`int_(1)^(3)(x-1)(x-2)(x-3)dx=`

Options

A. `3`
B. 2
C. 1
D. 0

Answer

To solve the integral \( I = \int_{1}^{3} (x-1)(x-2)(x-3) \, dx \), we can use the property of definite integrals that states:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

### Step 1: Apply the property
Here, \( a = 1 \) and \( b = 3 \). Therefore, \( a + b = 4 \). We can rewrite the integral as:

\[ I = \int_{1}^{3} (x-1)(x-2)(x-3) \, dx = \int_{1}^{3} (4-x-1)(4-x-2)(4-x-3) \, dx \]

### Step 2: Simplify the expression
Now, we simplify the expression inside the integral:

\[ 4 - x - 1 = 3 - x, \quad 4 - x - 2 = 2 - x, \quad 4 - x - 3 = 1 - x \]

So, we can rewrite the integral as:

\[ I = \int_{1}^{3} (3-x)(2-x)(1-x) \, dx \]

### Step 3: Expand the product
Next, we expand the product \( (3-x)(2-x)(1-x) \):

\[ (3-x)(2-x) = 6 - 5x + x^2 \]

Now, multiply this result by \( (1-x) \):

\[ (6 - 5x + x^2)(1 - x) = 6 - 6x - 5x + 5x^2 + x^2 - x^3 = 6 - 11x + 6x^2 - x^3 \]

### Step 4: Integrate term by term
Now we can integrate each term from 1 to 3:

\[ I = \int_{1}^{3} (6 - 11x + 6x^2 - x^3) \, dx \]

Calculating the integral term by term:

\[ = \left[ 6x - \frac{11x^2}{2} + 2x^3 - \frac{x^4}{4} \right]_{1}^{3} \]

### Step 5: Evaluate at the limits
Now we evaluate at the limits:

1. For \( x = 3 \):

\[ = 6(3) - \frac{11(3^2)}{2} + 2(3^3) - \frac{(3^4)}{4} \]
\[ = 18 - \frac{99}{2} + 54 - \frac{81}{4} \]
\[ = 18 + 54 - \frac{99}{2} - \frac{81}{4} \]
\[ = 72 - \frac{198}{4} - \frac{81}{4} = 72 - \frac{279}{4} = \frac{288 - 279}{4} = \frac{9}{4} \]

2. For \( x = 1 \):

\[ = 6(1) - \frac{11(1^2)}{2} + 2(1^3) - \frac{(1^4)}{4} \]
\[ = 6 - \frac{11}{2} + 2 - \frac{1}{4} \]
\[ = 8 - \frac{11}{2} - \frac{1}{4} = \frac{32 - 22 - 1}{4} = \frac{9}{4} \]

### Step 6: Combine results
Now, we combine the results:

\[ I = \left( \frac{9}{4} - \frac{9}{4} \right) = 0 \]

### Final Result
Thus, the value of the integral is:

\[ \int_{1}^{3} (x-1)(x-2)(x-3) \, dx = 0 \]