Question

大学数学作业

大学数学作业求理论知识。

Answer 1

6(5) ∫|x-x^2|dx = ∫(x^2-x)dx + ∫(x-x^2)dx + ∫(x^2-x)dx = [x^3/3-x^2/2] + [x^2/2-x^3/3] + [x^3/3-x^2/2] = 5/6 + 1/6 + 5/6 = 11/6; (7) 原式 = (1/2)∫d(1+x^2)/√(1+x^2) = [√(1+x^2)] = √2 - 1; (9) 原式 = ∫d(x-1/2)/[(x-1/2)^2+3/4] = (2/√3)[atctan(2x-1)/√3] = (2/√3)(π/3 + π/3) = 4π/(3√3). (11) 原式 = 2∫2>√[(cosx)^3 - (cosx)^5]dx = 2∫2>sinx(cosx)^(3/2)dx = -2∫2>(cosx)^(3/2)dcosx = -2[(2/5)(cosx)^(5/2)]2> = 4/5. 7. 两边对 x 求导得 f(x)/x^2 =1/√x, 则 f(x) = x^(3/2), 代入原式， 6 + ∫dt/√t = 2√x, 即 6 + [2√t] = 2√x, 6 + 2√x - 2√a = 2√x, 则 a = 9. 8. 两边对 x 求导得 f(x) + xf'(x) = 3x^2+f(x), 则 x ≠ 0 时，f'(x) = 3x, f(x) = (3/2)x^2 + C, 代入原式得 (3/2)x^3 + Cx = x^3 + ∫[(3/2)t^2+ C]dt = x^3 + [(1/2)t^3+ Ct] = x^3 + (1/2)x^3+ Cx - 1/2 - C 则 C = 1/2, f(x) = (3/2)x^2 + 1/2.

追问

大佬，您是不是发错了？