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Question
Prove by induction method that for all n≥ 1
∫xnexdx=n!ex[xnn!−xn−1(n−1)!+nn−2!(n−2)!+.....+(−1)n]
∫xnexdx=n!ex[xnn!−xn−1(n−1)!+nn−2!(n−2)!+.....+(−1)n]
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Solution
p(1) = ∫xex dx = x.ex−∫e2. 1dx = ex (x - 1)
R.H.S. = ex(x1!−10!)=ex (x - 1)
Thus p(1) holds good. Assume p(n)
∴ p(n + 1) = ∫xn+1ex dx
= xn+1.ex -(n + 1) ∫xnex dx
= xn+1.ex - (n - 1).n!ex[xnn!−xn−1(n−1)!+xn−2(n−1)!+...]
= (n + 1)! ex[xn−1(n+1)!−xnn!+xn−2(n−1)!−...]
Thus p(n + 1) also holds good.
R.H.S. = ex(x1!−10!)=ex (x - 1)
Thus p(1) holds good. Assume p(n)
∴ p(n + 1) = ∫xn+1ex dx
= xn+1.ex -(n + 1) ∫xnex dx
= xn+1.ex - (n - 1).n!ex[xnn!−xn−1(n−1)!+xn−2(n−1)!+...]
= (n + 1)! ex[xn−1(n+1)!−xnn!+xn−2(n−1)!−...]
Thus p(n + 1) also holds good.
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