Question

Using integration, find the area of the region bounded by the line \[ y = 5x + 2, \] the \( x \)-axis, and the ordinates \( x = -2 \) and \( x = 2 \).

Hint:

To find the area between a curve and the \( x \)-axis, use \( A = \int_a^b f(x) dx \) and ensure proper limits.

Answer 1

To solve the problem, we are required to find the area enclosed between the line \( y = 5x + 2 \), the x-axis, and the vertical lines (ordinates) \( x = -2 \) and \( x = 2 \) using integration.

1. Understanding the Problem:
We are given the linear function:
\( y = 5x + 2 \)
We want to compute the area between this curve and the x-axis from \( x = -2 \) to \( x = 2 \).

2. Identifying the Intersection with the X-axis:
Set \( y = 0 \) to find the x-intercept:
\[ 0 = 5x + 2 \Rightarrow x = -\frac{2}{5} \]
This means the line cuts the x-axis at \( x = -\frac{2}{5} \). So between \( x = -2 \) and \( x = -\frac{2}{5} \), the line is below the x-axis, and from \( x = -\frac{2}{5} \) to \( x = 2 \), the line is above the x-axis.

3. Splitting the Integral:
We split the total area into two parts and take the absolute value because area is always positive:
\[ \text{Area} = \int_{-2}^{-\frac{2}{5}} |5x + 2|\,dx + \int_{-\frac{2}{5}}^{2} |5x + 2|\,dx \]
Between \( x = -2 \) and \( x = -\frac{2}{5} \), \( 5x + 2 < 0 \). Between \( x = -\frac{2}{5} \) and \( x = 2 \), \( 5x + 2 > 0 \).

4. Evaluating the First Integral:
\[ \int_{-2}^{-\frac{2}{5}} -(5x + 2)\,dx = \int_{-2}^{-\frac{2}{5}} (-5x - 2)\,dx \]
\[ = \left[-\frac{5}{2}x^2 - 2x\right]_{-2}^{-\frac{2}{5}} = \left(-\frac{5}{2}\left(\frac{4}{25}\right) - 2\left(-\frac{2}{5}\right)\right) - \left(-\frac{5}{2}(4) - 2(-2)\right) \]
\[ = \left(-\frac{2}{5} + \frac{4}{5}\right) - (-10 + 4) = \frac{2}{5} - (-6) = \frac{2}{5} + 6 = \frac{32}{5} \]

5. Evaluating the Second Integral:
\[ \int_{-\frac{2}{5}}^{2} (5x + 2)\,dx = \left[\frac{5}{2}x^2 + 2x\right]_{-\frac{2}{5}}^{2} \]
\[ = \left(\frac{5}{2}(4) + 2(2)\right) - \left(\frac{5}{2}\left(\frac{4}{25}\right) + 2\left(-\frac{2}{5}\right)\right) = (10 + 4) - \left(\frac{2}{5} - \frac{4}{5}\right) = 14 - (-\frac{2}{5}) = 14 + \frac{2}{5} = \frac{72}{5} \]

6. Total Area:
Add both areas:
\[ \text{Area} = \frac{32}{5} + \frac{72}{5} = \frac{104}{5} \text{ square units} \]

Final Answer:
The required area is \( \frac{104}{5} \) square units.