积分题求给力(带图)

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积分题求给力(带图)积分题求给力(带图)积分题求给力(带图)∫x√(2x-x^2)*dx=∫x*√[1-(x-1)^2]*dx设(x-1)=sint,x=1+sint,dx=cost*dt.积分限变为

积分题求给力(带图)
积分题求给力(带图)

积分题求给力(带图)
∫x√(2x - x^2) *dx
=∫x*√[1 - (x - 1)^2] *dx
设 (x - 1) = sint,x = 1 + sint,dx = cost *dt.积分限变为 [-π/2,+π/2].则原积分变换为:
=∫(1+sint)*√[1-(sint)^2] * cost *dt
=∫(1+sint)* (cost)^2 *dt
=∫(cost)^2 *dt + ∫(cost)^2 * (sint * dt)
=1/2 * ∫[cos(2t) + 1] *dt - ∫(cost)^2 *d(cost)
=1/4 * ∫cos(2t)*d(2t) + 1/2*∫dt -1/3 * (cost)^3
=1/8 * [cos(2t)]^2 + 1/2 *t - 1/3 *(cost)^3
={1/8 * [cosπ]^2 + 1/2 * π/2 - 1/3 *[cos(π/2)]^3} - {1/8 *[cos(-π)]^2 + 1/2*(-π/2) - 1/3*[cos(-π/2)]^3}
={1/8 * 1 + π/4) - 1/3 * 0} - {1/8 * 1 - π/4 - 1/3 * 0}
=π/2