HELP! Area Using Integrals

Question

Determine the area that is bounded by the following curve and the x-axis on the interval below. (Round your answer to three decimal places.) y = x^2 − 16, −6 ≤ x ≤ 2

 

 

 

Answer 1

Hi Symmeatra,

 

If you sketch out the curve you're given, you'll see that it's a upward parabola that's shifted down by 16, with x-intercepts at -4 and +4. Because the sign of the graph changes at -4, and this is in your interval, you need to split it up, find the areas and add them together after.

 

∫_-6^-4 x²-16 dx

= 1/3 x³ - 16x|_-6^-4

= [1/3(-4)³ - 16(-4)] - [1/3(-6)³ - 16(-6)]

= 18.667

 

∫_-4² x²-16 dx

= 1/3 x³ - 16x|_-4²

= [1/3(2)³ - 16(2)] - [1/3(-4)³ - 16(-4)]

= -72

 

The total area is 18.667 + 72 = 90.667