Evaluate :
`int_0^2 (2x^2+3x+1)dx` by second Fundamental Theorem.

Answer

To evaluate the integral \( \int_0^2 (2x^2 + 3x + 1) \, dx \) using the second Fundamental Theorem of Calculus, we will follow these steps:

### Step 1: Identify the function to integrate
The function we want to integrate is:
\[ f(x) = 2x^2 + 3x + 1 \]

### Step 2: Find the antiderivative of \( f(x) \)
To find the antiderivative \( F(x) \), we integrate each term of \( f(x) \):
1. The antiderivative of \( 2x^2 \) is \( \frac{2}{3}x^3 \).
2. The antiderivative of \( 3x \) is \( \frac{3}{2}x^2 \).
3. The antiderivative of \( 1 \) is \( x \).

Thus, the antiderivative \( F(x) \) is:
\[ F(x) = \frac{2}{3}x^3 + \frac{3}{2}x^2 + x \]

### Step 3: Evaluate \( F(x) \) at the bounds
Now, we will evaluate \( F(x) \) at the upper limit \( x = 2 \) and the lower limit \( x = 0 \):
1. Calculate \( F(2) \):
\[ F(2) = \frac{2}{3}(2^3) + \frac{3}{2}(2^2) + 2 \]
\[ = \frac{2}{3}(8) + \frac{3}{2}(4) + 2 \]
\[ = \frac{16}{3} + 6 + 2 \]
\[ = \frac{16}{3} + \frac{18}{3} + \frac{6}{3} = \frac{40}{3} \]

2. Calculate \( F(0) \):
\[ F(0) = \frac{2}{3}(0^3) + \frac{3}{2}(0^2) + 0 = 0 \]

### Step 4: Apply the second Fundamental Theorem of Calculus
According to the second Fundamental Theorem of Calculus:
\[ \int_a^b f(x) \, dx = F(b) - F(a) \]
Substituting our values:
\[ \int_0^2 (2x^2 + 3x + 1) \, dx = F(2) - F(0) = \frac{40}{3} - 0 = \frac{40}{3} \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^2 (2x^2 + 3x + 1) \, dx = \frac{40}{3} \]

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