) / (x2 - x1).
In this case, since both y1 and y2 are 0, the slope becomes:
slope = (0 - 0) / (x - (-x))
= 0 / (2x)
= 0
Since the slope is 0, the equation of the line will be y = 0, which is the equation of the x-axis.
Step 3: Calculate the area of the rectangle.
The area of the rectangle can be calculated by finding the product of the length and width.
The length of the rectangle is the distance between the x-coordinates of the vertices, which is 2x.
The width of the rectangle will be the y-coordinate of any point on the curve, which in this case is 0.
Therefore, the area of the rectangle, A(x), is given by A(x) = length * width = 2x * 0 = 0.
a. Find A(1).
Substituting x = 1 into the formula, we have A(1) = 2 * 1 * 0 = 0.
b. What is the greatest value of A(x)? Justify your answer.
Since the area of the rectangle A(x) is always zero, there is no maximum value. The greatest value of A(x) is 0, which occurs for any positive value of x.
c. What is the average value of A(x) on the interval 0 < x < 2?
To find the average value of A(x) on the interval 0 < x < 2, we need to calculate the definite integral of A(x) from 0 to 2 and then divide it by the length of the interval.
Integrating A(x) from 0 to 2:
∫[0 to 2] A(x) dx = ∫[0 to 2] 0 dx = 0.
Dividing the integral by the length of the interval:
Average value = (1 / (2 - 0)) * ∫[0 to 2] A(x) dx = (1 / 2) * 0 = 0.
Therefore, the average value of A(x) on the interval 0 < x < 2 is 0.