Evaluate `int_(0)^(infty) e^(-x)x^(3)dx`.

Answer

To evaluate the integral \( I = \int_{0}^{\infty} e^{-x} x^{3} \, dx \), we will use integration by parts multiple times.

### Step 1: Set up integration by parts
We will use the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = x^3 \) (which we will differentiate)
- \( dv = e^{-x} \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now we differentiate \( u \) and integrate \( dv \):
- \( du = 3x^2 \, dx \)
- \( v = -e^{-x} \)

### Step 3: Apply integration by parts
Now we apply the integration by parts formula:
\[ I = \left[ -x^3 e^{-x} \right]_{0}^{\infty} + \int_{0}^{\infty} 3x^2 e^{-x} \, dx \]

### Step 4: Evaluate the boundary term
Now we evaluate the boundary term \( \left[ -x^3 e^{-x} \right]_{0}^{\infty} \):
- As \( x \to \infty \), \( e^{-x} \) approaches 0 faster than \( x^3 \) approaches infinity, so \( -x^3 e^{-x} \to 0 \).
- As \( x \to 0 \), \( -x^3 e^{-x} \to 0 \).

Thus, the boundary term evaluates to 0:
\[ \left[ -x^3 e^{-x} \right]_{0}^{\infty} = 0 \]

### Step 5: Simplify the integral
Now we have:
\[ I = 0 + \int_{0}^{\infty} 3x^2 e^{-x} \, dx = 3 \int_{0}^{\infty} x^2 e^{-x} \, dx \]

### Step 6: Repeat integration by parts
We need to evaluate \( \int_{0}^{\infty} x^2 e^{-x} \, dx \). We will apply integration by parts again:
Let:
- \( u = x^2 \)
- \( dv = e^{-x} \, dx \)

Then:
- \( du = 2x \, dx \)
- \( v = -e^{-x} \)

Applying integration by parts again:
\[ \int x^2 e^{-x} \, dx = \left[ -x^2 e^{-x} \right]_{0}^{\infty} + \int 2x e^{-x} \, dx \]

### Step 7: Evaluate the boundary term again
The boundary term \( \left[ -x^2 e^{-x} \right]_{0}^{\infty} \) evaluates to 0 for the same reasons as before.

### Step 8: Simplify the integral
Thus:
\[ \int_{0}^{\infty} x^2 e^{-x} \, dx = 0 + 2 \int_{0}^{\infty} x e^{-x} \, dx \]

### Step 9: Evaluate \( \int_{0}^{\infty} x e^{-x} \, dx \)
We apply integration by parts one more time:
Let:
- \( u = x \)
- \( dv = e^{-x} \, dx \)

Then:
- \( du = dx \)
- \( v = -e^{-x} \)

Applying integration by parts:
\[ \int x e^{-x} \, dx = \left[ -x e^{-x} \right]_{0}^{\infty} + \int e^{-x} \, dx \]

The boundary term \( \left[ -x e^{-x} \right]_{0}^{\infty} \) evaluates to 0. Thus:
\[ \int_{0}^{\infty} x e^{-x} \, dx = 0 + \left[ -e^{-x} \right]_{0}^{\infty} = 0 - (-1) = 1 \]

### Step 10: Combine results
Now we can combine our results:
\[ \int_{0}^{\infty} x^2 e^{-x} \, dx = 2 \cdot 1 = 2 \]
Thus:
\[ I = 3 \cdot 2 = 6 \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\infty} e^{-x} x^{3} \, dx = 6 \]