Sketch the region enclosed by the given curves. y = |9x|, y = x² - 10

Solution:

Given, the curves are y = |9x|, y = x² - 10

y = |9x| is a huge V passing through (-1, 9), origin is min point and (1, 9).

y = x² - 10 is the classical parabola y = x² but translated -10points downwards so that the absolute min point is (0, -10).

Both curves are symmetric with respect to vertical y axis,

For x > 0, the intersection point is

9x = x² - 10

On rearranging,

x² - 9x - 10 = 0

x² -10x + x - 10 = 0

x(x - 10) + 1(x - 10) = 0

(x + 1)(x - 10) = 0

x = 10 or -1

When x = 10, y = 9(10) = 90

When x = -1, y = 9(-1) = -9

Area is the integral of the difference = \(\int_{-1}^{10}(\left | 9x \right |-(x^{2}-10))dx\)

= \(\int_{-1}^{10}(9x-x^{2}+10)dx\)

= \([9\frac{x^{2}}{2}-\frac{x^{3}}{3}+10x]_{-1}^{10}\)

= \([9\frac{(10)^{2}}{2}-\frac{(10)^{3}}{3}+10(10)]-[9\frac{(-1)^{2}}{2}-\frac{(-1)^{3}}{3}+10(-1)]\)

= \([\frac{(900)}{2}-\frac{(1000)}{3}+100]-[\frac{9}{2}+\frac{1}{3}-10]\)

= \([\frac{2700-2000+600}{6}]-[\frac{27+2-60}{6}]\)

= \([\frac{1300}{6}]-[\frac{-31}{6}]\)

= \([\frac{1331}{6}]\)

Therefore, the area is 1331/6 square units.

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Sketch the region enclosed by the given curves. y = |9x|, y = x² - 10

Summary:

The region enclosed by the given curves. y = |9x|, y = x² - 10 is 1331/6 square units.