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Question: Prove that by the induction method that for all \(n\ge 1\) , \(\int{{{x}^{n}}{{e}^{x}}dx=n!{{e}^{x...

Prove that by the induction method that for all n≥1n\ge 1 ,
∫xnexdx=n!ex[xnn!−xn−1(n−1)!+xn−2(n−2)!−...........+(−1)n]\int{{{x}^{n}}{{e}^{x}}dx=n!{{e}^{x}}\left[ \dfrac{{{x}^{n}}}{n!}-\dfrac{{{x}^{n-1}}}{(n-1)!}+\dfrac{{{x}^{n-2}}}{(n-2)!}-...........+{{(-1)}^{n}} \right]}

Explanation

Solution

Hint:Let P (n) be a statement involving the natural number n such that P (1) is true and P (k+1) is true, whenever P (k) is true. Then P (n) is true for all n∈Nn\in N. This is known as the Principal of Mathematical Induction.

Complete step-by-step answer:
Suppose that we have to prove that a certain property of an expression in n,(n∈N)\left( n\in N \right) is true for all n, then we have to go through the following steps.
Let P(n)=∫xnexdx=n!ex[xnn!−xn−1(n−1)!+xn−2(n−2)!−...........+(−1)n],∀n∈NP(n)=\int{{{x}^{n}}{{e}^{x}}dx=n!{{e}^{x}}\left[ \dfrac{{{x}^{n}}}{n!}-\dfrac{{{x}^{n-1}}}{(n-1)!}+\dfrac{{{x}^{n-2}}}{(n-2)!}-...........+{{(-1)}^{n}} \right]},\forall n\in N
Step I (Foundation)
Denote the statement stating the property by P (n). Verify that P (1) is true.
Put n = 1 in the L.H.S. and integrate by using by parts, we get
L.H.S. = ∫xexdx=xex−∫1⋅exdx=xex−ex=ex(x−1)\int{x{{e}^{x}}dx}=x{{e}^{x}}-\int{1\cdot {{e}^{x}}dx=x{{e}^{x}}-{{e}^{x}}={{e}^{x}}(x-1)}
R.H.S. =ex(x1!−10!)=ex(x−1){{e}^{x}}\left( \dfrac{x}{1!}-\dfrac{1}{0!} \right)={{e}^{x}}(x-1)
Therefore, L.H.S = R.H.S.
P(n) is true for n = 1.
Step II (Assumption)
Assume that P (k) is true and write the result for P (k).
Let us consider that P (n) is true for n = k.
∫xkexdx=k!ex[xkk!−xk−1(k−1)!+xk−2(k−2)!−...........+(−1)k]..........(1)\int{{{x}^{k}}{{e}^{x}}dx=k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]}..........(1)
Step III (Succession)
Prove that P (k+1) is true, when P (k) is true.
To prove that ∫xk+1exdx=(k+1)!ex[xk+1(k+1)!−xk(k)!+xk−1(k−1)!−...........+(−1)k+1].........(2)\int{{{x}^{k+1}}{{e}^{x}}dx=(k+1)!{{e}^{x}}\left[ \dfrac{{{x}^{k+1}}}{(k+1)!}-\dfrac{{{x}^{k}}}{(k)!}+\dfrac{{{x}^{k-1}}}{(k-1)!}-...........+{{(-1)}^{k+1}} \right]}.........(2)
Consider the R.H.S.
R. H. S. =∫xk+1exdx=\int{{{x}^{k+1}}{{e}^{x}}dx}
Integrate with respect to x by using integrating by parts ∫uvdx=u∫vdx−∫[dudx∫vdx]dx\int{uvdx}=u\int{v}dx-\int{\left[ \dfrac{du}{dx}\int{v}dx \right]}dx , we get
R. H. S. =xk+1∫ex−∫[ddxxk+1∫exdx]dx={{x}^{k+1}}\int{{{e}^{x}}}-\int{\left[ \dfrac{d}{dx}{{x}^{k+1}}\int{{{e}^{x}}dx} \right]dx}
We have ∫exdx=ex and ddxxk+1=(k+1)xk\int{{{e}^{x}}dx={{e}^{x}}\text{ and }\dfrac{d}{dx}{{x}^{k+1}}=}(k+1){{x}^{k}}
R. H. S. =xk+1ex−∫(k+1)xkexdx={{x}^{k+1}}{{e}^{x}}-\int{(k+1){{x}^{k}}{{e}^{x}}dx}
R. H. S. =xk+1ex−(k+1)∫xkexdx={{x}^{k+1}}{{e}^{x}}-(k+1)\int{{{x}^{k}}{{e}^{x}}dx}
Put the value of the integral∫xkexdx=k!ex[xkk!−xk−1(k−1)!+xk−2(k−2)!−...........+(−1)k]\int{{{x}^{k}}{{e}^{x}}dx=k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]}, we get
R. H. S =xk+1ex−(k+1)k!ex[xkk!−xk−1(k−1)!+xk−2(k−2)!−...........+(−1)k]={{x}^{k+1}}{{e}^{x}}-(k+1)k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]
R. H. S =xk+1ex−(k+1)k!ex[xkk!−xk−1(k−1)!+xk−2(k−2)!−...........+(−1)k]={{x}^{k+1}}{{e}^{x}}-(k+1)k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]
We have (k+1)k!=(k+1)!(k+1)k!=(k+1)!
R. H. S =xk+1ex−(k+1)!ex[xkk!−xk−1(k−1)!+xk−2(k−2)!−...........+(−1)k]={{x}^{k+1}}{{e}^{x}}-(k+1)!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]
Take (k+1)!ex(k+1)!{{e}^{x}}common, we get
R. H. S =(k+1)!ex[xk+1(k+1)!−xkk!+xk−1(k−1)!−xk−2(k−2)!+...........+(−1)k+1]=(k+1)!{{e}^{x}}\left[ \dfrac{{{x}^{k+1}}}{(k+1)!}-\dfrac{{{x}^{k}}}{k!}+\dfrac{{{x}^{k-1}}}{(k-1)!}-\dfrac{{{x}^{k-2}}}{(k-2)!}+...........+{{(-1)}^{k+1}} \right]
R. H. S. = L. H. S.
Therefore, P (n) is true for n=k+1n=k+1.
Step IV (Induction)
From all steps above by the principal of mathematical induction, P(n) is true for all n∈Nn\in N.
Hence ∫xnexdx=n!ex[xnn!−xn−1(n−1)!+xn−2(n−2)!−...........+(−1)n],∀n∈N\int{{{x}^{n}}{{e}^{x}}dx=n!{{e}^{x}}\left[ \dfrac{{{x}^{n}}}{n!}-\dfrac{{{x}^{n-1}}}{(n-1)!}+\dfrac{{{x}^{n-2}}}{(n-2)!}-...........+{{(-1)}^{n}} \right]},\forall n\in N

Note: When doing Integration by parts, we know that LIATE can be a useful guide most of the time. For those not familiar, LIATE is a guide to help you decide which term to differentiate and which term to integrate.