
Robert R. answered 03/21/15
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The equation for a parabola that is concave up with a vertex above the x-axis has the equation:
y = a(x - h)2 + k where a, h and k are constants,
with a > 0 since the parabola is concave up and k > 0 since it's vertex is above the x-axis
if h > 0, then the parabola has its vertex in the 1st quadrant
if h < 0, then the parabola has its vertex in the 2nd quadrant
There are 2 possible things I can think of that the problem is asking for:
1. It's asking to find the area between the concave up parabola and a horizontal line y = c, where c is a constant > k (from above). In that case, you could use Arcimedes' formula. The "height" of the parabola would be c - k and the "base" would be the difference in x-values between the 2 points where the parabola intersects the horizontal line.
2. It's asking for the area underneath the concave up parabola above the x-axis. Since this is a not a concave down parabola with a < 0, you can't use Arcimedes' formula to find the area under the parabola.
There would need to be a range of x values defined, i.e. [x1, x2] where x1 < x2, in order to get the area under the concave up parabola, and in that case, the solution is the integral from x1 to x2 of a(x - h)2 + k with respect to x.
That works out to [(a/3)(x - h)3 + kx] dx from x2 down to x1
which equals (a/3)(x2 - h)3 +kx2 - [(a/3)(x1 - h)3 +kx1]
which simplifies to (a/3)[(x2 - h)3 - (x1 - h)3] + k(x2 - x1)
The problem as described doesn't mention a horizontal line y = c (possibility 1) or a range of x values (possibility 2) that I think are necessary for it to make sense. I would ask the teacher for clarification.