tailieunhanh - Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 91

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 91. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Integration by Parts The Product Rule in Reverse 881 The new integral is no simpler than the original but it is no more difficult either. We do integration by parts again. Then Let u e2x du 2e2x dv cos 3x dx dx v sin 3x 3 e2x sin 3x dx --e2x cos 3x - 3 3 e2x sin 3 sin 3x 2e2x dx e2x sin 3x dx - -- e2x cos 3x - e2x sin 3 9 e2x sin 3x dx . The original integral and the integral on the right are identical. We can solve algebraically for it. If you like you can denote the original integral by I and solve for I. e2x sin 3x dx e2x 3 22 cos 3x 9 e sin 3x e2 sin 3x dx ---------e 13 2x cos 3x ----e 13 2x sin 3x C I I REMARKS 1. Notice that in Equation we are missing a C This is because we never actually computed v du. We ve got to insert a C. 2. One of the key ingredients that makes this method work is that if sin x either is differentiated or integrated twice then we obtain sin x again and similarly with ex. 3. This integral could have been solved similarly by letting u sin 3x and du e2x dx and repeating this. It will not however be productive to let u sin 3x the first time and then u e2x the second time. EXERCISE Find e2x sin 3x dx by letting u be the trigonometric function. Then try using integration by parts twice but letting u e2x the first time and u sin 3x the second time in order to verify that this is unproductive. EXERCISE The integral f sin2 x dx can be computed most easily using the trigonometric identity sin2 x 2 1 - cos 2x . However it can also be solved using the techniques of Example . Do the latter. Check your answer by differentiating. In Example and Exercise we saw that integration by parts can be used to reduce the complexity of an integral. There are numerous reduction formulas that can be derived by integration by parts. Exercise below leads you through one such derivation. EXERCISE Derive the reduction formula sin x dx cos x sin 1 x ----------- n sin 2 x dx where n is an integer n 2 as .

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