Answer :
To find the area enclosed between the curves [tex]\( y = x^2 \)[/tex] and [tex]\( y = x + 2 \)[/tex], we need to follow several steps: finding the points of intersection, setting up the integral to find the area between these curves, and then evaluating that integral.
### Step 1: Find the Points of Intersection
First, we need to determine where the curves intersect by setting their equations equal to each other:
[tex]\[ x^2 = x + 2 \][/tex]
Rearrange this equation to form a standard quadratic equation:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
Factorize the quadratic equation:
[tex]\[ (x - 2)(x + 1) = 0 \][/tex]
The solutions to this equation are the [tex]\( x \)[/tex]-values of the points of intersection:
[tex]\[ x = 2 \quad \text{and} \quad x = -1 \][/tex]
### Step 2: Set Up the Integral for the Enclosed Area
Next, we identify which curve is above the other within the interval [tex]\([-1, 2]\)[/tex]. For [tex]\( y = x + 2 \)[/tex] and [tex]\( y = x^2 \)[/tex]:
- For [tex]\( x \)[/tex] between -1 and 2, the line [tex]\( y = x + 2 \)[/tex] lies above the parabola [tex]\( y = x^2 \)[/tex].
The area between the two curves from [tex]\( x = -1 \)[/tex] to [tex]\( x = 2 \)[/tex] can be found by integrating the difference between the upper function [tex]\( y = x + 2 \)[/tex] and the lower function [tex]\( y = x^2 \)[/tex]:
[tex]\[ A = \int_{-1}^{2} ((x + 2) - x^2) \, dx \][/tex]
### Step 3: Evaluate the Integral
We integrate the function with respect to [tex]\( x \)[/tex]:
[tex]\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx \][/tex]
Separate the integral:
[tex]\[ A = \int_{-1}^{2} x \, dx + \int_{-1}^{2} 2 \, dx - \int_{-1}^{2} x^2 \, dx \][/tex]
Evaluate each integral individually:
1. [tex]\(\int_{-1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{2} = \left( \frac{2^2}{2} \right) - \left( \frac{(-1)^2}{2} \right) = \frac{4}{2} - \frac{1}{2} = 2 - 0.5 = 1.5\)[/tex]
2. [tex]\(\int_{-1}^{2} 2 \, dx = \left[ 2x \right]_{-1}^{2} = 2(2) - 2(-1) = 4 + 2 = 6\)[/tex]
3. [tex]\(\int_{-1}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{2} = \left( \frac{2^3}{3} \right) - \left( \frac{-1^3}{3} \right) = \frac{8}{3} - \left( -\frac{1}{3} \right) = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3\)[/tex]
Combine these results to find the net area:
[tex]\[ A = 1.5 + 6 - 3 = 4.5 \][/tex]
So, the area enclosed between the curves [tex]\( y = x^2 \)[/tex] and [tex]\( y = x + 2 \)[/tex] is:
[tex]\[ \boxed{\frac{9}{2}} \][/tex]
### Step 1: Find the Points of Intersection
First, we need to determine where the curves intersect by setting their equations equal to each other:
[tex]\[ x^2 = x + 2 \][/tex]
Rearrange this equation to form a standard quadratic equation:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
Factorize the quadratic equation:
[tex]\[ (x - 2)(x + 1) = 0 \][/tex]
The solutions to this equation are the [tex]\( x \)[/tex]-values of the points of intersection:
[tex]\[ x = 2 \quad \text{and} \quad x = -1 \][/tex]
### Step 2: Set Up the Integral for the Enclosed Area
Next, we identify which curve is above the other within the interval [tex]\([-1, 2]\)[/tex]. For [tex]\( y = x + 2 \)[/tex] and [tex]\( y = x^2 \)[/tex]:
- For [tex]\( x \)[/tex] between -1 and 2, the line [tex]\( y = x + 2 \)[/tex] lies above the parabola [tex]\( y = x^2 \)[/tex].
The area between the two curves from [tex]\( x = -1 \)[/tex] to [tex]\( x = 2 \)[/tex] can be found by integrating the difference between the upper function [tex]\( y = x + 2 \)[/tex] and the lower function [tex]\( y = x^2 \)[/tex]:
[tex]\[ A = \int_{-1}^{2} ((x + 2) - x^2) \, dx \][/tex]
### Step 3: Evaluate the Integral
We integrate the function with respect to [tex]\( x \)[/tex]:
[tex]\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx \][/tex]
Separate the integral:
[tex]\[ A = \int_{-1}^{2} x \, dx + \int_{-1}^{2} 2 \, dx - \int_{-1}^{2} x^2 \, dx \][/tex]
Evaluate each integral individually:
1. [tex]\(\int_{-1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{2} = \left( \frac{2^2}{2} \right) - \left( \frac{(-1)^2}{2} \right) = \frac{4}{2} - \frac{1}{2} = 2 - 0.5 = 1.5\)[/tex]
2. [tex]\(\int_{-1}^{2} 2 \, dx = \left[ 2x \right]_{-1}^{2} = 2(2) - 2(-1) = 4 + 2 = 6\)[/tex]
3. [tex]\(\int_{-1}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{2} = \left( \frac{2^3}{3} \right) - \left( \frac{-1^3}{3} \right) = \frac{8}{3} - \left( -\frac{1}{3} \right) = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3\)[/tex]
Combine these results to find the net area:
[tex]\[ A = 1.5 + 6 - 3 = 4.5 \][/tex]
So, the area enclosed between the curves [tex]\( y = x^2 \)[/tex] and [tex]\( y = x + 2 \)[/tex] is:
[tex]\[ \boxed{\frac{9}{2}} \][/tex]
