The area of the region bounded by the parabola `y=x^(2)-4x+5` and the straight line `y=x+1` is

Options

A. `1//2`
B. 2
C. 3
D. `9//2`

Answer

To find the area of the region bounded by the parabola \( y = x^2 - 4x + 5 \) and the line \( y = x + 1 \), we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the parabola and the line intersect. To do this, we set the equations equal to each other:

\[ x^2 - 4x + 5 = x + 1 \]

Rearranging gives:

\[ x^2 - 4x - x + 5 - 1 = 0 \implies x^2 - 5x + 4 = 0 \]

### Step 2: Solve the quadratic equation
We can factor the quadratic equation:

\[ (x - 4)(x - 1) = 0 \]

Thus, the solutions are:

\[ x = 4 \quad \text{and} \quad x = 1 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 1 \) to \( x = 4 \) is given by the integral of the upper curve minus the lower curve. Here, the parabola is above the line in this interval:

\[ A = \int_{1}^{4} \left( (x^2 - 4x + 5) - (x + 1) \right) \, dx \]

This simplifies to:

\[ A = \int_{1}^{4} \left( x^2 - 4x + 5 - x - 1 \right) \, dx = \int_{1}^{4} \left( x^2 - 5x + 4 \right) \, dx \]

### Step 4: Calculate the integral
Now we compute the integral:

\[ A = \int_{1}^{4} (x^2 - 5x + 4) \, dx \]

Calculating the integral:

\[ = \left[ \frac{x^3}{3} - \frac{5x^2}{2} + 4x \right]_{1}^{4} \]

### Step 5: Evaluate the definite integral
Now we evaluate at the bounds:

1. For \( x = 4 \):

\[ = \frac{4^3}{3} - \frac{5 \cdot 4^2}{2} + 4 \cdot 4 = \frac{64}{3} - \frac{80}{2} + 16 = \frac{64}{3} - 40 + 16 \]

Converting 40 to a fraction:

\[ = \frac{64}{3} - \frac{120}{3} + \frac{48}{3} = \frac{64 - 120 + 48}{3} = \frac{-8}{3} \]

2. For \( x = 1 \):

\[ = \frac{1^3}{3} - \frac{5 \cdot 1^2}{2} + 4 \cdot 1 = \frac{1}{3} - \frac{5}{2} + 4 \]

Converting \( \frac{5}{2} \) and 4 to fractions:

\[ = \frac{1}{3} - \frac{15}{6} + \frac{24}{6} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} \]

Converting \( \frac{3}{2} \):

\[ = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{18}{6} = \frac{19}{6} \]

### Step 6: Find the area
Now substitute back into the area formula:

\[ A = \left( \frac{-8}{3} \right) - \left( \frac{19}{6} \right) = \frac{-16}{6} - \frac{19}{6} = \frac{-35}{6} \]

Since area cannot be negative, we take the absolute value:

\[ A = \frac{35}{6} \]

### Final Answer
The area of the region bounded by the parabola and the line is:

\[ \frac{35}{6} \text{ square units} \]