Determine the area of the region bounded by the given set of curves y=x^2+2, y=sin(x), x=2.
Assuming\ it\ is\ also\ bounded\ by\ x=0,\
∫02x2+2−sin(x) dx\int_{0}^{2}{x^2+2-\sin{\left(x\right)}\ dx}∫02x2+2−sin(x) dx
x33+2x+cos(x) (0, 2)\frac{x^3}{3}+2x+\cos{\left(x\right)}\ \ \left(0,\ 2\right)3x3+2x+cos(x) (0, 2)
83+4+cos(2c)−1\frac{8}{3}+4+\cos{\left(2^c\right)}-138+4+cos(2c)−1
5.2505 square units5.2505\ square\ units5.2505 square units
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