The area (in square units) in the first quadr | vedclass

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(D) The required area is bounded by the curve $y=x^2+2$ and the line $y=x+1$ between $x=0$ and $x=3$.Since $x^2+2 \geq x+1$ for all $x \in [0, 3]$,the

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$\frac{15}{2}$

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(D) The required area is bounded by the curve $y=x^2+2$ and the line $y=x+1$ between $x=0$ and $x=3$.Since $x^2+2 \geq x+1$ for all $x \in [0, 3]$,the area is given by:$	ext{Area} = \int_0^3 [(x^2+2) - (x+1)] \, dx$$= \int_0^3 (x^2 - x + 1) \, dx$$= \left[ \frac{x^3}{3} - \frac{x^2}{2} + x ight]_0^3$$= \left( \frac{3^3}{3} - \frac{3^2}{2} + 3 ight) - (0)$$= \left( 9 - \frac{9}{2} + 3 ight)$$= 12 - 4.5 = 7.5 = \frac{15}{2} 	ext{ sq. units}$.

The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$,is

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