MATH 152 Midterm: MATH 152 TAMU 2010c Exam 2a Solutions
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Exam ii version a solutions: c as n , e n 0, so an . , so the terms of the sequence alternate between approaching. 3 (odd-numbered terms), which means the: d the denominator contains a repeating lin- ear factor and an irreducible quadratic factor, so the form of the partial fraction decompo- sition is. Cx + d x2 + 2x + 3. 2 : e let x = 2 sin . When x = 2, = sin 1(cid:18) 2. = /2 (2 cos )(2 cos ) d = 4 /2. /3 p4 4 sin2 (2 cos d ) /3: b the function is unbounded at x = 0, so we rewrite the integral as 0. The type-2 improper integral a x2 dx + 3. 1 x2 dx. verges if and only if p < 1, so both integrals diverge: e since sin2 x 1, a improper integral only if p > 1, so comparison with .
