Evaluate the improper integral I=∫_0^(∞) x^(3/2) e^-x dx and give the numerical value correct to three decimal places.

Evaluate the improper integral $I = \int_{0}^{\infty} x^{\frac{3}{2}} e^{- x} d x$ and give the numerical value correct to three decimal places.

hard 4 marks Class 12 · Mathematics · Integrals

Answer

The value of the integral is

1.329

(to three decimal places).

The correct answer is The value of the integral is $1.329$ (to three decimal places). — Round to three decimal places: $1.329$ .

Recognise the integral as a Gamma function: $\int_{0}^{\infty} x^{n} e^{- x} d x = Γ (n + 1)$ .
Here $n = \frac{3}{2}$ , so the integral equals $Γ (\frac{5}{2})$ .
Use the recurrence $Γ (z + 1) = z Γ (z)$ with $Γ (\frac{1}{2}) = π$ :
First, $Γ (\frac{3}{2}) = \frac{1}{2} π$ .
Then, $Γ (\frac{5}{2}) = \frac{3}{2} Γ (\frac{3}{2}) = \frac{3}{2} \cdot \frac{1}{2} π = \frac{3}{4} π$ .
Compute numerically: $π \approx 1.77245385$ , so $\frac{3}{4} π \approx 0.75 \times 1.77245385 = 1.3293404$ .
Round to three decimal places: $1.329$ .

Common mistake: Using

Γ (n) = (n - 1)!

for non‑integer

n

, which leads to an incorrect value.

Concept tested: Gamma function for improper integrals