The area (in sq. units) of the region A = {(x | vedclass

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(B) The region is defined by $0 \le x \le 3$,$0 \le y \le 4$,and $y \le x^2 + 3x$. First,find the intersection of $y = x^2 + 3x$ and $y = 4$: $x^2 + 3

Vedclass · Accepted Answer

$\frac{59}{6}$

Vedclass · Answer

(B) The region is defined by $0 \le x \le 3$,$0 \le y \le 4$,and $y \le x^2 + 3x$. First,find the intersection of $y = x^2 + 3x$ and $y = 4$: $x^2 + 3x = 4 \implies x^2 + 3x - 4 = 0 \implies (x+4)(x-1) = 0$. Since $x \ge 0$,the intersection point is $x = 1$. For $0 \le x \le 1$,the region is bounded by $y = x^2 + 3x$ and the $x$-axis. Area $A_1 = \int_0^1 (x^2 + 3x) dx = [\frac{x^3}{3} + \frac{3x^2}{2}]_0^1 = \frac{1}{3} + \frac{3}{2} = \frac{2+9}{6} = \frac{11}{6}$. For $1 \le x \le 3$,the region is bounded by $y = 4$ and the $x$-axis. Area $A_2 = \int_1^3 4 dx = [4x]_1^3 = 4(3-1) = 8$. Total Area = $A_1 + A_2 = \frac{11}{6} + 8 = \frac{11+48}{6} = \frac{59}{6}$.

The area (in sq. units) of the region $A = \{(x,y) \in R \times R | 0 \le x \le 3, 0 \le y \le 4, y \le x^2 + 3x\}$ is

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