作业帮定积分求教
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/28 23:31:51
定积分求教
定积分求教
定积分求教
y = √[|x(x - 2)|]
= √[x(x - 2)] 或 √[x(2 - x)]
y1 = √[x(x - 2)]、定义域x - 2 ≥ 0 => x ≥ 2、x ≤ 0、∴x ≤ 0或x ≥ 2
y2 = √[x(2 - x)]、定义域2 - x ≥ 0 => x ≤ 2、x ≥ 0、∴0 ≤ x ≤ 2
设K = ∫(1→3) √[|x(x - 2)|] dx
= ∫(1→2) y dx + ∫(2→3) y dx
= ∫(1→2) √[x(2 - x)] dx + ∫(2→3) √[x(x - 2)] dx
= ∫(1→2) √[1 - (x - 1)²] dx + ∫(2→3) √[(x - 1)² - 1] dx
= A + B
A = ∫(1→2) √[1 - (x - 1)²] dx、令x - 1 = siny、dx = cosy dy
= ∫(0→π/2) cos²y dy = (1/2)∫(0→π/2) (1 + cos2y) dy = (1/2)[y + (1/2)sin2y] |(0→π/2)
= (1/2)(π/2) = π/4
B = ∫(2→3) √[(x - 1)² - 1] dx、令x - 1 = secy、dx = secytany dy
= ∫(0→π/3) secytan²y dy
= ∫(0→π/3) secy(sec²y - 1) dy
= ∫(0→π/3) (sec³y - secy) dy
= (1/2)secytany |(0→π/3) - (1/2)ln(secy + tany) |(0→π/3)
= √3 - (1/2)ln(2 + √3)
∴原式K = π/4 + √3 - (1/2)ln(2 + √3)