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
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!
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