Yahoo Answers 06-19-14

a) Determine the area between the curves y = x2+4 and y = 12-x2 b) Draw a simple sketch to illustrate the area in question. Task 1... x^2+4 is a parabola opening up and 12-x^2 opens down find the intersection to find the interval to integrate over x^2+4=12-x^2 2x^2=8 x^2=4 x=±2 integrate from -2 to 2 ⌠x^2+4 dx=x^3/3+4x ⌠12-x^2 dx=12x-x^3/3 (12x-x^3/3 from -2 to 2)-(x^3/3+4x from -2 to 2) 12(2)-8/3-(12(-2)-(-8/3))=24-8/3+24-8/3=48-16/3 8/3+8+8/3+8=16+16/3 48-16/3-16-16/3=32-32/3=96/3-32/3=64/3