Sketch the region enclosed by the given curves.y = |3x|, y = x2 − 4 Find its area.?

Well,

first of all, welcome to this forum from (maybe the only) french guy in the zone !! ?

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

y = x2 − 4 is the classical parabola y=x^2 but translated -4 points downwards so that the absolute min point is (0, -4)

curves are both symmetric wrt to vertical y axis,

for x > 0, the intersection point is so that :

3x = x^2 - 4

x^2 - 3x - 4 = 0

(x+1)(x-4) = 0 ==> x = 4 and y = |3x| = x^2 - 4 for x=4 therefore : y = 12

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

area is the integral of the difference

∫ (x € [-4 , 4] ) ( |3x| - (x^2 - 4)) dx = 2 ∫(x € [0,4]) (3x - x^2 + 4)) dx

= 2 * [3x^2 /2] - x^3 /3 + 4x](between 0 and 4)

= 2 [ 38 - 64/3 + 16]

= (2/3) [ 383 - 64 + 163]

= (2/3) * [ 72 - 64 + 48]

= (2/3) * 56

= 112/3

= 37 1/3

hope it' ll help !!

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Based on the graph you know that they intersect at (-4, 12) and (4, 12). So just follow the usual protocol for these problems and do the integral from -4 to 4 of the top function minus the bottom function.

Since you have an absolute value you technically have 2 different graphs. On the negative side of the graph you have the function -3x. On the positive side you have the function 3x. Therefore to solve the integral by hand you must split the integral into 2 parts. One that calculates the area enclosed by -3x and x^2-4 and another that calculates the area enclosed by 3x and x^2-4. You should get a final area of 37.33.

Alternatively, if you look at the graph you can see that the region enclosed by y = |3x| and y = x^2 − 4 is symmetrical about the y axis. This means that the area in the positive region (x from 0 to 4) is equivalent to the area in the negative region of the graph (x from -4 to 0). So instead of calculating an integral from -4 to 4, you can calculate and integral of half of the region (say from 0 to 4) and double that area.

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