let `f (x) = int _(0) ^(x) e ^(x-y) f'(y) dy - (x ^(2) -x+1)e ^(x)`
Find the number of roots of the equation `f (x) =0.`

Answer

To solve the problem, we need to find the number of roots of the equation \( f(x) = 0 \) where

\[ f(x) = \int_0^x e^{x-y} f'(y) \, dy - (x^2 - x + 1)e^x. \]

### Step 1: Set up the equation
We start by setting \( f(x) = 0 \):

\[ 0 = \int_0^x e^{x-y} f'(y) \, dy - (x^2 - x + 1)e^x. \]

Rearranging gives us:

\[ \int_0^x e^{x-y} f'(y) \, dy = (x^2 - x + 1)e^x. \]

### Step 2: Factor out \( e^x \)
Since \( e^{x-y} = e^x e^{-y} \), we can factor \( e^x \) out of the integral:

\[ \int_0^x e^{x-y} f'(y) \, dy = e^x \int_0^x e^{-y} f'(y) \, dy. \]

Thus, we can rewrite our equation as:

\[ e^x \int_0^x e^{-y} f'(y) \, dy = (x^2 - x + 1)e^x. \]

### Step 3: Cancel \( e^x \)
Since \( e^x \) is never zero, we can safely divide both sides by \( e^x \):

\[ \int_0^x e^{-y} f'(y) \, dy = x^2 - x + 1. \]

### Step 4: Differentiate both sides
Now we differentiate both sides with respect to \( x \):

Using the Fundamental Theorem of Calculus on the left side, we get:

\[ e^{-x} f'(x) = 2x - 1. \]

### Step 5: Solve for \( f'(x) \)
Multiplying both sides by \( e^x \) gives us:

\[ f'(x) = (2x - 1)e^x. \]

### Step 6: Integrate to find \( f(x) \)
Now we integrate \( f'(x) \):

\[ f(x) = \int (2x - 1)e^x \, dx. \]

Using integration by parts, let \( u = 2x - 1 \) and \( dv = e^x \, dx \):

\[ du = 2 \, dx, \quad v = e^x. \]

Thus,

\[ f(x) = (2x - 1)e^x - \int 2e^x \, dx = (2x - 1)e^x - 2e^x + C, \]

where \( C \) is the constant of integration.

Simplifying gives:

\[ f(x) = (2x - 3)e^x + C. \]

### Step 7: Find \( f(0) \)
To find \( C \), we evaluate \( f(0) \):

\[ f(0) = (2(0) - 3)e^0 + C = -3 + C. \]

### Step 8: Set \( f(0) \) in the original equation
From our earlier work, we know \( f(0) = -1 \):

\[ -3 + C = -1 \implies C = 2. \]

### Step 9: Final form of \( f(x) \)
Thus, we have:

\[ f(x) = (2x - 3)e^x + 2. \]

### Step 10: Set \( f(x) = 0 \)
Now we set \( f(x) = 0 \):

\[ (2x - 3)e^x + 2 = 0. \]

Rearranging gives:

\[ (2x - 3)e^x = -2. \]

Since \( e^x > 0 \) for all \( x \), the left side is positive for \( x > \frac{3}{2} \) and negative for \( x < \frac{3}{2} \). Thus, we need to find where:

\[ 2x - 3 = -\frac{2}{e^x}. \]

### Step 11: Analyze the function
The left side \( 2x - 3 \) is a linear function that crosses zero at \( x = \frac{3}{2} \). The right side \( -\frac{2}{e^x} \) is always negative.

### Step 12: Finding the roots
The function \( f(x) \) will cross the x-axis at two points: one before \( x = \frac{3}{2} \) and one after. Thus, we conclude that there are **two roots** for the equation \( f(x) = 0 \).

### Conclusion
The number of roots of the equation \( f(x) = 0 \) is **2**.