By applying definite integrals and integration rules, the area bounded between the curves f(x) = x² and g(x) = -(x+2)² + 4 is equal to 8 square units.
How to determine the between the two integrals
The points of intersection between the two curves are (x, y) = (0, 0) and (x, y) = (-2, 4). The curve is f(x) = x² and upper curve is g(x) = -(x+2)² + 4. The area between the two curves is defined by this definite integral:
[tex]A = \int\limits^{0}_{-2} {f(x) - g(x)} \, dx[/tex]
[tex]A = \int\limits^{0}_{-2} {f(x)} \, dx - \int\limits^{0}_{-2} {g(x)} \, dx[/tex]
[tex]A = \int\limits^0_{-2} {[-(x+2)^{2}+4]} \, dx -\int\limits^0_{-2} {x^{2}} \, dx[/tex]
[tex]A = -\frac{(x+2)^{3}}{3}|_{-2}^{0} + 4\cdot x|_{-2}^{0}-\frac{x^{3}}{3}|_{-2}^{0}[/tex]
[tex]A = - \frac{2^{3}}{3} + 0 + 4\cdot [0 - (-2)] - 0 + \frac{(-2)^{3}}{3}[/tex]
[tex]A = -\frac{8}{3} + 8 -\frac{8}{3}[/tex]
A = 8
By applying definite integrals and integration rules, the area bounded between the curves f(x) = x² and g(x) = -(x+2)² + 4 is equal to 8 square units.
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