Sketch the region bounded by the graphs of the algebraic functions & find the area of the region:
f(x)=10/x, x=0 , y=2, y=10
I started idk how to keep going y = 10/x for x -->x = 10/y
Huh?
Just sketch the graph. It's a smooth curve coming down from far up the y-axis, bending smoothly at about x=3, and continuing on out, getting closer to the x-axis. Any grapher can show this.
Your region of interest is a small triangular-ish strip beside the y-axis between y=2 and y=10.
So, the area is the Integral of 10/x from x=10/10 to x=10/2
Intgeral of 10/x dx is just 10 lnx
So, evaluate 10lnx[1,5]
= 10(ln5 - ln1] = 10ln5 = 16.09
To sketch the region bounded by the graphs of the functions and find the area, follow these steps:
Step 1: Sketch the graph of the function f(x) = 10/x. Since x cannot be equal to 0, the graph will have a vertical asymptote at x = 0. The function will be symmetrical about the line y = x, and as x approaches positive or negative infinity, the graph will get closer to the x-axis.
Step 2: Plot the points of intersection between the graph and the lines x = 0, y = 2, and y = 10. For x = 0, the point is (0, ∞), for y = 2, the point is (5, 2), and for y = 10, the point is (0.5, 10).
Step 3: Draw the three lines that bound the region, connecting the points of intersection on the graph.
Step 4: Now, calculate the area of the region. The desired region is enclosed by two curves and two vertical lines.
To find the area, integrate the difference between the upper function and the lower function with respect to x.
Since f(x) = 10/x is the upper function and x = 0 is the lower function, the area, A, is given by:
A = ∫[0.5, 5] (10/x) dx.
To evaluate this integral, we can use the property that the integral of 1/x is ln|x|. Thus, integrating 10/x will give 10ln|x|.
A = 10[ln|x|] evaluated from 0.5 to 5.
A = 10[ln(5) - ln(0.5)].
Finally, calculate the value of A to find the area of the region bounded by the graph.
To sketch the region bounded by the graphs of the given algebraic functions, we first need to plot the graphs of these functions on a coordinate plane.
Let's start by graphing the function f(x) = 10/x. Since this function has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0, it will have a shape similar to a hyperbola.
To plot the graph of f(x), choose some values of x and calculate the corresponding y values. For example, let's use x = -1, 1, 2, 5, and 10. Evaluating f(x) for these x-values, we get:
f(-1) = -10
f(1) = 10
f(2) = 5
f(5) = 2
f(10) = 1
Plotting these points on a graph, we can see that the graph of f(x) passes through the points (-1, -10), (1, 10), (2, 5), (5, 2), and (10, 1). Draw a smooth curve that connects these points.
Now, let's find the region bounded by the graphs of f(x), x = 0, y = 2, and y = 10. Looking at the graph, we can see that the region is enclosed by the x-axis (x = 0), the vertical line x = 0, the horizontal line y = 2, and the graph of f(x).
To find the area of this region, we need to calculate the definite integral of f(x) with respect to x, between the appropriate limits.
Since the region is bounded by the vertical line x = 0 and the graph of f(x), we need to find the definite integral of f(x) from x = 1 to x = 10 (as it is bounded by the points (1, 10) and (10, 1)).
Using the definite integral formula, the area of the region is given by:
Area = ∫(from x=1 to x=10) [f(x)] dx
Substituting the equation for f(x), we get:
Area = ∫(from x=1 to x=10) (10/x) dx
To integrate this function, we can use the logarithmic property of integration. So, the area can be calculated as:
Area = [10ln|x|] (from x=1 to x=10)
Evaluating this expression for the limits, we get:
Area = [10ln|10|] - [10ln|1|]
= 10ln(10) - 0
= 10ln(10)
Therefore, the area of the region bounded by the graphs of the functions f(x), x = 0, y = 2, and y = 10 is 10ln(10) square units.