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20b_03f_fs

# 20b_03f_fs - Math 20B 1(a By parts e 3 4 1 Final Exam...

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Math 20B Final Exam Solutions Fall 2003 1. (a) By parts: integraldisplay e 1 x 3 ln x dx = ( x 4 / 4) ln x vextendsingle vextendsingle vextendsingle e 1 integraldisplay e 1 x 3 4 dx = parenleftBig ( x 4 / 4) ln x x 4 / 16 vextendsingle vextendsingle vextendsingle e 1 = 3 e 4 / 16+1 / 16 . (b) Since (2 + cos x ) /x 1 /x and integraltext π/ 2 0 (1 /x ) dx diverges, so does the given integral. (c) Let 1 + 2 u = t to obtain integraltext t 1 0 ( dt/ 2) = t 1 1 / 22 + C = (1 + 2 u ) 1 1 / 22 + C . (d) With x = sin t and dx = cos t dt , integraldisplay 1 / 2 0 x 2 1 x 2 dx = integraldisplay π/ 4 0 sin 2 t dt = integraldisplay π/ 4 0 1 cos 2 t 2 dt = t (sin 2 t ) / 2 2 vextendsingle vextendsingle vextendsingle vextendsingle π/ 4 0 = π/ 8 1 / 4 . (e) By partial fractions, integraldisplay 2 1 1 u + u 2 du = integraldisplay 2 1 parenleftbigg 1 u 1 1 + u parenrightbigg du = parenleftBig ln | u | − ln | 1 + u | vextendsingle vextendsingle vextendsingle 2 1 = (ln 2 ln 3) ( ln 2) = ln(4 / 3) . (f) With e t = u , integraltext e t + e t dt = integraltext e u du = e u + C = e e t + C . 2. (a) w = 1 + i 1 3 i 1 + 3 i 1 + 3 i = 2 + 4 i 10 = 1 5 + 2 i 5 . and w = 1 5 2 i 5 .

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20b_03f_fs - Math 20B 1(a By parts e 3 4 1 Final Exam...

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