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The area of the region bounded by the curve $y = 5x + 2$, the x-axis, and the ordinates $x = -2$ and $x = 2$ can be found using integration. The area is given by the definite integral of the function $y$ with respect to $x$ from $x = -2$ to $x = 2$.
The area $A$ is given by: $$A = \int_{-2}^{2} |5x + 2| \, dx$$ Since $5x+2$ changes sign in the interval $[-2, 2]$, we need to find where $5x+2 = 0$. This occurs at $x = -\frac{2}{5}$. We need to split the integral at this point.
We split the integral into two parts: $$A = \int_{-2}^{-\frac{2}{5}} -(5x + 2) \, dx + \int_{-\frac{2}{5}}^{2} (5x + 2) \, dx$$
$$\int_{-2}^{-\frac{2}{5}} -(5x + 2) \, dx = -\left[\frac{5x^2}{2} + 2x\right]_{-2}^{-\frac{2}{5}}$$ $$= -\left[\left(\frac{5(-\frac{2}{5})^2}{2} + 2(-\frac{2}{5})\right) - \left(\frac{5(-2)^2}{2} + 2(-2)\right)\right]$$ $$= -\left[\left(\frac{5(\frac{4}{25})}{2} - \frac{4}{5}\right) - \left(\frac{5(4)}{2} - 4\right)\right]$$ $$= -\left[\left(\frac{2}{5} - \frac{4}{5}\right) - (10 - 4)\right]$$ $$= -\left[-\frac{2}{5} - 6\right] = \frac{2}{5} + 6 = \frac{2+30}{5} = \frac{32}{5}$$
$$\int_{-\frac{2}{5}}^{2} (5x + 2) \, dx = \left[\frac{5x^2}{2} + 2x\right]_{-\frac{2}{5}}^{2}$$ $$= \left[\left(\frac{5(2)^2}{2} + 2(2)\right) - \left(\frac{5(-\frac{2}{5})^2}{2} + 2(-\frac{2}{5})\right)\right]$$ $$= \left[\left(\frac{5(4)}{2} + 4\right) - \left(\frac{5(\frac{4}{25})}{2} - \frac{4}{5}\right)\right]$$ $$= \left[(10 + 4) - \left(\frac{2}{5} - \frac{4}{5}\right)\right]$$ $$= \left[14 - \left(-\frac{2}{5}\right)\right] = 14 + \frac{2}{5} = \frac{70+2}{5} = \frac{72}{5}$$
$$A = \frac{32}{5} + \frac{72}{5} = \frac{104}{5}$$
The area of the region is $\frac{104}{5}$ square units.
Final Answer: 104/5
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