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 (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from 0 to 3 of 1/√(x + 1) dx

∫ from 0 to 3 of 1/√(x + 1) dx

Accepted Answer

Welcome back everyone. In this problem, we want to determine the minimum number of subintervals needed so that Simpson's rule approximates the integral of 1 divided by the square root of x + 3 with respect to x between the bones of 1 and 4 with an error of magnitude less than 104. A says it's 2, B4, C 22, and D says it's 24. Now what can we do that can help us to figure out this minimum number of subintervals? What do we know? Well, recall. OK, that for an integral, a definite integral between the limits of A and B of F of X with respect to x, OK, then the error bound formula. For Simpson's rule basically states that the absolute value of our error is going to be less than or equal to b minus a 5th divided by 180 n 4th multiplied. Let me write that 4 a little bit more clearly here, multiplied by the maximum value of X between A and B, OK, of the 4th derivative of F of X. So if we can first figure out the 4th derivative of fx, we should be able to find the max value, plug it into our error bound formula, and thus solve for the number of subintervalsn. So let's go ahead and do that. Let's first start by trying to compute the 4th derivative of f of x. Now we know here from our problem that F of X is equal to 1 divided by the square root of x + 3, OK, or. We could rewrite it here as x + 3 1/2. In that case, then the first derivative of FFX is going to be a negative half of. X + 3 to the power of -3 halves. The second derivative, OK, is going to be equal to -122 multiplied by -3 halves multiplied by x + 3 to the power of -5 halves multiplied by the derivative of x + 3, which is 1. And that is going to be equal to 3/4 of x + 3 to the power of -5 halves. Next or third derivative is going to be equal to 3/4 multiplied by -5 halves multiplied by x + 3 to the -7 halves multiplied again by the derivative of x + 3 which is 1, and this equals -15/8 of x + 3 to the -7 halves. Finally, then, our 4th derivative is going to be equal to -15/8. Multiplied by -7 halves multiplied by x + 3 to the -9 halves multiplied by the derivative of x + 3, which is 1, and this is going to be equal to 105/16 of x plus 3 to the -9 halves. So now we have the fourth derivative of f of x. Next, we need to find the maximum value, OK, we need to find the maximum value of the absolute value of the 4th derivative of F of X, OK, on the interval from 1 to 4. And now if we look at the 4th derivative we can tell here that it's a decreasing function on our interval, so the maximum value will be at x equals 1. Therefore, our maximum value will be 105/16 multiplied by 1 + 3 to the -9 halves. 1 + 3 is 4, and here. We could rewrite this then as 105 divided by 16 multiplied by 512. That would be the value of 4 to -9 halves, and this equals 105 divided by 8,192. So now we have our max value of the 4th derivative of F of X, so we can plug it into the error bound formula because remember we said that we want, we want to find an error of magnitude less than 10-4. So this means then. That the absolute value for error, OK, which is less than or equal to 4 minus 15th divided by 180 n 4th multiplied by our maximum value 1. 105, sorry. Divided by 8,192 is going to be less than 10 to -4. So now let's see if we can use this to help us to solve for n, OK. Now in this case, that means 335 divided by 180 multiplied by 105 divided by 8,192. Multiplied by 10 to 4 is going to be less than n 4. That means then that n 4th is going to be greater than this expression. And if we find a 4th root of both sides, that means n, OK, is going to be greater than the 4th root. The fourth root of 3 to 5th divided by 180 multiplied by 105 divided by 8,192 multiplied by 104. Now if we think about it here, if we calculate this expression, we know that that is approximately equal to 3.63. So n is approximately equal to 3.63, and the smallest even integer that is greater than 3.63 is 4. Therefore, OK. The minimum number. Of sub intervals. To guarantee. That the error. Is less than 104. Is going to be 4. 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 (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)∫ from 0 to 3 of 1/√(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 (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from 0 to 3 of 1/√(x + 1) dx