Question

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

Answer 1

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 $n\in N$. 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,$\left( n\in N \right)$ is true for all n, then we have to go through the following steps.
Let $P(n)=\int{{{x}^{n}}{{e}^{x}}dx=n!{{e}^{x}}\left[ \dfrac{{{x}^{n}}}{n!}-\dfrac{{{x}^{n-1}}}{(n-1)!}+\dfrac{{{x}^{n-2}}}{(n-2)!}-...........+{{(-1)}^{n}} \right]},\forall n\in N$
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. = $\int{x{{e}^{x}}dx}=x{{e}^{x}}-\int{1\cdot {{e}^{x}}dx=x{{e}^{x}}-{{e}^{x}}={{e}^{x}}(x-1)}$
R.H.S. =${{e}^{x}}\left( \dfrac{x}{1!}-\dfrac{1}{0!} \right)={{e}^{x}}(x-1)$
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.
$\int{{{x}^{k}}{{e}^{x}}dx=k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]}..........(1)$
Step III (Succession)
Prove that P (k+1) is true, when P (k) is true.
To prove that $\int{{{x}^{k+1}}{{e}^{x}}dx=(k+1)!{{e}^{x}}\left[ \dfrac{{{x}^{k+1}}}{(k+1)!}-\dfrac{{{x}^{k}}}{(k)!}+\dfrac{{{x}^{k-1}}}{(k-1)!}-...........+{{(-1)}^{k+1}} \right]}.........(2)$
Consider the R.H.S.
R. H. S. $=\int{{{x}^{k+1}}{{e}^{x}}dx}$
Integrate with respect to x by using integrating by parts $\int{uvdx}=u\int{v}dx-\int{\left[ \dfrac{du}{dx}\int{v}dx \right]}dx$ , we get
R. H. S. $={{x}^{k+1}}\int{{{e}^{x}}}-\int{\left[ \dfrac{d}{dx}{{x}^{k+1}}\int{{{e}^{x}}dx} \right]dx}$
We have $\int{{{e}^{x}}dx={{e}^{x}}\text{ and }\dfrac{d}{dx}{{x}^{k+1}}=}(k+1){{x}^{k}}$
R. H. S. $={{x}^{k+1}}{{e}^{x}}-\int{(k+1){{x}^{k}}{{e}^{x}}dx}$
R. H. S. $={{x}^{k+1}}{{e}^{x}}-(k+1)\int{{{x}^{k}}{{e}^{x}}dx}$
Put the value of the integral$\int{{{x}^{k}}{{e}^{x}}dx=k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]}$, we get
R. H. S \[={{x}^{k+1}}{{e}^{x}}-(k+1)k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]\]
R. H. S \[={{x}^{k+1}}{{e}^{x}}-(k+1)k!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]\]
We have $(k+1)k!=(k+1)!$
R. H. S \[={{x}^{k+1}}{{e}^{x}}-(k+1)!{{e}^{x}}\left[ \dfrac{{{x}^{k}}}{k!}-\dfrac{{{x}^{k-1}}}{(k-1)!}+\dfrac{{{x}^{k-2}}}{(k-2)!}-...........+{{(-1)}^{k}} \right]\]
Take $(k+1)!{{e}^{x}}$common, we get
R. H. S \[=(k+1)!{{e}^{x}}\left[ \dfrac{{{x}^{k+1}}}{(k+1)!}-\dfrac{{{x}^{k}}}{k!}+\dfrac{{{x}^{k-1}}}{(k-1)!}-\dfrac{{{x}^{k-2}}}{(k-2)!}+...........+{{(-1)}^{k+1}} \right]\]
R. H. S. = L. H. S.
Therefore, P (n) is true for $n=k+1$.
Step IV (Induction)
From all steps above by the principal of mathematical induction, P(n) is true for all $n\in N$.
Hence $\int{{{x}^{n}}{{e}^{x}}dx=n!{{e}^{x}}\left[ \dfrac{{{x}^{n}}}{n!}-\dfrac{{{x}^{n-1}}}{(n-1)!}+\dfrac{{{x}^{n-2}}}{(n-2)!}-...........+{{(-1)}^{n}} \right]},\forall n\in N$

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.