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# ch06_02 - 514 CHAPTER 6 Integration Techniques 6-6 46 sec x...

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514 CHAPTER 6 . . Integration Techniques 6-6 46. sec x dx and sec 2 x dx 47. Find 2 0 f ( x ) dx , where f ( x ) = x / ( x 2 + 1) if x 1 x 2 / ( x 2 + 1) if x > 1 48. Find 2 2 f ( x ) dx , where f ( x ) = xe x 2 if x < 0 x 2 e x 3 if x 0 49. Rework example 1.5 by rewriting the integral as 4 x + 4 2 x 2 + 4 x + 10 dx 3 2 x 2 + 4 x + 10 dx and complet- ing the square in the second integral. EXPLORATORY EXERCISES 1. Find 1 1 + x 2 dx , x 1 + x 2 dx , x 2 1 + x 2 dx and x 3 1 + x 2 dx . Generalize to give the form of x n 1 + x 2 dx for any positive integer n , as completely as you can. 2. Find x 1 + x 4 dx , x 3 1 + x 4 dx and x 5 1 + x 4 dx . Generalize to give the form of x n 1 + x 4 dx for any odd positive integer n . 3. Use a CAS to find xe x 2 dx , x 3 e x 2 dx and x 5 e x 2 dx . Verify that each antiderivative is correct. Generalize to give the form of x n e x 2 dx for any odd positive integer n . 4. In many situations, the integral as we’ve defined it must be extended to the Riemann-Stieltjes integral con- sidered in this exercise. For functions f and g , let P be a regular partition of [ a , b ] and define the sums R ( f , g , P ) = n i = 1 f ( c i )[ g ( x i ) g ( x i 1 )]. The integral b a f ( x ) dg ( x ) equals the limit of the sums R ( f , g , P ) as n → ∞ , if the limit exists and equals the same number for all evaluation points c i . (a) Show that if g exists, then b a f ( x ) dg ( x ) = b a f ( x ) g ( x ) dx . (b) If g ( x ) = 1 a x c 2 c < x b for some constant c with a < c < b , evaluate b a f ( x ) dg ( x ). (c) Find a function g ( x ) such that 1 0 1 x dg ( x ) exists. 6.2 INTEGRATION BY PARTS At this point, you will have recognized that there are many integrals that cannot be evaluated using our basic formulas or integration by substitution. For instance, x sin x dx cannot be evaluated with what you presently know. We improve this situation in the current section by introducing a powerful tool called integration by parts . HISTORICAL NOTES Brook Taylor (1685–1731) An English mathematician who is credited with devising integration by parts. Taylor made important contributions to probability, the theory of magnetism and the use of vanishing lines in linear perspec- tive. However, he is best known for Taylor’s Theorem (see section 8.7), in which he generalized results of Newton, Halley, the Bernoullis and others. Personal tragedy (both his wives died during childbirth) and poor health limited the mathematical output of this brilliant mathematician. Wehaveobservedthateverydifferentiationrulegivesrisetoacorrespondingintegration rule. So, for the product rule: d dx [ f ( x ) g ( x )] = f ( x ) g ( x ) + f ( x ) g ( x ) , integrating both sides of this equation gives us d dx [ f ( x ) g ( x )] dx = f ( x ) g ( x ) dx + f ( x ) g ( x ) dx . Ignoring the constant of integration, the integral on the left-hand side is simply f ( x ) g ( x ). Solving for the second integral on the right-hand side then yields f ( x ) g ( x ) dx = f ( x ) g ( x ) f ( x ) g ( x ) dx .

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