Question

lim(X→0)【[∫上限X下线0(sin2)tdt]/xcosx】要过程

定积分：
∫上限9下限4√x(1+√x)dx
∫上限2下限-1(x^2-1)dx
∫上限2下限1(1/x+x)dx
∫上限π/4下限0(sinx+cosx)dx
∫上限2下限1(√x+1/x^2)dx

Answer 1

lim[x→0] {∫[0，x] sin²t/dt} / (xcosx)，应用洛必达法则
= lim[x→0] sin²x / (cosx - xsinx)
= (0)² / (1 - 0)
= 0

∫[4，9] √x(1+√x) dx
= ∫[4，9] (√x+x) dx
= (2/3)x^(3/2) + x²/2
= [(2/3)(9)^(3/2) + 9²/2] - [(2/3)(4)^(3/2) + 4²/2]
= 271/6

∫[-1，2] (x²-1) dx
= x³/3 - x
= [2³/3 - 2] - [(-1)³/3 - (-1)]
= 0

∫[1，2] (1/x+x) dx
= ln|x| + x²/2
= [ln2 + 2²/2] - [ln1 + 1²/2]
= 3/2 + ln2

∫[0，π/4] (sinx+cosx) dx
= -cosx + sinx
= [-cos(π/4) + sin(π/4)] - [-cos(0) + sin(0)]
= [-√2/2 + √2/2] - [-1 + 0]
= 1

∫[1，2] (√x + 1/x²) dx
= ∫[1，2] [x^(1/2) + x^(-2)] dx
= (2/3)x^(3/2) - 1/x
= [(2/3)(2)^(3/2) - 1/2] - [(2/3)(1)^(3/2) - 1]
= (4√2)/3 - 1/6