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Question Number 170532 by MikeH last updated on 26/May/22

Let I_n =∫x^n e^(−x) dx, n = 0,1,2,... (i) Show that I_n = −x^n e^(−x) +nI_(n−1) (ii) Show that ∫_0 ^∞ x^n e^(−x) dx = n!

LetIn=∫xne−xdx,n=0,1,2,...(i)ShowthatIn=−xne−x+nIn−1(ii)Showthat∫0∞xne−xdx=n!

Answered by FelipeLz last updated on 26/May/22

(i) I_n = ∫x^n e^(−x) dx x^n = u ⇒ u′ = (d/dx)[x^n ] = nx^(n−1) e^(−x) = v^′ ⇒ v = ∫e^(−x) dx = −e^(−x) ∫uv′dx = uv−∫u′vdx I_n = x^n (−e^(−x) )−∫nx^(n−1) (−e^(−x) )dx I_n = −x^n e^(−x) +n∫x^(n−1) e^(−x) dx I_n = −x^n e^(−x) +nI_(n−1) (ii) ∫x^n e^(−x) dx = −x^n e^(−x) +n∫x^(n−1) e^(−x) dx ∫x^n e^(−x) dx = −x^n e^(−x) −nx^(n−1) e^(−x) +n(n−1)∫x^(n−2) e^(−x) dx ∫x^n e^(−x) dx = −x^n e^(−x) −nx^(n−1) e^(−x) −n(n−1)x^(n−2) e^(−x) +n(n−1)(n−2)∫x^(n−3) e^(−x) dx ∫x^n e^(−x) dx = −x^n e^(−x) −nx^(n−1) e^(−x) −n(n−1)x^(n−2) e^(−x) −n(n−1)(n−2)x^(n−3) e^(−x) +n(n−1)(n−2)(n−3)∫x^(n−4) e^(−x) dx ⋮ ∫x^n e^(−x) dx = −e^(−x) Σ_(k=0) ^n ((n!)/((n−k)!))x^(n−k) ∫_0 ^∞ x^n e^(−x) dx = −Σ_(k=0) ^n [((n!)/((n−k)!))∙lim_(x→∞) ((x^(n−k) /e^x ))]+Σ_(k=0) ^(n−1) [((n!)/((n−k)!))∙(0^(n−k) /e^0 )]+((n!)/e^0 ) ∫_0 ^∞ x^n e^(−x) dx = −Σ_(k=0) ^n [((n!)/((n−k)!))∙0]+Σ_(k=0) ^(n−1) [((n!)/((n−k)!))∙0]+((n!)/1) = n!

(i)In=∫xne−xdxxn=u⇒u′=ddx[xn]=nxn−1e−x=v′⇒v=∫e−xdx=−e−x∫uv′dx=uv−∫u′vdxIn=xn(−e−x)−∫nxn−1(−e−x)dxIn=−xne−x+n∫xn−1e−xdxIn=−xne−x+nIn−1(ii)∫xne−xdx=−xne−x+n∫xn−1e−xdx∫xne−xdx=−xne−x−nxn−1e−x+n(n−1)∫xn−2e−xdx∫xne−xdx=−xne−x−nxn−1e−x−n(n−1)xn−2e−x+n(n−1)(n−2)∫xn−3e−xdx∫xne−xdx=−xne−x−nxn−1e−x−n(n−1)xn−2e−x−n(n−1)(n−2)xn−3e−x+n(n−1)(n−2)(n−3)∫xn−4e−xdx⋮∫xne−xdx=−e−x∑nk=0n!(n−k)!xn−k∫0∞xne−xdx=−∑nk=0[n!(n−k)!⋅limx→∞(xn−kex)]+∑n−1k=0[n!(n−k)!⋅0n−ke0]+n!e0∫0∞xne−xdx=−∑nk=0[n!(n−k)!⋅0]+∑n−1k=0[n!(n−k)!⋅0]+n!1=n!

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