Question

Assertion :$I_n={\int }_0^{\infty }{x}^n{e}^{-x}dx(n$ is a positive integer$)=n!$ Reason: $I_n=n{I}_{n-1}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let $I_n={\int }_0^{\infty }{x}^n{e}^{-x}dx$$={[x^n.\frac{e^{-x}}{(-1)}]}_0^{\infty }-{\int }_0^{\infty }n{x}^{n-1}.\frac{e^{-x}}{(-1)}dx$

$=n{\int }_0^{\infty }{x}^{n-1}{e}^{-x}dx=n{I}_{n-1}$

$\therefore I_n=n.I_{n-1}$ ...(1)

Changing $n$ to $n-1,I_{n-1}=(n-1)I_{n-2}$

Substituting the value of $I_{n-1}$ in (1), we get

$I_n=n(n-1)I_{n-2}$

Generalizing from (1) and (2), we have

$I_n=[n(n-1)...$ to $n$ factors $]I_{n-n}$

$=n!I_0=n!{\int }_0^{\infty }{x}^0{e}^{-x}dx=n!{\int }_0^{\infty }{e}^{-x}dx$ $=n!{\left[\frac{e^{-x}}{-1}\right]}_0^{\infty }=n!.$