The area bounded by y = -x^2 + 2x + 3 and y = | vedclass

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(B) Given the curve $y = -x^2 + 2x + 3$ and the line $y = 0$.To find the points of intersection,set $y = 0$:$-x^2 + 2x + 3 = 0$$x^2 - 2x - 3 = 0$$(x -

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$\frac{32}{3}$

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(B) Given the curve $y = -x^2 + 2x + 3$ and the line $y = 0$.To find the points of intersection,set $y = 0$:$-x^2 + 2x + 3 = 0$$x^2 - 2x - 3 = 0$$(x - 3)(x + 1) = 0$So,$x = -1$ and $x = 3$.The required area $A$ is given by the integral:$A = \int_{-1}^{3} (-x^2 + 2x + 3) dx$$A = \left[ -\frac{x^3}{3} + x^2 + 3x \right]_{-1}^{3}$$A = \left( -\frac{27}{3} + 9 + 9 \right) - \left( -\frac{(-1)^3}{3} + (-1)^2 + 3(-1) \right)$$A = (-9 + 9 + 9) - \left( \frac{1}{3} + 1 - 3 \right)$$A = 9 - \left( \frac{1 - 6}{3} \right) = 9 - \left( -\frac{5}{3} \right) = 9 + \frac{5}{3} = \frac{27 + 5}{3} = \frac{32}{3}$ square units.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

The area bounded by $y = -x^2 + 2x + 3$ and $y = 0$ is

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