To find the area of the region bounded by the curve \( y = (x^2 + 2)^2 + 2x \) between the ordinates \( x = 0 \) and \( x = 2 \), we will follow these steps:
### Step 1: Evaluate the function at the limits
First, we need to evaluate the function at the endpoints \( x = 0 \) and \( x = 2 \).
1. **At \( x = 0 \)**:
\[
y = (0^2 + 2)^2 + 2 \cdot 0 = (2)^2 + 0 = 4
\]
So, the point is \( (0, 4) \).
2. **At \( x = 2 \)**:
\[
y = (2^2 + 2)^2 + 2 \cdot 2 = (4 + 2)^2 + 4 = (6)^2 + 4 = 36 + 4 = 40
\]
So, the point is \( (2, 40) \).
### Step 2: Set up the integral for the area
The area \( A \) under the curve from \( x = 0 \) to \( x = 2 \) can be calculated using the definite integral:
\[
A = \int_{0}^{2} y \, dx = \int_{0}^{2} \left( (x^2 + 2)^2 + 2x \right) dx
\]
### Step 3: Expand the integrand
We need to expand \( (x^2 + 2)^2 \):
\[
(x^2 + 2)^2 = x^4 + 4x^2 + 4
\]
Thus, the integrand becomes:
\[
y = x^4 + 4x^2 + 4 + 2x = x^4 + 4x^2 + 2x + 4
\]
### Step 4: Integrate the function
Now, we can integrate:
\[
A = \int_{0}^{2} (x^4 + 4x^2 + 2x + 4) \, dx
\]
Calculating the integral term by term:
\[
\int x^4 \, dx = \frac{x^5}{5}, \quad \int 4x^2 \, dx = \frac{4x^3}{3}, \quad \int 2x \, dx = x^2, \quad \int 4 \, dx = 4x
\]
Thus, we have:
\[
A = \left[ \frac{x^5}{5} + \frac{4x^3}{3} + x^2 + 4x \right]_{0}^{2}
\]
### Step 5: Evaluate the definite integral
Now we will evaluate this from \( 0 \) to \( 2 \):
\[
A = \left( \frac{2^5}{5} + \frac{4 \cdot 2^3}{3} + 2^2 + 4 \cdot 2 \right) - \left( \frac{0^5}{5} + \frac{4 \cdot 0^3}{3} + 0^2 + 4 \cdot 0 \right)
\]
Calculating the upper limit:
\[
= \frac{32}{5} + \frac{32}{3} + 4 + 8
\]
To simplify, we need a common denominator, which is 15:
\[
= \frac{32 \cdot 3}{15} + \frac{32 \cdot 5}{15} + \frac{60}{15} + \frac{120}{15}
\]
\[
= \frac{96 + 160 + 60 + 120}{15} = \frac{436}{15}
\]
### Step 6: Final area calculation
Thus, the area \( A \) is:
\[
A = \frac{436}{15}
\]
### Final Answer
The area of the region bounded by the curve \( y = (x^2 + 2)^2 + 2x \) between the ordinates \( x = 0 \) and \( x = 2 \) is \( \frac{436}{15} \).
