To determine the area bounded by the line \( y = 5x + 2 \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \), integration is employed.
1. Problem Definition:
The given linear function is \( y = 5x + 2 \). The objective is to calculate the area between this function and the x-axis over the interval \( x \in [-2, 2] \).
2. X-axis Intercept Identification:
The x-intercept is found by setting \( y = 0 \):
\[ 0 = 5x + 2 \Rightarrow x = -\frac{2}{5} \]
The line intersects the x-axis at \( x = -\frac{2}{5} \). Consequently, the line lies below the x-axis for \( x \in [-2, -\frac{2}{5}] \) and above the x-axis for \( x \in [-\frac{2}{5}, 2] \).
3. Integral Decomposition:
The total area is computed by summing the absolute values of two integrals, as area is inherently positive:
\[ \text{Area} = \int_{-2}^{-\frac{2}{5}} |5x + 2|\,dx + \int_{-\frac{2}{5}}^{2} |5x + 2|\,dx \]
Within the interval \( x \in [-2, -\frac{2}{5}] \), \( 5x + 2<0 \). Within the interval \( x \in [-\frac{2}{5}, 2] \), \( 5x + 2>0 \).
4. Evaluation of the First Integral:
For the interval where \( 5x + 2<0 \):
\[ \int_{-2}^{-\frac{2}{5}} -(5x + 2)\,dx = \int_{-2}^{-\frac{2}{5}} (-5x - 2)\,dx \]
\[ = \left[-\frac{5}{2}x^2 - 2x\right]_{-2}^{-\frac{2}{5}} = \left(-\frac{5}{2}\left(-\frac{2}{5}\right)^2 - 2\left(-\frac{2}{5}\right)\right) - \left(-\frac{5}{2}(-2)^2 - 2(-2)\right) \]
\[ = \left(-\frac{5}{2}\left(\frac{4}{25}\right) + \frac{4}{5}\right) - \left(-\frac{5}{2}(4) + 4\right) = \left(-\frac{2}{5} + \frac{4}{5}\right) - (-10 + 4) = \frac{2}{5} - (-6) = \frac{2}{5} + 6 = \frac{32}{5} \]
5. Evaluation of the Second Integral:
For the interval where \( 5x + 2>0 \):
\[ \int_{-\frac{2}{5}}^{2} (5x + 2)\,dx = \left[\frac{5}{2}x^2 + 2x\right]_{-\frac{2}{5}}^{2} \]
\[ = \left(\frac{5}{2}(2)^2 + 2(2)\right) - \left(\frac{5}{2}\left(-\frac{2}{5}\right)^2 + 2\left(-\frac{2}{5}\right)\right) = \left(\frac{5}{2}(4) + 4\right) - \left(\frac{5}{2}\left(\frac{4}{25}\right) - \frac{4}{5}\right) = (10 + 4) - \left(\frac{2}{5} - \frac{4}{5}\right) = 14 - (-\frac{2}{5}) = 14 + \frac{2}{5} = \frac{72}{5} \]
6. Total Area Calculation:
The sum of the two areas yields the total area:
\[ \text{Area} = \frac{32}{5} + \frac{72}{5} = \frac{104}{5} \text{ square units} \]
Conclusion:
The calculated area is \( \frac{104}{5} \) square units.