高等数学C答案(2016-2017学年第一学期期末A卷) 1.(1) dxe2+1'dr2(e2+1)3 1(3)x+ln2 14 001 (7)p0时f(a)>0得到x>0时 f(x)>0.,又由f(0)=0得到x>0时f(x)> 3ex+1 3.(1)f(x) f"(x) 驻点为r1=lm 3+√5 由f"(x1)0得到f在x1取到极大值,在 取到极小值 (2)由lim(f(x)-x)=-1,lim(f(x)-x)=0得到f(x)有渐近线 =x-1,y=x.又有垂直渐近线x=0 4.(1)根据Roll定理,由f(1)=f(2)=f(3)=f(4)=0得到存 在51∈(1,2),52∈(2,3),53∈(3,4),使得∫(5)=0(i=1,2,3);存在 n∈(51,2),m2∈(52,53),使得∫"(m)=0(=1,2);存在(∈(mh,m2)使得 (2)在行列式的第j列提出(x-j)(j=1,2,3,4),得到 ∫(x)=(x-1)(x-2)(x-3)(x-4) 1 x2+2x+22 x2+3x+32 x2+4x+4 x3+x2+x+1x3+2x2+22x+23x3+3x2+32x+3x3+4x2+42x+43 第三行乘以(-x)加到第四行上,第二行乘以(-x)加到第三行上,第一行
p êÆ C Y (2016-2017 Æc1ÆÏÏ" A ò) 1. (1) dy dx = e t e t + 1 , d 2 y dx 2 = 1 (et + 1)3 (2) a = 1, b = −1 (3) π 4 + 1 2 ln 2 (4) π 2 2 (5) 3 2 √3 2 (6) 1 −1 4 0 0 1 −1 2 −7 (7) p 0 f 00(x) > 0 x > 0 f 0 (x) > 0, qd f(0) = 0 x > 0 f(x) > 0. 3. (1) f 0 (x) = e 2x − 3ex + 1 (ex − 1)2 , f 00(x) = e x (ex + 1) (ex − 1)3 . 7: x1 = ln 3 − √ 5 2 , x2 = ln 3 + √ 5 2 . d f 00(x1) 0 f 3 x1 4, 3 x2 4. (2) d lim x→−∞ (f(x) − x) = −1, lim x→+∞ (f(x) − x) = 0 f(x) kìC y = x − 1, y = x. qkRìC x = 0. 4. (1) â Rolle ½n, d f(1) = f(2) = f(3) = f(4) = 0 3 ξ1 ∈ (1, 2), ξ2 ∈ (2, 3), ξ3 ∈ (3, 4), ¦ f 0 (ξi) = 0 (i = 1, 2, 3); 3 η1 ∈ (ξ1, ξ2), η2 ∈ (ξ2, ξ3), ¦ f 00(ηi) = 0 (i = 1, 2); 3 ζ ∈ (η1, η2) ¦ f 000(ζ) = 0. (2) 31ª1 j JÑ (x − j) (j = 1, 2, 3, 4), f(x) = (x − 1)(x − 2)(x − 3)(x − 4)· 0 BBB@ 1 1 1 1 x + 1 x + 2 x + 3 x + 4 x 2 + x + 1 x 2 + 2x + 22 x 2 + 3x + 32 x 2 + 4x + 42 x 3 + x 2 + x + 1 x 3 + 2x 2 + 22x + 23 x 3 + 3x 2 + 32x + 33 x 3 + 4x 2 + 42x + 43 1 CCCA . 1n1¦± (−x) \1o1þ, 11¦± (−x) \1n1þ, 11 1
乘以(-x)加到第二行上,得到 111 ∫(x)=(x-1) 1)(x-2)(x-3)(x-a1234 (x-1)(x-2)(x-3)(x-4) dt= tant+C (2)分部积分 arctan r dr arctan- 1+x21-x2 r=sint arctan r Sin r arctan r dt 1 sin't /1-3 d(cos t r arctan r +C. 6.(1)对非负整数 a?2cdz=/(kT +a)sinda (k丌+x)(1+cos(2x)d 2k+1) 所以 (2k+1)2 sina dr (2)记S0=/ sintppdt,S1=/ tainted,则对非负整数k, (k+1)r (+列如甲Pd=k8+8
¦± (−x) \11þ, f(x) = (x − 1)(x − 2)(x − 3)(x − 4) 1 1 1 1 1 2 3 4 1 22 3 3 4 2 1 23 3 3 4 3 = 12(x − 1)(x − 2)(x − 3)(x − 4). . 5. (1) Z dx (1 − x 2 ) 3/2 x=sin t = Z 1 cos2 t dt = tan t + C = x √ 1 − x 2 + C (2) ©ÜÈ© Z arctan x (1 − x 2 ) 3/2 dx = arctan x x √ 1 − x 2 − Z 1 1 + x 2 x √ 1 − x 2 dx x=sin t = x arctan x √ 1 − x 2 − Z sin t 1 + sin2 t dt = x arctan x √ 1 − x 2 + Z 1 2 − cos2 t d(cost) = x arctan x √ 1 − x 2 + 1 2 √ 2 ln √ 2 + √ 1 − x 2 √ 2 − √ 1 − x 2 + C. 6. (1) éKê k, Z (k+1)π kπ x sin2 x dx = Z π 0 (kπ + x) sin2 x dx = 1 2 Z π 0 (kπ + x)(1 + cos(2x)) dx = π 2 (2k + 1) 4 , ¤± Z nπ 0 x sin2 x dx = Xn−1 k=0 π 2 (2k + 1) 4 = π 2n 2 4 . (2) P S0 = Z π 0 |sin t| p dt, S1 = Z π 0 t|sin t| p dt, KéKê k, Z (k+1)π kπ x|sin x| p dx = Z π 0 (kπ + x)|sin x| p dx = kπS0 + S1, 2
设x=(n+a)丌(0≤a<1),则有不等式 on(n-1)So +nsi tlsintlPdt/tlsintIp (n+1)2r2 =m+122-≤ (n+1)r sin tOdt 1 n(n+1)丌So+(n+1)S1 由夹逼性得 t sint dt lim 存在
x = (n + α)π (0 ≤ α < 1), KkØ ª 1 2 n(n − 1)πS0 + nS1 (n + 1)2π 2 = Z nπ 0 t|sin t| p dt (n + 1)2π 2 ≤ Z x 0 t|sin t| p dt x 2 ≤ Z (n+1)π 0 t|sin t| p dt n2π 2 = 1 2 n(n + 1)πS0 + (n + 1)S1 n2π 2 , dY%5 lim x→+∞ Z x 0 t|sin t| dt x 2 = S0 2π 3. 3