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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

(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 $A:B = 14:14 = 1:1$.

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:

Similar Questions

The area bounded by the circle $x^{2}+y^{2}=16$ and the lines $x=0$ and $x=2$ is

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

The area of the region (in square units) bounded by the curve ${x^2} = 4y$,the line $x = 2$,and the $x$-axis is:

The area of the curve $xy^2 = a^2(a - x)$ bounded by the $y$-axis is

Find the area of the smaller region bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module