If `f'(x)=x e^(x)` and `f(0)=1` then find f(x).

Answer

To find the function \( f(x) \) given that \( f'(x) = x e^x \) and \( f(0) = 1 \), we will follow these steps:

### Step 1: Integrate \( f'(x) \)
We start by integrating \( f'(x) \) to find \( f(x) \):
\[ f(x) = \int f'(x) \, dx = \int x e^x \, dx \]

### Step 2: Use Integration by Parts
To integrate \( x e^x \), we use integration by parts, where we let:
- \( u = x \) (thus \( du = dx \))
- \( dv = e^x dx \) (thus \( v = e^x \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ \int x e^x \, dx = x e^x - \int e^x \, dx \]

### Step 3: Integrate \( e^x \)
Now we integrate \( e^x \):
\[ \int e^x \, dx = e^x \]
So, substituting back, we get:
\[ \int x e^x \, dx = x e^x - e^x + C \]

### Step 4: Write \( f(x) \)
Thus, we can express \( f(x) \):
\[ f(x) = x e^x - e^x + C \]

### Step 5: Use the Initial Condition
Now we use the initial condition \( f(0) = 1 \) to find \( C \):
\[ f(0) = 0 \cdot e^0 - e^0 + C = 1 \]
This simplifies to:
\[ 0 - 1 + C = 1 \]
\[ C - 1 = 1 \implies C = 2 \]

### Step 6: Final Expression for \( f(x) \)
Substituting \( C \) back into the expression for \( f(x) \):
\[ f(x) = x e^x - e^x + 2 \]

### Conclusion
The function \( f(x) \) is:
\[ f(x) = x e^x - e^x + 2 \]