To find the area bounded by the curve [tex]\( y = 3x^2 + 2x + 1 \)[/tex], the [tex]\( x \)[/tex]-axis, and the lines [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex], we need to calculate the definite integral of the function [tex]\( y \)[/tex] from [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex].
1. Function to Integrate:
The given curve is [tex]\( y = 3x^2 + 2x + 1 \)[/tex].
2. Integration Limits:
We are integrating between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex].
3. Definite Integral Setup:
The area under the curve [tex]\( y = 3x^2 + 2x + 1 \)[/tex] from [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex] is found by evaluating the definite integral:
[tex]\[
\int_{1}^{2} (3x^2 + 2x + 1) \, dx
\][/tex]
4. Integrating the Function:
To find the integral [tex]\( \int (3x^2 + 2x + 1) \, dx \)[/tex]:
- Integral of [tex]\( 3x^2 \)[/tex] is [tex]\( x^3 \)[/tex].
- Integral of [tex]\( 2x \)[/tex] is [tex]\( x^2 \)[/tex].
- Integral of [tex]\( 1 \)[/tex] is [tex]\( x \)[/tex].
Therefore,
[tex]\[
\int (3x^2 + 2x + 1) \, dx = x^3 + x^2 + x
\][/tex]
5. Evaluating the Definite Integral:
We need to evaluate this antiderivative from [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]\[
\left[ x^3 + x^2 + x \right]_{1}^{2}
\][/tex]
First, evaluate at the upper limit [tex]\( x = 2 \)[/tex]:
[tex]\[
(2^3 + 2^2 + 2) = (8 + 4 + 2) = 14
\][/tex]
Next, evaluate at the lower limit [tex]\( x = 1 \)[/tex]:
[tex]\[
(1^3 + 1^2 + 1) = (1 + 1 + 1) = 3
\][/tex]
6. Finding the Area:
Subtract the lower limit result from the upper limit result:
[tex]\[
14 - 3 = 11
\][/tex]
Therefore, the area bounded by the curve [tex]\( y = 3x^2 + 2x + 1 \)[/tex], the [tex]\( x \)[/tex]-axis, and the lines [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex] is [tex]\( 11 \)[/tex]. Thus, the correct answer is [tex]\( \boxed{11} \)[/tex].
