sol for exam 2 take home

U dv here treat x as a constant by integration by

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Unformatted text preview: e indefinite integral tx e−t dt . u dv Here, treat x as a constant. By “integration by parts”, we have tx e−t dt u tx · dv = − e−t u − e−t v x tx−1 dt v − du . In other words, tx e−t dt − e−t + x tx−1 dt = = tx · − e−t +C − tx e−t + C . The corresponding definite integral: tx e−t dt t=0 +∞ +∞ +∞ + t=0 − e−t 10 x tx−1 dt = − tx e−t t=0 . That is, +∞ +∞ t=0 tx e−t dt − x tx−1 e−t dt t=0 +∞ − tx e−t = = − t=0 tx e−t lim − t − +∞ → − 0x e−0 0 = 0 0. Here, the reason why the limit lim tx e−t t − +∞ → equals 0 is as follows: First, when x is a positive integer, then tx is a polynomial in t, and hence this is “Review of Lectures – XIV”, Formula A. Second, when x is not a positive integer, then choose any positive integer m which is bigger than x. Then 0 < tx e−t < tm e−t holds for an arbitrary real number t with t > 1...
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