Q6Mathematics I

Question

Evaluate:

\int_{0}^{\infty} x^{6} e^{- 2 x} d x

by using beta-gamma function.

Answer 1

To accurately evaluate $I = \int_{0}^{\infty} x^{6} e^{- 2 x} d x$ , we introduce the algebraic substitution $t = 2 x$ . This implies $x = t /2$ and the differential $d x = d t /2$ . The infinite boundaries remain unchanged.

I = \int_{0}^{\infty} (\frac{t}{2})^{6} e^{- t} \frac{d t}{2} = \frac{1}{128} \int_{0}^{\infty} t^{6} e^{- t} d t

The resulting integral exactly matches the formal definition of the Gamma function $Γ (7)$ . Since $Γ (7) = 6! = 720$ , the calculation strictly simplifies to $\frac{720}{128} = \frac{45}{8}$ .

Q6Mathematics I

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