Bathyouth
幼苗
共回答了23个问题采纳率:91.3% 举报
设切点为N(n, e⁻ⁿ)
y = e^(-x)
y' = [e^(-x)](-x)' = -e^(-x)
x = n, y' = -e⁻ⁿ
切线y - e⁻ⁿ = -e⁻ⁿ(x - n)
切线原点: 0 - e⁻ⁿ = -e⁻ⁿ(0 - n)
n = -1
N(-1, e)
切线: y - e = -e(x + 1), y = -ex
该曲线过原点的切线和y轴所围图形的面积S = ∫₋₁⁰[e^(-x) - (-ex)]dx
= [-e^(-x) + ex²/2]|₋₁⁰
= (-1+ 0) - (-e + e/2)
= (e/2) -1
1年前
追问
6