Area between two curves:
y=9-x^2
y=x^2 - 7
Show that the area is (128√2)/3
So...
y=9-x^2
y'=-2x
I(x) = 9x - (1/3)x^3
y=x^2 - 7
y'=2x
I(x) = (1/3)x^3 - 7x
9 - x^2 = x^2 - 7
x = √8
Correct so far? What next? How can I make it to find the correct area? Thanks!
So then the -√8 and √8...
I plug them into each integral
I(x) = 9x - (1/3)x^3
I(√8) = 9(√8) - (1/3)(√8)^3 = 17.9
I(-√8) = 9(-√8) - (1/3)(-√8)^3 = -17.9
I(x) = (1/3)x^3 - 7x
I(√8) = (1/3)(√8)^3 - 7(√8) = -12.2
I(-√8) = (1/3)(-√8)^3 - 7(-√8) = 12.2
How do I arrive at the area of 128√2 / 3 ?
Area = b-a
Area = which one is b and which is a?