∫(1,0)dx∫[(1-x),0)]12e^-(3x+4y)dy就详细解答

1个回答

  • ∫(0→1) dx ∫(0→1 - x) 12e^[- (3x + 4y)] dy

    = ∫(0→1) dx ∫(0→1 - x) 12e^[- (3x + 4y)] (- 1/4)d[- (3x + 4y)]

    = - 3∫(0→1) e^[- (3x + 4y)] |(0→1 - x) dx

    = - 3∫(0→1) e^[- (3x + 4(1 - x)] - e^[- (3x)] dx

    = - 3∫(0→1) e^(x - 4) dx + 3∫(0→1) e^(- 3x) dx

    = - 3e^(x - 4) + 3(- 1/3)e^(- 3x) |(0→1)

    = [- 3e^(1 - 4) - e^(- 3)] - [- 3e^(- 4) - 1]

    = 1 + 3/e⁴ - 4/e³

    ≈ 0.855799