2018-2019学年北师大版选修2-2 微积分基本定理 课时作业
2018-2019学年北师大版选修2-2    微积分基本定理  课时作业第2页

C.f(x)=-4x+2 D.f(x)=-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=5,0①

  ∫_0^1▒ xf(x)dx=∫_0^1▒ (ax2+bx)dx

  =(1/3 ax^3+1/2 bx^2 ) "|" _0^1=1/3a+1/2b=17/6.0② 学 ]

  联立①②,解得a=4,b=3,

  故f(x)=4x+3.

答案:A

6.∫_0^(π/2)▒ sin2x/2dx等于(  )

A.π/4 B.π/2-1 C.2 D.(π"-" 2)/4

解析:∫_0^(π/2)▒ sin2x/2dx=∫_0^(π/2)▒ (1"-" cosx)/2dx=1/2(x-sin x)"|" _0^(π/2)=(π"-" 2)/4.

答案:D

7.∫_0^2▒ (3x2+k)dx=10,则k=     .

解析:∫_0^2▒ (3x2+k)dx=(x3+kx)"|" _0^2=10,则k=1.

答案:1

8.∫_2^4▒ 2xln 2dx=     . Z

解析:∫_2^4▒ 2xln 2dx=2x"|" _2^4=24-22=12.

答案:12

9.若f(x)在R上可导,f(x)=x2+2f'(2)x+3,则∫_0^3▒ f(x)dx=     . Z

解析:∵f'(x)=2x+2f'(2),∴f'(2)=4+2f'(2).

  ∴f'(2)=-4.∴f(x)=x2-8x+3.

  ∴∫_0^3▒ f(x)dx=∫_0^3▒ (x2-8x+3)dx

  =(1/3 x^3 "-" 4x^2+3x) "|" _0^3=-18.

答案:-18

10.求下列定积分:

(1)∫_("-" a)^a▒ √(x^2 )dx(a>0); 学 ]

(2)∫_1^2▒ (t+2)dx.