Find the area of the region under the curve y=7-4x^2 and above the x-axis

Answer 1

The graph should look like this:

y = 7-4x²

0 = 7-4x²

4x²= 7

x²= 7/4

x= ±√7 / 2 ≈ ± 1.323

The area of the region (A) is the definite integral ∫_a^b f(x) dx. In this case, a = -√7 / 2, b= √7 / 2,

f(x) = 7- 4x²

A = ∫^b_a f(x) dx = ∫^√7/2_-√7/2 (7- 4x²) dx

=[7x - (4/3) x³]^√7/2_-√7/2

= 7(√7/2) - (4/3)(√7/2)³ - [7(-√7/2) - (4/3) (-√7/2)³]

= 7√7 / 2 - (4/3)(7√7 /8) - [-7√7 /2 - (4/3) (-7√7 /8]

= 7√7 / 2 - 7√7 /6 - [-7√7 /2 + 7√7 /6]

= 21√7 /6 - 7√7 /6 - [-21√7 /6 + 7√7 /6]

= 7√7 /3 - [-7√7 /3]

= 14√7 /3 ≈ 12.35 sq. units

Answer 2

Zero y = 7 – 4x² by writing 7 = 4x² or x equals ±√7/2. A graph will show y above the x-axis for

-√7/2 < x < √7/2.

Then evaluate ∫_{(from -√7/2 to √7/2)}(7 – 4x²)dx equal to 7x – (4/3)x³_{(from -√7/2 to √7/2)}.

Then write [7(√7/2) – (4/3)(√7/2)³] – [7(-√7/2) – (4/3)(-√7/2)³] which reduces to 14√7/2 – (8/3)(√7/2)³ or

7√7 – (8/3)(7√7/8) or 7√7 – 7√7/3 or 14√7/3 equal to 12.34683945 square units.