would someone mind checking if my answer is correct? Antidifferentiation?

Question

Just not sure if I'm correct. Thanks

1. Find the area bounded by the curves y= I X^2-4 I and y= X^2 -4

My answer: 8/3

Answer 1

Required area is 2 x area between y = x² - 4 and the x - axis from x = -2 to 2

So area = 2|∫[-2 to 2] (x² - 4) dx|
= 2 * | [x³/3 - 2x²] from,-2 to 2 |
= 2 * | [(8/3 - 8) - (-8/3 - 8)] |
= 32/3

Answer 2

You're not. First we recast the absolute value function in terms of a piecewise function - note that x²-4≤0 for all x in the range [-2, 2], and x²-4>0 otherwise, so the first function is:
y={x²-4 if x<2 or x>2, 4-x² if -2≤x≤2}

This function is clearly the same as the second function for all x not in [-2, 2], and so the area between the functions everywhere else is zero. To find the area between the two functions between -2 and 2, we take the definite integral:

[-2, 2]∫4-x²-(x²-4) dx
Which is:
[-2, 2]∫8-2x² dx
8x-2/3x³ from -2 to 2
16-16/3 - (-16+16/3)
32-32/3
64/3

So your area is 64/3, not 8/3.

Answer 3

Both above answers are wrong. Pascal has the right idea but I think he made a mistake.

It's easy to figure this out if you notice that the area under the curve y=x^2-4 from -2 to 0 is equal to the area from 0 to 2. So, this area works out to be: |x^3/3-4x| from -2 to 0 multiplied by 2 is 32/3. The remaining area is equal to 16 - 32/3 which evaluates to 16 square units. How is this possible? Well, shift the parabola so that the turning point is at the origin. Then you will notice that it fits into the rectangle created by vertices (-2,0) (-2,4) (2,0) and (2,4) whose are is 16. Since the area under the parabola will now correspond to twice the area under the curve from -2 to 0, it makes sense to deduce that the difference will be the area under the shifted parabola.

So I think the answer is 16 square units.