Consider the function `f(x)=x^(3)-8x^(2)+20x-13`
Area enclosed by `y=f(x)` and the coordinate axes is

Options

A. `65//12`
B. `13//12`
C. `71//12`
D. None of these

Answer

To find the area enclosed by the function \( f(x) = x^3 - 8x^2 + 20x - 13 \) and the coordinate axes, we will follow these steps:

### Step 1: Find the roots of the function
To determine the area enclosed by the curve and the axes, we first need to find the points where the function intersects the x-axis. This is done by solving the equation \( f(x) = 0 \).

We can start by checking for rational roots using the Rational Root Theorem. Testing \( x = 1 \):

\[ f(1) = 1^3 - 8(1^2) + 20(1) - 13 = 1 - 8 + 20 - 13 = 0 \]

Since \( f(1) = 0 \), \( x = 1 \) is a root. We can factor \( f(x) \) as \( (x - 1)g(x) \).

### Step 2: Perform polynomial long division
Now, we will divide \( f(x) \) by \( (x - 1) \) to find \( g(x) \).

Using synthetic division or polynomial long division, we divide \( x^3 - 8x^2 + 20x - 13 \) by \( x - 1 \):

1. Divide the leading term: \( x^3 \div x = x^2 \)
2. Multiply: \( x^2(x - 1) = x^3 - x^2 \)
3. Subtract: \( (-8x^2 + x^2) = -7x^2 \)
4. Bring down the next term: \( -7x^2 + 20x \)
5. Divide: \( -7x^2 \div x = -7x \)
6. Multiply: \( -7x(x - 1) = -7x^2 + 7x \)
7. Subtract: \( (20x - 7x) = 13x \)
8. Bring down: \( 13x - 13 \)
9. Divide: \( 13x \div x = 13 \)
10. Multiply: \( 13(x - 1) = 13x - 13 \)
11. Subtract: \( 0 \)

Thus, we have:

\[ f(x) = (x - 1)(x^2 - 7x + 13) \]

### Step 3: Solve the quadratic equation
Next, we need to find the roots of \( x^2 - 7x + 13 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(13)}}{2(1)} = \frac{7 \pm \sqrt{49 - 52}}{2} = \frac{7 \pm \sqrt{-3}}{2} \]

Since the discriminant is negative, \( x^2 - 7x + 13 \) has no real roots. Therefore, the only real root is \( x = 1 \).

### Step 4: Determine the area under the curve
To find the area enclosed by the curve and the axes, we need to evaluate the integral from \( x = 0 \) to \( x = 1 \):

\[ \text{Area} = \int_0^1 |f(x)| \, dx \]

Since \( f(x) \) is negative between \( x = 0 \) and \( x = 1 \), we will take the absolute value:

\[ \text{Area} = -\int_0^1 (x^3 - 8x^2 + 20x - 13) \, dx \]

### Step 5: Calculate the integral
Calculating the integral:

\[ \int (x^3 - 8x^2 + 20x - 13) \, dx = \frac{x^4}{4} - \frac{8x^3}{3} + 10x^2 - 13x \]

Evaluating from 0 to 1:

\[ = \left[ \frac{1^4}{4} - \frac{8(1^3)}{3} + 10(1^2) - 13(1) \right] - \left[ 0 \right] \]

Calculating this gives:

\[ = \left[ \frac{1}{4} - \frac{8}{3} + 10 - 13 \right] \]

Finding a common denominator (12):

\[ = \left[ \frac{3}{12} - \frac{32}{12} + \frac{120}{12} - \frac{156}{12} \right] = \left[ \frac{3 - 32 + 120 - 156}{12} \right] = \frac{-65}{12} \]

Taking the absolute value:

\[ \text{Area} = \frac{65}{12} \]

### Final Answer
The area enclosed by \( y = f(x) \) and the coordinate axes is:

\[ \boxed{\frac{65}{12}} \text{ square units} \]