Find area of the region under the curve y=9−2x2 and above the x-axis.
area=?
To find the area under the curve y = 9 - 2x^2 and above the x-axis, we can use integration.
First, let's sketch the graph of the curve to visualize the region we are interested in.
The curve y = 9 - 2x^2 is a downward-opening parabola with its vertex at the point (0, 9). It intersects the x-axis at x = -3 and x = 3.
To find the area under the curve and above the x-axis, we need to set up an integral.
The region under the curve y = 9 - 2x^2 and above the x-axis can be represented by the integral:
∫[a, b] (9 - 2x^2) dx
where 'a' and 'b' represent the x-values at which the curve intersects the x-axis.
In this case, considering that the curve intersects the x-axis at x = -3 and x = 3, the integral becomes:
∫[-3, 3] (9 - 2x^2) dx
Now, let's integrate:
∫[-3, 3] (9 - 2x^2) dx = [9x - (2/3)x^3] |[-3, 3]
Evaluating this definite integral, we subtract the value of the integral at the lower bound (x = -3) from the value at the upper bound (x = 3):
[9(3) - (2/3)(3)^3] - [9(-3) - (2/3)(-3)^3]
Simplifying further:
[27 - 18] - [-27 + 18] = 27
Therefore, the area under the curve y = 9 - 2x^2 and above the x-axis is 27 square units.