1.设a>0,a≠1,若∫_0^2▒ axdx=-2ax"|" _0^2,则a的值为(　　)

A.e-2 B.e2 C.e^("-" 1/2) D.e^(1/2)

解析:∵∫_0^2▒ axdx=1/lnaax"|" _0^2=a^2/lna-1/lna=-2ax"|" _0^2=-2a2+2,∴(a2-1)(1/lna+2)=0.

　　又∵a>0且a≠1,∴ln a=-1/2.∴a=e^("-" 1/2).

答案:C

2.导学号88184048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.4x+3 B.3x+4 C.-4x+2 D.-3x+4

解析:设f(x)=ax+b(a≠0),则∫_0^1▒ f(x)dx=∫_0^1▒ (ax+b)dx=(1/2 ax^2+bx) "|" _0^1=1/2a+b=5.

　　∫_0^1▒ xf(x)dx=∫_0^1▒ (ax2+bx)dx

　　=(1/3 ax^3+1/2 bx^2 ) "|" _0^1=1/3a+1/2b=17/6,解方程组{■(1/2 a+b=5"," @1/3 a+1/2 b=17/6 "," )┤得{■(a=4"," @b=3"." )┤故f(x)=4x+3.

答案:A

3.已知函数f(x)的图像是折线段ABC,其中A(0,0),B(1/2 "," 1),C(1,0),则函数y=xf(x)(0≤x≤1)的图像与x轴围成的图形面积为　　　　　.

解析:由已知条件得f(x)={■(2x"," 0≤x≤1/2 "," @"-" 2x+2"," 1/2

　　则xf(x)={■(2x^2 "," 0≤x≤1/2 "," @"-" 2x^2+2x"," 1/2

　　故所求面积S=∫_0^1▒ xf(x)dx

　　=∫_0^(1/2)▒ 2x2dx+∫_(1/2)^1▒〖"(-" 2〗 x^2+2x")" dx

　　=2/3x3"|" _0^(1/2)+("-" 2/3 x^3+x^2 ) "|" _(1/2)^1=2/3×(1/2)^3+("-" 2/3+1)-["-" 2/3×(1/2)^3+(1/2)^2 ]=1/4.

答案:1/4

4.已知f(x)=ax2+bx+c(a≠0),且f(-1)=2,f'(0)=0,∫_0^1▒ f(x)dx=-2,求a,b,c的值.

解∵f(-1)=2,

　　∴a-b+c=2.

　　又f'(x)=2ax+ b,

　　∴f'(0)=b=0.

　　∵∫_0^1▒ f(x)dx

　　=∫_0^1▒ (ax2+bx+c)dx

　　=(1/3 ax^3+1/2 bx^2+cx) "|" _0^1

　　=1/3a+1/2b+c,

　　∴1/3a+1/2b+c=-2.

解方程组{■(a"-" b+c=2"," @b=0"," @1/3 a+1/2 b+c="-" 2"," )┤得{■(a=6"," @b=0"," @c="-" 4"." )┤