所以S2

答案B

2∫_0^5π▒ (ex-sin x)dx的值为(　　)

A.e5π-1 B.e5π-2 C.e5π-3 D.e5π-4

解析∫_0^5π▒ (ex-sin x)dx=∫_0^5π▒ exdx-∫_0^5π▒ sin xdx

　　=ex"|" _0^5π+cos x"|" _0^5π=e5π-e0+cos 5π-cos 0

　　=e5π-1-1-1=e5π-3.

答案C

3若f(x)=x2+2∫_0^1▒ f(x)dx,则∫_0^1▒ f(x)dx=(　　)

A.-1 B.-1/3 C.1/3 D.1

答案B

4若f(x)={■("-" e^x "," x>1"," @"|" x"|," x≤1)┤(e为自然对数的底数),则∫_0^2▒ f(x)dx=(　　)

A.1/2+e2-e B.1/2+e

C.1/2+e-e2 D.-1/2+e-e2

解析∫_0^2▒ f(x)dx=∫_0^1▒ |x|dx+∫_1^2▒ (-ex)dx

　　=∫_0^1▒ xdx-∫_1^2▒ exdx=1/2x2"|" _0^1-ex"|" _1^2=1/2+e-e2.

答案C

★5若f(x)是一次函数,且∫_0^1▒ f(x)dx=5,∫_0^1▒ xf(x)dx=17/6,则f(x)为(　　)

A.4x+3 B.3x+4

C.-4x+2 D.-3x+4 学 ]

解析由题知f(x)是一次函数,设f(x)=ax+b(a≠0),

　　则∫_0^1▒ f(x)dx=∫_0^1▒ (ax+b)dx

　　=∫_0^1▒ axdx+∫_0^1▒ bdx=1/2a+b=5,

　　∫_0^1▒ xf(x)dx=∫_0^1▒ x(ax+b)dx

　　=∫_0^1▒ (ax2)dx+∫_0^1▒ bxdx=1/3a+1/2b=17/6.

　　由{■(1/2 a+b=5"," @1/3 a+1/2 b=17/6 "," )┤解得a=4,b=3,

　　故f(x)=4x+3. 学 ]

答案A

6若∫_0^1▒ (2ax2-a2x)dx=1/6,则实数a的值是　　　　.

答案1/3或1