1041微甲06-10班期末考解答和评分标准1.(14%).PDFVIP

1041微微微甲甲甲06-10班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準 1. (14%) (a) (7%) Find the indefinite integrals ˆ √ 1 dx (5%) and ˆ 1 dx (2%). (You may use the integral formula x2 + 1 x + 2 of ˆ sec θ dθ.) (b) (4%) Find the value of the constant a for which the improper integral ˆ ∞ √ 1 − a dx converges. 0 x2 + 1 x + 2 (c) (3%) Evaluate the improper integral for this a. Solution: (a)(7pts) [part1–(5pts)] ˆ √ 1 dx x2 + 1 2 let x = tan θ ⇒ dx=sec θ dθ (2pts) ˆ √ 1 dx x2 + 1 =ˆ sec2 θ dθ sec θ =ˆ sec θ dθ =ln | sec θ + tan θ | + c =ln | x2 + 1 + x | + c [part2–(2pts)] ˆ 1 dx x + 2 =ln | x + 2 | + c (b)(4pts) ˆ ∞ ( √ 1 − a )dx 0 x2 + 1 x + 2 = lim ˆ t ( √ 1 − a )dx t→∞ 0 x2 + 1 x + 2 consider the indefinite integral ˆ ( √ 1 − a )dx x2 + 1 x + 2 =ln | x2 + 1 + x | −a ln | x + 2 | + c Page 1 of 10 Let I=ˆ ∞ ( √ 1 − a )dx 0 x2 + 1 x + 2 I=ˆ ∞ ( √ 1 − a )dx 0 x2 + 1 x + 2 = lim ˆ t ( √ 1 − a )dx t→∞ 0 x2 + 1 x + 2 = lim ( ln(t2 + 1 + t) − ln 1 − a ln (t + 2) + a ln 2 ) t→∞ √t2 + 1 + t = lim ln( a ) + a ln 2 —————————————— (2pts) t→∞ (t + 2) √t2 + 1 + t Thus we only need to consider lim ln t→∞