1. **Problem Statement:** Evaluate the integral $$\int_0^\infty x^n e^{-x} \, dx$$ for $n=0,1,2,3$. 2. **Formula and Important Rule:** This integral is known as the Gamma function for positive integers, defined as $$\Gamma(n+1) = \int_0^\infty x^n e^{-x} \, dx = n!$$ where $n!$ is the factorial of $n$. 3. **Step-by-step Evaluation:** - For $n=0$: $$\int_0^\infty x^0 e^{-x} \, dx = \int_0^\infty e^{-x} \, dx = \left[-e^{-x}\right]_0^\infty = 1$$ - For $n=1$: $$\int_0^\infty x^1 e^{-x} \, dx = 1! = 1$$ - For $n=2$: $$\int_0^\infty x^2 e^{-x} \, dx = 2! = 2$$ - For $n=3$: $$\int_0^\infty x^3 e^{-x} \, dx = 3! = 6$$ 4. **Prediction for arbitrary positive integer $n$:** We predict that $$\int_0^\infty x^n e^{-x} \, dx = n!$$ 5. **Proof by Mathematical Induction:** - **Base case:** For $n=0$, the integral equals $1 = 0!$, which is true. - **Inductive step:** Assume for some $k \geq 0$, $$\int_0^\infty x^k e^{-x} \, dx = k!$$ - We need to prove for $k+1$: Using integration by parts, let $u = x^{k+1}$ and $dv = e^{-x} dx$. Then $du = (k+1) x^k dx$ and $v = -e^{-x}$. So, $$\int_0^\infty x^{k+1} e^{-x} \, dx = \left[-x^{k+1} e^{-x}\right]_0^\infty + (k+1) \int_0^\infty x^k e^{-x} \, dx$$ The boundary term is zero because $x^{k+1} e^{-x} \to 0$ as $x \to \infty$ and is zero at $x=0$. Therefore, $$\int_0^\infty x^{k+1} e^{-x} \, dx = (k+1) k! = (k+1)!$$ This completes the induction. **Final answer:** $$\int_0^\infty x^n e^{-x} \, dx = n!$$ for all non-negative integers $n$. --- **Summary:** The integral evaluates to the factorial of $n$, which is a fundamental result connecting integrals and factorials via the Gamma function.