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Question: If \(I_{n} = \int_{0}^{\infty}{e^{- x}x^{n - 1}dx,}\) then \(\int_{0}^{\infty}e^{- \lambda x}x^{n - ...

If In=∫0∞e−xxn−1dx,I_{n} = \int_{0}^{\infty}{e^{- x}x^{n - 1}dx,} then ∫0∞e−λxxn−1dx\int_{0}^{\infty}e^{- \lambda x}x^{n - 1}dx is equal to

A

λIn\lambda I_{n}

B

1λIn\frac{1}{\lambda}I_{n}

C

Inλn\frac{I_{n}}{\lambda^{n}}

D

λnIn\lambda^{n}I_{n}

Answer

Inλn\frac{I_{n}}{\lambda^{n}}

Explanation

Solution

Put, λx=t,\lambda x = t, λdx=dt,\lambda dx = dt, we get,

∫0∞e−λxxn−1dx=1λn∫0∞e−ttn−1dt\int_{0}^{\infty}e^{- \lambda x}x^{n - 1}dx = \frac{1}{\lambda^{n}}\int_{0}^{\infty}e^{- t}t^{n - 1}dt =1λn∫0∞e−xxn−1dx=Inλn\frac{1}{\lambda^{n}}\int_{0}^{\infty}{e^{- x}x^{n - 1}}dx = \frac{I_{n}}{\lambda^{n}}