Hello! I am trying to create something to study from, so if someone could give the correct answers and explanations I would be very grateful!
1. Measurements of a lake’s width were taken at 15-foot intervals, as shown:
x: 0, 15, 30, 45, 60, 75, 90, 105, 120
f(x): 0, 15, 18, 20, 19, 23, 24, 22, 12
Estimate (120 on top, 0 on bottom) ∫f(x)dx with n = 4, using each of the following methods.
A. Left-hand approximation
B. Right-hand approximation
C. Trapezoidal approximation
D. Midpoint
2. Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f shown below is composed of two semicircles.
(graph) gyazo.com/3520e36ca7d1770b2690116ca742e7e5
A. Determine the domain of h(x).
B. Find h'(5/2)
C. At what x is h(x) a maximum? Show all the analysis that leads to your conclusion.
since you want 4 subintervals to cover the distance, each will have width=30. So,
∫[0,120] f(x) ≈
4
∑ 30f(xk)
k=1
where xk is the left, middle, or right value for the subinterval.
The trapezoidal rule, or course, averages the left and right values of f(xk), since those are the bases of the trapezoids.
For the graph,
clearly the domain of f(t) = h'(x) is [0,6], since f is continuous there.
But the domain of h(x) is not the same, since for 0 < t < 1/2, 2t-1 < 0, so f(2t-1) is undefined there. Similarly, for t>7/2, 2t-1 > 6, so f is undefined there.
So, the domain of h(x) is [1/2, 7/2]
at t=5/2, f(t) is a circle with radius 2, with center at (2,0). So,
f^2(5/2) + (1/2)^2 = 2^2
f(5/2) = √15/2 = h'(5/2)
This one is kinda tricky, since h(x) is not just ∫[0,x] f(t) dt
To see the graph of h(x), go to desmos.com (a very handy tool!) and enter two lines of text:
f(t) = {0<=t<=4: sqrt(4-(t-2)^2), 4<t<=6: sqrt(1-(t-5)^2)}
h(x) = int0^2x-1v f(t) dt
it will do typesetting for you, and the ^ in the integral means hit the up-arrow so it changes from the lower limit to the upper limit, and the "v" indicates the down-arrow, so it drops down to the line for the integrand.
h has a max when h' = f = 0, and h" = f' < 0
That is, at t=4
Thank you so much for the help!
Do you think you could expand a bit on the Left-hand approximation Right-hand approximation, Trapezoidal approximation, and Midpoint?
Also, is there any way of doing part 2C without graphing? Again, thank you so much for using up your time and effort to help me!
Sure! I'd be happy to help you with these questions.
1. To estimate the integral ∫f(x)dx using different methods, let's break down the process for each approximation:
A. Left-hand approximation:
To use the left-hand approximation, you divide the interval [0, 120] into equal subintervals of width h = (120 - 0) / n, where n is the number of subintervals. In this case, n = 4, so h = (120 - 0) / 4 = 30.
Then, for each subinterval, you evaluate f(x) at the left endpoint of the subinterval and multiply by h. So, you calculate:
Σ[f(x_i) * h], where x_i = 0 + (i * h) for i = 0 to 3.
B. Right-hand approximation:
Similar to the left-hand approximation, you divide the interval [0, 120] into equal subintervals of width h = (120 - 0) / n. Then, for each subinterval, you evaluate f(x) at the right endpoint of the subinterval and multiply by h. So, you calculate:
Σ[f(x_i) * h], where x_i = h + (i * h) for i = 0 to 3.
C. Trapezoidal approximation:
Using the trapezoidal approximation, you divide the interval [0, 120] into equal subintervals of width h = (120 - 0) / n. Then, for each subinterval, you calculate the area of the trapezoid formed by the function values at the two endpoints. Finally, you sum up these areas to get the estimate of the integral.
D. Midpoint approximation:
For the midpoint approximation, you divide the interval [0, 120] into equal subintervals of width h = (120 - 0) / n. Then, for each subinterval, you evaluate f(x) at the midpoint of the subinterval and multiply by h. So, you calculate:
Σ[f(x_i+1/2) * h], where x_i+1/2 = (x_i + x_i+1) / 2 and x_i = 0 + (i * h) for i = 0 to 3.
2. Let's address the questions about the functions in the graph:
A. To determine the domain of h(x), you need to look at the graph. In this case, since the given graph shows two semicircles, the domain of h(x) would be the same as the x-values that lie within those semicircles.
B. To find h'(5/2), you need to take the derivative of the function h(x) with respect to x and then evaluate it at x = 5/2. Since the graph shows semicircles, you can identify that h(x) is a combination of two different functions, one for each semicircle. Derive each semicircle function separately and then, at x = 5/2, plug in the value to find the derivative.
C. To determine at what x h(x) is a maximum, you need to find the critical points of h(x) by finding where the derivative h'(x) equals zero or is undefined. You also need to consider the endpoints of the given closed interval [0, 6]. Then, you evaluate h(x) at the critical points and the endpoints to determine which value gives the maximum.
I hope this explanation helps! If you have any further questions or need more clarification, feel free to ask.