In Exercises 11–22, estimate the minimum number of subinterval...

Question

In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (a) the Trapezoidal Rule (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from 0 to 2 of sin(x + 1) dx

∫ from 0 to 2 of sin(x + 1) dx

Accepted Answer

Welcome back everyone. In this problem, determine the minimum number of subintervals needed so that the trapezoidal rule approximates the integral of the sin of x + 2 with respect to x between the limits of 1 and 3 with an error of magnitude less than 10 to -4. A says 7, B 8, C 81, and D 82. Now for us to determine the minimum number of subintervals required, OK, then we'll need to use what we know about the error of magnitude, since that's the condition here. What do we know? Well, recall that the error bond. For the trapezoidal rule basically says that the absolute value of our error or the magnitude of the error is going to be less than or equal to b minus a cubed divided by 12 n 2 multiplied by the maximum value between a and b of the magnitude of the second derivative of f of x. And that is for the integral of fx with respect to x between the bones of A and B. Now our expression is already in that form, so we can tell that a equals 1 here and b equals 3 and F of x is the sign of x + 2. So we'll need to first find the second derivative of f of x, use that to help us find the maximum value of its magnitude, and then plug that into our error bound to help us solve for the number of subintervals needed n. So let's go ahead and do that. So for starters, given FFX. Equals the sin of x + 2. OK. Then that means the first derivative of FX is going to be equal to the cosine of x + 2. And the second derivative of FX will be equal to the negative sin of x + 2. Therefore, the magnitude of the second derivative of fx is going to be equal to the sin of x + 2 or the magnitude of the sin of x + 2, OK. So no, this is what we need to find the maximum of, OK? No Here So now we're solving for the maximum. Let's just make a note here, solving for the maximum value. Of the magnitude of the sin of x + 2 for the interval for x between 1 and 3. When we're going to find this maximum value, we first have to ask ourselves, what do we know about the sine function. Well, we know that sine theta is a standard sine function. Its values always lie between -1 and 1 for all real values of theta. In other words, the absolute value of sine theta is less than or equal to 1 for all values of theta. Now, the interval for x is from 1 to 3, so the argument inside the sine function. OK, X + 2 is going to be in the interval from 1 + 2 to 3 + 2. OK, we're basically adding 2 to either side of the interval or moving our interval by 2, which gives us a new interval from 3 to 5. And know on the interval from 3 to 5, the sin of x + 2 will take values between -1 and 1, and we want to find the maximum of the magnitude of the sin of x + 2. Now the sine function reaches its maximum magnitude of 1 at points where the argument is an odd multiple of a half of pi, OK. So at the maximum magnitude. OK. Then our argument x + 2 will be an odd multiple of a half of pi, so it will be equal to 12 of pi plus k pi, where k is an integer. So we need to check if any of these points fall within the interval from 3 to 5, and fork being equal to 1, then we'll get a half of pi plus pi, which equals to 3 halves of pi, or approximately 4.712, and that value lies in 35, OK, which lies in our interval. So since it lies in our interval at x + 2 equals 3 halves of pi, the sine function, the sin of x + 2 equals -1, so the absolute value is 1, OK. Therefore, OK, the maximum absolute value. And on the interval from 1 to 3. Of the, the, the, well, the maximum absolute value of the sin of x + 2 on the interval is exactly 1, so the maximum value here of our second derivative. OK, the maximum value of our second derivative of F of X is also going to be 1. So since we have the maximum value of our second derivative, now we can use it into our errorborne formula because that means that here our error borne 3 minus 1 to 3rd divided by 12 n 2 multiplied by 1, our maximum value will be less than 10 4. If we do some math here, then this is going to be 2. Uh, it's gonna be 2 cubed, OK, and if we do, do some more simplifying here, that will be 8, and 8 divided by 12 is 2/3. So really this will be 2 divided by 3 and squared, that will be less than 10-4. If we rearrange our equation, then, or sorry, or inequality, then we'll have n2 being greater than 2 divided by 3 multiplied by 10 to 4. And then if we square root both sides, then we'll get n to be greater than 2 divided by 3 multiplied by 10 to the 4th and that value is approximately 81.65. Now we can't have 81.65 subintervals and remember that here. That means we'll need to round up to the next whole number to ensure that the number of subintervals is sufficient to keep the trapezoidal rule error within the desired limit. Thus we can say that the minimum number. Of sub intervals required. Is 82. When we look back at our answer choices, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (a) the Trapezoidal Rule (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)∫ from 0 to 2 of sin(x + 1) dx

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In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (a) the Trapezoidal Rule (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from 0 to 2 of sin(x + 1) dx