Question

use the limit process to find the area of the region between the graph of the function and the x-axis over the given interval. y = 64 − x3, [2, 4]

Answer 1

The area of the region between the function y = 64 - x^3 and the x-axis over the interval [2, 4] is found by integrating the function, resulting in 72 square units.

Step-by-step explanation:

To find the area of the region between the graph of the function y = 64 - x^3 and the x-axis over the interval [2, 4], we use the limit process, also known as integrating the function. We need to calculate the definite integral of the function from x = 2 to x = 4.

Step-by-step Calculation:

Set up the integral: ∫24(64 - x^3) dx.

Evaluate the integral by finding the antiderivative: The antiderivative of 64 is 64x, and the antiderivative of x^3 is ⅔x^4.

Apply the limits of integration to find the area: 64(4) - ⅔(4)^4 - (64(2) - ⅔(2)^4).

Simplify the expression to get the final answer: 256 - ⅔(256) - (128 - ⅔(16)) = 256 - 64 - 128 + 8 = 72 square units.

The area of the region between the function and the x-axis over the given interval is 72 square units.

Answer 2

The area of the region between the graph of the function y = 64 - x^3 and the x-axis over the interval [2, 4] is calculated using the definite integral, resulting in an area of 4 square units.

Step-by-step explanation:

To find the area of the region between the graph of the function y = 64 − x^3 and the x-axis over the interval [2, 4], we use the definite integral of the function from the lower limit of integration (2) to the upper limit of integration (4).

The process involves calculating the integral of the function, which gives us the total area under the curve from x = 2 to x = 4. This area calculation can be thought of as summing up infinitesimally small rectangles under the curve, with width dx and height f(x). This sum is represented by the integral of f(x) from x1 to x2, as shown in Figure 7.8.

The definite integral is therefore:

∫24 (64 - x^3) dx

Carrying out this integration gives us:

[½(64x) - ¼(x^4)] from 2 to 4

Plugging in the upper limit:

½(64*4) - ¼(4^4) = 128 - 64 = 64

And then the lower limit:

½(64*2) - ¼(2^4) = 64 - 4 = 60

Finally, we subtract the value of the integral at the lower limit from the value at the upper limit to find the total area:

64 - 60 = 4 square units