242hw6solns - Math 242 Homework 6 Solutions(to the by hand...

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Math 242 Homework 6 Solutions (to the “by hand” questions) Questions 1 through 5 should be done by hand. For questions 6 through 8, you can use Maple as much or as little as you like. 1. Evaluate Z sin(ln x ) dx . (May be a little tricky.) Solution. Let I = Z sin(ln x ) dx , and integrate by parts with u = sin(ln x ) and dv = 1 dx . Then du = cos(ln x ) · 1 x dx and v = x . We then have I = x sin(ln x ) - Z cos(ln x ) dx. Integrate by parts again, with u = cos(ln x ) and dv = 1 dx . Then du = - sin(ln x ) · 1 x dx and v = x . This gives I = x sin(ln x ) - x cos(ln x ) - Z - sin(ln x ) dx I = x sin(ln x ) - x cos(ln x ) - Z sin(ln x ) dx 2 I = x sin(ln x ) - x cos(ln x ) I = 1 2 x sin(ln x ) - x cos(ln x ) + C. 2. Evaluate Z 0 x x 2 + 2 dx or show that it is divergent. Solution. Consider Z M 0 x x 2 + 2 dx . Substitute u = x 2 +2, so du = 2 x dx 1 2 du = x dx . The integral becomes 1 2 Z M 2 +2 2 1 u du = 1 2 ln( M 2 + 2) - ln 2 . This approaches as M → ∞ , i.e. the integral diverges. 1
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3. Use Newton’s Method to find a good approximation to 9999 1 / 4 . Solution. We want x that satisfies x 4 = 9999, so define f ( x ) = x 4 - 9999. Then f 0 ( x ) = 4 x 3 . A good first approximation is x 1 = 10. The next approximation (computable by hand) is x 2 = 10 - f (10) f 0 (10) = 10 - 10 4 - 9999 4(10) 3 = 10 - 1 4000 = 9 . 99975 . 4. Find the length of the curve y = ln(sec x ) between x = 0 and x = π/ 4. Solution. Note dy dx = 1 sec x · sec x
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