Question
Hi, I need help with the problem: "Use a definite integral to find the area of the region under the curve y=6-4x^2 and above the x-axis". I keep getting stuck at 6x-(4x^3/3). Thanks!!
Answers
Steve
you need to do a definite integral. So, what are the limits? The curve intersects the x-axis at x = ±?(3/2)
so now all you have to do is evaluate 6x - 4/3 x^3 at those points. Due to symmetry, you can just take twice the value from 0 to ?(3/2)
f(?(3/2)) = 6?(3/2) - 4/3 (3/2)^(3/2) = 2?6
So, the total area is 4?6
http://www.wolframalpha.com/input/?i=2%E2%88%AB%5B0,%E2%88%9A(3%2F2)%5D+(6+-+4x%5E2)+dx
so now all you have to do is evaluate 6x - 4/3 x^3 at those points. Due to symmetry, you can just take twice the value from 0 to ?(3/2)
f(?(3/2)) = 6?(3/2) - 4/3 (3/2)^(3/2) = 2?6
So, the total area is 4?6
http://www.wolframalpha.com/input/?i=2%E2%88%AB%5B0,%E2%88%9A(3%2F2)%5D+(6+-+4x%5E2)+dx