Sketch the region enclosed by the given curves. y = |3x|, y = x² - 4

Solution:

Given, the curves are y = |3x|, y = x² - 4

y = |3x| is a huge V passing through (-1, 3), origin is min point and (1, 3).

y = x² - 4 is the classical parabola y = x² but translated -4points downwards so that the absolute min point is (0, -4).

Both curves are symmetric with respect to vertical y axis,

For x > 0, the intersection point is

⇒ 3x = x² - 4

On rearranging,

⇒ x² - 3x - 4 = 0

⇒ x² - 4x + x - 4 = 0

⇒ x(x - 4) + 1(x - 4) = 0

⇒ (x + 1)(x - 4) = 0

⇒ x = 4 or -1

When x = 4, y = 3(4) = 12

When x = -1, y = 3(-1) = -3

The intersection point is (4, 12) symmetrically the other is (-4, 12).

Area is the integral of the difference = \(\int_{-4}^{4}(\left | 3x \right |-(x^{2}-4))dx\)

= \(\int_{-4}^{4}(3x-x^{2}+4)dx\)

= \(2\int_{0}^{4}(3x-x^{2}+4)dx\)

= \(2[3(\frac{x^{2}}{2})-\frac{x^{3}}{3}+4x]_{0}^{4}\)

= \(2[3(\frac{(4)^{2}}{2})-\frac{(4)^{3}}{3}+4(4)-0]\)

= \(2[3(\frac{16}{2})-\frac{64}{3}+16]\)

= \(2[3(8)-\frac{64}{3}+16]\)

= \(\frac{2}{3}[72-64+48]\)

= \(\frac{2}{3}[56]\)

= 112/3

Therefore, the area is 112/3 square units.

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Sketch the region enclosed by the given curves. y = |3x|, y = x² - 4

Summary:

The region enclosed by the given curves. y = |3x|, y = x² - 4 is 112/3 square units.