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(A) The area bounded by the curves $y=x^{2}+2, \,y=x, \,x=0,$ and $x=3$ is given by the integral of the difference between the upper curve and the low

Vedclass · Accepted Answer

$\frac{21}{2}$

Vedclass · Answer

(A) The area bounded by the curves $y=x^{2}+2, \,y=x, \,x=0,$ and $x=3$ is given by the integral of the difference between the upper curve and the lower curve within the given limits.Area $= \int_{0}^{3} ((x^{2}+2) - x) \, dx$$= \int_{0}^{3} (x^{2} - x + 2) \, dx$$= \left[ \frac{x^{3}}{3} - \frac{x^{2}}{2} + 2x ight]_{0}^{3}$$= \left( \frac{3^{3}}{3} - \frac{3^{2}}{2} + 2(3) ight) - (0)$$= \left( \frac{27}{3} - \frac{9}{2} + 6 ight)$$= 9 - 4.5 + 6$$= 10.5 = \frac{21}{2} 	ext{ sq. units}$

Find the area of the region bounded by the curves $y=x^{2}+2, \,y=x,\, x=0$ and $x=3$.

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