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Calculus Challenge 2001 R 1. Evaluate the integral 1 dxx3 The method of partial fractions and completing the square yields u? 32 x ? x3 = ( x ? x2 ?x ) = ( x ? u2 34 ) where u = x ? . This leads to R 3dx = (ln(x + 1) ? ln(u + ) + p tan? ( pu )) = (ln( p x ) + 3 x2 ?x 2 px ? xp? p p this ?we p?obtain p R 13tandx3 (= ( p). From ? 3tan ( )) = ( + ) = . X 0 1 +1 1 3 1 +1 1 2 0 2 +1 1 3 +1 +1 1+ 1 3 1 2 1 +1 2 1 2 + 3 4 3 2 2 3 1 3 3 3 +1 1 3 1 +1 1 3 1 3 3 2 1 1 2 1 6 3 2 9 2. Show that xe ex for all positive real numbers x and determine all number x for which equality holds. 1 1 1 The inequality is equivalent to x x e e . Let f (x) = x x . Then f 0 (x) = 1 x x ( ?xln2 x ). This is positive if 0 < x < e and is negative if e < x. Therefore, f (x) has an absolute maximum at x = e and so f (x) f (e) for all x > 0 and equality holds if, and only if, x = e. Thus xe ex for all x > 0 and equality holds if, and only if, x = e. 1 ( ) 3. Let > ?1 and > ?1. Calculate limn!1 n?R nn Pn k 1 dx limn!1 n? nn = limn!1 Pkn=1 nnk = R01 xx dx = 1 +2 + 1 +2 + 1 +2 + 1 +2 + + + k=1 ( ) ( ) 0 + + +1 +1 ln 4. Find the sum of the following series: P1 n en n P nxn = P1 ln n = ln(2) P1 ne?n = ln(2)f ( ) where f (x) = 1 n n nP e e n )0 = x( 0 = x 2 for jxj < 1. Thus the value of the series x( 1 x ) n ?x ?x e?1 ln eln is ?e?1 2 = e? 2 . (2n ) =1 1 (2 ) =1 1 =0 (1 (2) ) ( 1 (2) 1) =1 (1 ) =1 R 5. Find the antiderivative: pedx a ?ex x The substitution sin = ee a22 reduces the integral to R a cosd = 2e ?2a R cscd = ?2e ?2a lnjcot+cscj+C = ?2e ?2a ln( pea?exx e 2 sincos e2 p ?a ?a a C = xe 2 ? 2e 2 ln(e 2 + ea ? ex ) + C 2 ? n converge or diverge? Justify your answer. 6. Does the series P1 n ne This converges because of the integral test. Using the substitution with integration by parts, Ru1=xex?ptogether x dx = R 1 2u e?u du = 2(?u e?u ? 3u e?u ? 6ue?u ? 6e?u )j1 = p =1 2 1 32 e 1 3 3 2 1 e a + 2 )+