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

Question

Dr. Abubacker · Accepted Answer

(c)
Given area bounded by the curve, $y = \sqrt{3x + 4},x$-axis and the line $x = - 1$ and $x = 4$ is $A$ and area bounded by the curve $y = \sqrt{3x + 4}$
$ie,\ y = \pm (3x + 4)^{1/2}\ \ x$-axis and the line $x = - 1$ and $x = 4\ is\ B$
$	herefore B = 2A$
[Since, it is the area of both sides about x-axis]
$Now,\ \ \ A:B = A:2\ A = 1:2$

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

Explanation

Reviewed by Dr. Abubacker

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