Use a graph to give a rough estimate of the area of the region that lies beneath the given curve. Then find the exact area.
y = x^−3, 2 ≤ x ≤ 5
You can see a nice graph with grid lines here:
http://rechneronline.de/function-graphs/
You may have to adjust the y scale some so it fits. You can use that to estimate the area. I'm sure you can then evaluate the integral.
As a Clown Bot, I must apologize as I am unable to create and show you a graph. However, I can provide a rough estimate and calculate the exact area for you.
Rough Estimate:
By visualizing the curve y = x^−3 within the given interval 2 ≤ x ≤ 5, we can see that it is a decreasing curve. Therefore, the area under the curve will be some sort of triangular shape.
Exact Calculation:
To find the exact area, we need to integrate the curve. Recall that the formula to find the area under a curve, in this case, is given by the definite integral:
A = ∫(from 2 to 5) x^−3 dx
Evaluating the integral, we get:
A = [-1/2x^−2] (from 2 to 5)
A = [-1/(2 * 5^2)] - [-1/(2 * 2^2)]
A = [-1/50] - [-1/8]
A = -0.02 - (-0.125)
A = 0.105
Therefore, the exact area under the curve y = x^−3, within the interval 2 ≤ x ≤ 5, is approximately 0.105 square units.
To estimate the area under the curve y = x^(-3) for 2 ≤ x ≤ 5, we can create a rough graph and outline the approximate area.
First, plot the curve y = x^(-3) on a coordinate plane.
Next, draw a rectangle with a width of 1 unit between each integer value within the given range. The height of each rectangle should be determined by the corresponding value of y = x^(-3).
Count the number of rectangles that intersect with the curve and estimate the area enclosed.
To find the exact area under the curve y = x^(-3) for 2 ≤ x ≤ 5, we need to evaluate the definite integral of the function over the given interval.
Integrating y = x^(-3) with respect to x gives us:
∫[2,5] (x^(-3)) dx
To perform this integration, we can simplify the expression by using the power rule for integration:
∫[2,5] (x^(-3)) dx = ∫[2,5] (1 / (x^3)) dx
Applying the power rule, we can rewrite the expression as:
= (-1/2) * (x^(-2)) |[2,5]
Evaluating the integral from 2 to 5 gives us:
= (-1/2) * ((5^(-2)) - (2^(-2)))
Simplifying further, we have:
= (-1/2) * (1 / 25 - 1 / 4)
= (-1/2) * (4 - 25)/100
= (-1/2) * (-21/100)
= 21/200
Thus, the exact area under the curve y = x^(-3) for 2 ≤ x ≤ 5 is 21/200 square units.
To give a rough estimate of the area of the region that lies beneath the curve y = x^(-3), 2 ≤ x ≤ 5, we can plot the graph and shade the area between the curve and the x-axis.
First, let's plot the graph:
1. Choose a scale for the x-axis and y-axis. Since the range of x is from 2 to 5, we can choose a scale of 1 unit for the x-axis and 1 unit for the y-axis.
2. Mark the x-axis from 2 to 5 in increments of 1 unit.
3. For each x-value, calculate the corresponding y-value using the equation y = x^(-3).
- When x = 2, y = (2^(-3)) = 1/8 = 0.125
- When x = 3, y = (3^(-3)) = 1/27 ≈ 0.037
- When x = 4, y = (4^(-3)) = 1/64 ≈ 0.016
- When x = 5, y = (5^(-3)) = 1/125 ≈ 0.008
4. Mark the corresponding points on the graph.
5. Connect the points with a smooth curve. Note that the curve will be decreasing as x increases.
Now that we have a rough estimate of the area beneath the curve, we can find the exact area using integration.
To find the exact area, we need to integrate the function y = x^(-3) over the interval [2, 5]. The integral represents the signed area bounded by the curve and the x-axis, and since the function is positive over the entire interval, we can calculate the absolute value of the integral to find the area.
The definite integral of x^(-3) with respect to x over the interval [2, 5] can be calculated as follows:
∫[2, 5] x^(-3) dx
= [(-1/2)x^(-2)] from 2 to 5
= [(-1/2)(5^(-2)) - (-1/2)(2^(-2))]
= [(-1/2)(1/25) - (-1/2)(1/4)]
= (-1/50) + (1/8)
= 13/200 ≈ 0.065
Therefore, the exact area under the curve y = x^(-3), 2 ≤ x ≤ 5, is approximately 0.065 square units.
estimate as a trapezoid:
height at x =2 is 1/8
height at x=5 = 1/125
average height = (1/8+1/125)/2 = 133/2000
appr area = (3)(133/2000)
= 399/2000
= appr .2
actual area
= ∫1/x^3 dx from 2 to 5
= ( - 1/(2x^2) ) from 2 to 5
= (-1/2)( 1/25 - 1/4)
= 21/200
= .105