Question

大学高数作业

1，3，5，需要画图

Answer 1

解答及图如下：

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Answer 2

6(5) ∫<-1, 2>|x-x^2|dx
= ∫<-1, 0>(x^2-x)dx + ∫<0, 1>(x-x^2)dx + ∫<1, 2>(x^2-x)dx
= [x^3/3-x^2/2]<-1, 0> + [x^2/2-x^3/3]<0, 1> + [x^3/3-x^2/2]<1, 2>
= 5/6 + 1/6 + 5/6 = 11/6;
(7) 原式 = (1/2)∫<0, 1>d(1+x^2)/√(1+x^2) = [√(1+x^2)]<0, 1> = √2 - 1;
(9) 原式 = ∫<0, 1>d(x-1/2)/[(x-1/2)^2+3/4]
= (2/√3)[atctan(2x-1)/√3]<0, 1> = (2/√3)(π/3 + π/3) = 4π/(3√3).
(11) 原式 = 2∫<0, π/2>√[(cosx)^3 - (cosx)^5]dx
= 2∫<0, π/2>sinx(cosx)^(3/2)dx = -2∫<0, π/2>(cosx)^(3/2)dcosx
= -2[(2/5)(cosx)^(5/2)]<0, π/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 + ∫<a, x>dt/√t = 2√x,
即 6 + [2√t]<a, x> = 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 + ∫<1, x>[(3/2)t^2+ C]dt
= x^3 + [(1/2)t^3+ Ct]<1, x> = x^3 + (1/2)x^3+ Cx - 1/2 - C
则 C = 1/2, f(x) = (3/2)x^2 + 1/2.