不答的不要捣乱22

3个回答

  • 5

    令t=2x+1

    ∫(0到1)f(2x+1)dx=1/2∫(0到1)f(2x+1)d(2x+1)=1/2∫(1到3)f(t)dt=1/2

    三大题

    第一小题

    1

    =(x^3-x^2/2+x) | 0到a

    =a^3-a^2/2+a

    2

    原式=arcsinx |(-1/2到1/2)

    =π/3

    3

    =4∫(0到π/2)sinxdx=4

    4

    (3x^4+3x^2+1)/(x^2+1)

    =(3x^2(x^2+1)+1)/(x^2+1)

    =3x^2+ 1/(x^2+1)

    所以积分=∫(-1到0)【3x^2+ 1/(x^2+1)】dx

    =x^3+arctanx |(-1到0)

    =1+π/4

    5

    ∫(tanx)^2 dx=∫ [(secx)^2-1]dx= tanx-x |(0到π/4)

    =1-π/4

    6

    令t=√x,x=t^2 dx=2tdt

    原积分=2∫(2到3)t^2(1+t)dt=2t^3/3+t^2 | (2到3)

    =53/3

    第二小题

    1

    令t=x-π/3

    =∫(0到π/6)costdt=1/2

    2

    1/(x^2+2x+2)=1/[1+(x+1)^2]

    令t=x+1

    原积分=∫(-2到0) 1/[1+(x+1)^2] d(x+1)

    =∫(-1到1) 1/[1+t^2] dt

    =arctant | (-1到1)

    =π/2