Find the area under the curve `y =(x^2 + 2)^2 + 2x` between the ordinates x =0 and x=2`

Options

A. `236/15` sq units
B. `136/14` sq units
C. `430/14` sq units
D. `436/14` sq units

Answer

To find the area under the curve \( y = (x^2 + 2)^2 + 2x \) between the ordinates \( x = 0 \) and \( x = 2 \), we will follow these steps:

### Step 1: Define the function and the limits
We are given the function:
\[ y = (x^2 + 2)^2 + 2x \]
We need to find the area under this curve from \( x = 0 \) to \( x = 2 \).

### Step 2: Calculate the values of the function at the limits
First, we calculate the value of \( y \) at \( x = 0 \):
\[ y(0) = (0^2 + 2)^2 + 2 \cdot 0 = 2^2 + 0 = 4 \]
Next, we calculate the value of \( y \) at \( x = 2 \):
\[ y(2) = (2^2 + 2)^2 + 2 \cdot 2 = (4 + 2)^2 + 4 = 6^2 + 4 = 36 + 4 = 40 \]

### Step 3: Set up the integral for the area
The area \( A \) under the curve from \( x = 0 \) to \( x = 2 \) can be found using the definite integral:
\[ A = \int_{0}^{2} \left( (x^2 + 2)^2 + 2x \right) \, dx \]

### Step 4: Expand the integrand
We need to simplify the integrand:
\[ (x^2 + 2)^2 = x^4 + 4x^2 + 4 \]
Thus, the integrand becomes:
\[ A = \int_{0}^{2} \left( x^4 + 4x^2 + 4 + 2x \right) \, dx = \int_{0}^{2} \left( x^4 + 4x^2 + 2x + 4 \right) \, dx \]

### Step 5: Integrate term by term
Now we can integrate each 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 6: Evaluate the integral at the limits
Now we evaluate this expression at \( x = 2 \) and \( x = 0 \):
\[ A = \left( \frac{2^5}{5} + \frac{4 \cdot 2^3}{3} + 2^2 + 4 \cdot 2 \right) - \left( \frac{0^5}{5} + \frac{4 \cdot 0^3}{3} + 0^2 + 4 \cdot 0 \right) \]
Calculating at \( x = 2 \):
\[ A = \left( \frac{32}{5} + \frac{32}{3} + 4 + 8 \right) \]

### Step 7: Find a common denominator and simplify
To combine these fractions, we find a common denominator, which is 15:
\[ \frac{32}{5} = \frac{96}{15}, \quad \frac{32}{3} = \frac{160}{15}, \quad 4 = \frac{60}{15}, \quad 8 = \frac{120}{15} \]
Thus,
\[ A = \frac{96 + 160 + 60 + 120}{15} = \frac{436}{15} \]

### Final Result
The area under the curve between \( x = 0 \) and \( x = 2 \) is:
\[ \boxed{\frac{436}{15}} \]