2　微积分基本定理

1.∫^π▒ _0(cos x+1)dx等于(　　)

A.1 B.0 C.π+1 D.π

解析:∫^π▒ _0(cos x+1)dx=∫^π▒ _0cos xdx+∫^π▒ _01dx

　　=sin x〖"|" ^π〗_0+x〖"|" ^π〗_0=π.

答案:D

2.∫_0^4▒ |x2-2x|dx=(　　)

A.64/3 B.0 C.8 D.16

解析:∵f(x)=|x2-2x|={■(2x"-" x^2 "," 0≤x≤2"," @x^2 "-" 2x"," 2

　　∴∫_0^4▒ |x2-2x|dx=∫_0^2▒ (2x-x2)dx+∫_2^4▒ (x2-2x)dx

　　=(x^2 "-" 1/3 x^3 ) "|" _0^2+(1/3 x^3 "-" x^2 ) "|" _2^4=8.

答案:C

3.若∫_1^a▒ (2x+1/x)dx=3+ln 2,则a的值是(　　)

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

解析:∫_1^a▒ (2x+1/x)dx=∫_1^a▒ 2xdx+∫_1^a▒ 1/xdx

　　=x2"|" _1^a+ln|x|"|" _1^a=a2-1+ln a=3+ln 2.

　　故a=2.

答案:D

4.若f(x)={■(f"(" x"-" 5")(" x>0")," @2^x+∫_0^(π/6)▒cos 3tdt"(" x≤0")," )┤则f(2 015)等于0(　　)

A.1 B.2 C.4/3 D.5/3

解析:∫_0^(π/6)▒ cos 3tdt=sin3t/3 "|" _0^(π/6)=1/3,f(2 015)=f(0)=20+1/3=4/3.

答案:C

5.已知f(x)是一次函数,且∫_0^1▒ f(x)dx=5,∫_0^1▒ xf(x)dx=17/6,则f(x)的解析式为(　　)

A.f(x)=4x+3 B.f(x)=3x+4