Use a definite integral to find area of the region under the curve y=7-4x^2 and above the x-axis.

Thanks in advance!

since the curve has x-intercepts at ±√7/2, we have

∫[-√7/2,√7/2] 7-4x^2 dx
= 7x - 4/3 x^3 [-√7/2,√7/2]
= 14√7 / 3

To find the area of the region under the curve y = 7 - 4x^2 and above the x-axis, we can use a definite integral.

Step 1: Determine the boundaries of the region. The curve intersects the x-axis when y = 0. So we need to find the x-values that make y = 7 - 4x^2 equal to 0.

0 = 7 - 4x^2
4x^2 = 7
x^2 = 7/4
x = ±√(7/4) = ±√(7)/2

The curve intersects the x-axis at x = √(7)/2 and x = -√(7)/2. To find the region in question, we consider the portion of the curve where x ranges from -√(7)/2 to √(7)/2.

Step 2: Set up the definite integral to calculate the area. Since the curve is above the x-axis, the function y = 7 - 4x^2 is positive for the given range of x.

Therefore, the area can be calculated as:

Area = ∫(from -√(7)/2 to √(7)/2) (7 - 4x^2) dx

Step 3: Evaluate the integral. To find the antiderivative of (7 - 4x^2), we can use the power rule of integration.

∫(7 - 4x^2) dx
= 7x - (4/3)x^3 + C

Where C is the constant of integration.

Now we can evaluate the definite integral:

Area = [7x - (4/3)x^3] (from -√(7)/2 to √(7)/2)
= [7(√(7)/2) - (4/3)(√(7)/2)^3] - [7(-√(7)/2) - (4/3)(-√(7)/2)^3]
= [(7√(7)/2) - (4/3)(7/8)] - [(-7√(7)/2) - (4/3)(-7/8)]
= (7√(7)/2 - 7/6) - (-7√(7)/2 + 7/6)
= 7√(7)/2 - 7/6 + 7√(7)/2 - 7/6
= 14√(7)/2 - 14/6
= 7√(7)/2 - 7/3.

Thus, the area of the region under the curve y = 7 - 4x^2 and above the x-axis is 7√(7)/2 - 7/3 square units.