If A is the area of the region bounded by the | vedclass

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(A) For area $A$,the curve is $y = \sqrt{3x + 4}$. The area is given by $\int_{-1}^{4} \sqrt{3x + 4} \, dx$. Evaluating this: $\left[ \frac{2}{3} \cdo

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$1:1$

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(A) For area $A$,the curve is $y = \sqrt{3x + 4}$. The area is given by $\int_{-1}^{4} \sqrt{3x + 4} \, dx$. Evaluating this: $\left[ \frac{2}{3} \cdot \frac{1}{3} (3x + 4)^{3/2} ight]_{-1}^{4} = \frac{2}{9} [ (3(4) + 4)^{3/2} - (3(-1) + 4)^{3/2} ] = \frac{2}{9} [ 16^{3/2} - 1^{3/2} ] = \frac{2}{9} [ 64 - 1 ] = \frac{2}{9} 	imes 63 = 14$. For area $B$,the curve is $y^2 = 3x + 4$,which implies $y = \pm \sqrt{3x + 4}$. The area bounded by the curve and the $x$-axis is the integral of $|y|$. Since $|y| = \sqrt{3x + 4}$ for the region bounded by the curve and the $x$-axis,the area $B$ is $\int_{-1}^{4} |\sqrt{3x + 4}| \, dx = \int_{-1}^{4} \sqrt{3x + 4} \, dx = 14$. Thus,$A = 14$ and $B = 14$. The ratio $A:B = 14:14 = 1:1$.

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

If $A$ is the area of the region bounded by the curve $y = \sqrt{3x + 4}$,the $x$-axis,and the lines $x = -1$ and $x = 4$,and $B$ is the area bounded by the curve $y^2 = 3x + 4$,the $x$-axis,and the lines $x = -1$ and $x = 4$,then $A:B$ is equal to:

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