The area of the region bounded by the line $y = 2x$ and the x-axis between $x = -2$ and $x = 2$ is

The area between the line $y = 2x$ and the x-axis from $x = -2$ to $x = 2$ needs to be found.

When $x = -2$: $y = 2(-2) = -4$ (below the x-axis)

When $x = 0$: $y = 2(0) = 0$ (crosses the x-axis)

When $x = 2$: $y = 2(2) = 4$ (above the x-axis)

The line is below the x-axis on the left side and above it on the right side.

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If calculating $\int_{-2}^{2} 2x \, dx$ directly:

$[x^2]_{-2}^{2}$

$= 4 - 4$

$= 0$

This gives 0 because the negative area (below x-axis) and positive area (above x-axis) cancel each other out. For finding area, the absolute value is needed - both regions should be added, not subtracted.

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The region splits into two parts at $x = 0$ where the line crosses the x-axis.

For $x = -2$ to $x = 0$ (below x-axis):

Since $y = 2x$ is negative here:

Area $= \int_{-2}^{0} |2x| \, dx$

$= \int_{-2}^{0} (-2x) \, dx$

$= [-x^2]_{-2}^{0}$

$= -(0)^2 - (-(−2)^2)$

$= 0 - (-4)$

$= 4$

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For $x = 0$ to $x = 2$ (above x-axis):

Since $y = 2x$ is positive here:

Area $= \int_{0}^{2} 2x \, dx$

$= [x^2]_{0}^{2}$

$= 2^2 - 0^2$

$= 4$

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Total Area $= 4 + 4 = 8$ square units