Use series to approximate the value of the integral from 0 to 1 of xsin(x^3)dx w
ID: 3287219 • Letter: U
Question
Use series to approximate the value of the integral from 0 to 1 of xsin(x^3)dx with an error of magnitude less than 10^-8Explanation / Answer
f(x) = integral from 0 to x of t * sin(t^3) dt This would mean we are estimating f(1). We know that the series for sin(x) is of infinite radius of convergence and: sin(x) = x / 1! - x^3 / 3! + x^5 / 5! - x^7 / 7! + ... sin(x^3) = x^3 / 1! - x^9 / 3! + x^15 / 5! - x^21 / 7! + ... x * sin(x^3) = x^4 / 1! - x^10 / 3! + x^16 / 5! - x^22 / 7! + ... By the fundamental theorem of Calculus, we have f'(x) = x * sin(x^3) for any x. Since power series differentiate term-by-term, they also antidifferentiate term-by-term, so for some constant of integration, we have: f(x) = C + x^5 / (5 * 1!) - x^11 / (11 * 3!) + x^17 / (17 * 5!) - x^23 / (23 * 7!) + ... At x = 0, f(x) is the integration from 0 to 0, which is 0, so: 0 = f(0) = C Thus: f(x) = x^5 / (5 * 1!) - x^11 / (11 * 3!) + x^17 / (17 * 5!) - x^23 / (23 * 7!) + ... f(1) = 1 / (5 * 1!) - 1 / (11 * 3!) + 1 / (17 * 5!) - 1 / (23 * 7!) + ... The Leibniz Alternating Series Test tells us that the error (difference) between a partial sum of this series and the true limit f(1) will be smaller than the absolute value of the next term in the series. Notice that each term is smaller than the last, so we just need to find a term that is smaller than 10^(-8). We can find such a term by calculating terms until we find one. We don't even necessarily need a calculator. For example, 1 / (23 * 7!) will not be sufficient, because we would need: 23 * 7! > 10^8 23 * 7 * 6 * 5 * 4 * 3 * 2 * 1 > 10 * 10 * 10 * 10 * 10 * 10 * 10 * 10 ===> 24 * 7 * 6 * 5 * 4 * 3 * 2 * 1 > 10 * 10 * 10 * 10 * 10 * 10 * 10 * 10 8 * 7 * 6 * 5 * 4 * 3 * 2 * 3 > 10 * 10 * 10 * 10 * 10 * 10 * 10 * 10 This is not true, because we have 8 factors on either side, and the ones on the left are smaller than the ones on the right. Given the disparity, I think the next term, i.e. 1 / (29 * 9!) will probably not suffice either. Let's try the next one, i.e. -1 / (35 * 11!). We need: 35 * 11! > 10^8 35 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 > 10 * 10 * 10 * 10 * 10 * 10 * 10 * 10 35 * 11 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 > 10 * 10 * 10 * 10 * 10 * 10 * 10 35 * 11 * 9 * 8 * 7 * 6 * 4 * 3 > 10 * 10 * 10 * 10 * 10 * 10 7 * 11 * 9 * 8 * 7 * 6 * 2 * 3 > 10 * 10 * 10 * 10 * 10 11 * 9 * 8 * 7 * 6 * 14 * 3 > 10 * 10 * 10 * 10 * 10 11 * 9 * 8 * 21 * 6 * 14 > 10 * 10 * 10 * 10 * 10 11 * 18 * 24 * 21 * 14 > 10 * 10 * 10 * 10 * 10 which is true because we have the same number of factors on either side, and each factor on the left is strictly greater than factors on the right. So, an approximation that with precision up to 10^(-8) is: 1 / (5 * 1!) - 1 / (11 * 3!) + 1 / (17 * 5!) - 1 / (23 * 7!) + 1 / (29 * 9!) ... (the next term is smaller than 10^(-8)) = 1198335731/6465951360 = 0.18533015 (8 dp)Related Questions
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