The area of the region bounded by the curve y | vedclass

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Question

(B) The required area is given by the integral of the upper curve minus the lower curve between the given limits.$	ext{Required Area} = \int_{0}^{3}

Vedclass · Accepted Answer

$\frac{21}{2} 	ext{ sq units}$

Vedclass · Answer

(B) The required area is given by the integral of the upper curve minus the lower curve between the given limits.$	ext{Required Area} = \int_{0}^{3} (y_{	ext{upper}} - y_{	ext{lower}}) \, dx$$	ext{Required Area} = \int_{0}^{3} ((x^2 + 2) - x) \, dx$$	ext{Required Area} = \int_{0}^{3} (x^2 - x + 2) \, dx$Integrating term by term:$= \left[ \frac{x^3}{3} - \frac{x^2}{2} + 2x ight]_{0}^{3}$$= \left( \frac{3^3}{3} - \frac{3^2}{2} + 2(3) ight) - (0 - 0 + 0)$$= \left( \frac{27}{3} - \frac{9}{2} + 6 ight)$$= 9 - 4.5 + 6$$= 10.5 = \frac{21}{2} 	ext{ sq units}$

The area of the region bounded by the curve $y = x^2 + 2$ and the lines $y = x$,$x = 0$,and $x = 3$ is

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