Worksheet1_1B_Solutions - Math 1B Worksheet 1 Solutions 1....

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Math 1B Worksheet 1 Solutions 1. Let m and n be constants. Evaluate R e mx sin nx dx . Let u = sin nx and dv = e mx dx . Then, we have du = n cos nx dx and v = e mx /m , and so Z e mx sin nx dx = 1 m e mx sin nx - n m Z e mx cos nx dx. Using integration by parts again, we set u = cos nx and dv = e mx dx . Thus, du = - n sin nx dx and v = e mx /m , and Z e mx sin nx dx = 1 m e mx sin nx - n m 2 e mx cos nx - n 2 m 2 Z e mx sin nx dx. Adding ( n 2 /m 2 R e mx sin nx dx to both sides yields ± 1 + m 2 n 2 ²Z e mx sin nx dx = 1 m e mx sin nx - n m 2 e mx cos nx. Rewriting (1 + n 2 /m 2 ) = ( m 2 + n 2 ) /m 2 and dividing finally gives us Z e mx sin nx dx = m m 2 + n 2 e mx sin nx - n m 2 + n 2 e mx cos nx. 2. In finding R xe x dx , we would choose u = x and dv = e x dx . Technically, we have many choices for v , namely v = e x + A for any constant A . Show that the answer you get for R xe x dx does not depend on A . Let
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Worksheet1_1B_Solutions - Math 1B Worksheet 1 Solutions 1....

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