§2　微积分基本定理

课后训练案巩固提升

A组

1.∫_("-" π/2)^(π/2)▒ (1+cos x)dx等于(　　)

A.π B.2 C.π-2 D.π+2

解析:∫_("-" π/2)^(π/2)▒ (1+cos x)dx=(x+sin x)"|" _("-" π/2)^(π/2)=(π/2+sin π/2)-["-" π/2+sin("-" π/2)]

　　=π+2.

答案:D

2.若∫_1^a▒ (2x+1/x)dx=3+ln 2(a>1),则a的值为(　　)

A.2 B.3 C.4 D.6

解析:∵∫_1^a▒ (2x+1/x)dx=(x2+ln x)"|" _1^a=a2+ln a-1,

　　∴a2+ln a-1=3+ln 2,则a=2.

答案:A

3.若S1= ∫_1^2▒ x2dx,S2=∫_1^2▒ 1/xdx,S3=∫_1^2▒ exdx,则S1,S2,S3的大小关系是(　　)

A.S1

C.S2

解析:S1=∫_1^2▒ x2dx=1/3x3"|" _1^2=7/3,S2=∫_1^2▒ 1/xdx=ln 2,S3=∫_1^2▒ exdx=e2-e,

　　∵e2-e=e(e-1)>e>7/3>ln 2,

　　∴S2

答案:B

4.设f(x)={■(x^2 "," x"∈[" 0"," 1")," @2"-" x"," x"∈[" 1"," 2"]," )┤则∫_0^2▒ f(x)dx=(　　)

A.3/4 B.4/5 C.5/6 D.6/5

解析:∫_0^2▒ f(x)dx=∫_0^1▒ x2dx+∫_1^2▒ (2-x)dx=1/3x3"|" _0^1+(2x"-" 1/2 x^2 ) "|" _1^2=5/6.

答案:C

5.设函数f(x)=xm+ax的导函数为f'(x)=2x+1,则∫_1^2▒ f(-x)dx的值等于(　　)

A.5/6 B.1/2 C.2/3 D.1/6

解析:∵f'(x)=2x+1,

　　∴f(x)=x2+x,

　　于是∫_1^2▒ f(-x)dx=∫_1^2▒ (x2-x)dx

=(1/3 x^3 "-" 1/2 x^2 ) "|" _1^2=5/6.