Evaluate the improper integral ∫_0^(∞) x^2.5 e^-x dx and give the numerical value up to three decimal places.

Evaluate the improper integral $\int_{0}^{\infty} x^{2.5} e^{- x} d x$ and give the numerical value up to three decimal places.

hard 4 marks Class 12 · Mathematics · Integrals

Answer

The value of the integral is

3.323

The correct answer is The value of the integral is $3.323$ . — Round to three decimal places: $3.323$ .

Recognise the integral as a Gamma‑function form: $\int_{0}^{\infty} x^{n} e^{- x} d x = Γ (n + 1)$ .
Identify $n = 2.5$ . Hence the integral equals $Γ (2.5 + 1) = Γ (3.5)$ .
Use the property $Γ (z + 1) = z Γ (z)$ . Apply repeatedly:
$Γ (3.5) = 2.5 Γ (2.5)$ .
$Γ (2.5) = 1.5 Γ (1.5)$ .
$Γ (1.5) = \frac{π}{2}$ (a standard value).
Combine: $Γ (3.5) = 2.5 \times 1.5 \times \frac{π}{2} = \frac{2.5 \times 1.5}{2} π = \frac{3.75}{2} π = 1.875 π$ .
Compute numerically: $π \approx 1.77245385$ .
Multiply: $1.875 \times 1.77245385 \approx 3.323$ .
Round to three decimal places: $3.323$ .

Common mistake: Treating the exponent 2.5 as an integer and applying the factorial formula directly.

Concept tested: Gamma function evaluation of improper integrals