Question

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 the area bounded by curve $y^2=3x+4$, x-axis and the lines $x=-1$ and $x=4$, then $A:B=$

• A. $1:1$
• B. $2:1$
• C. $1:2$
• D. None of these

Answer 1

The correct option is A

$1:1$

Explanation for the correct option:

Step 1. Find the value of $A:B$:

Given, $y=\sqrt{3x+4}$ …..(1)

$y^2=3x+4$ …..(2)

Step 2. Integrate equation (1) with respect to $x$:

$A={\int }_{-1}^4\sqrt{3}x+4xdx$ …..(3)

Let $3x+4=t^2$

$\Rightarrow$ $3dx=2tdt$

$\Rightarrow$ $dx=\frac{2t}{3}dt$

Step 3. Put the value of $dx$ in equation (3):

$\begin{array}{rcl}A& =& \frac{2}{3}{\int }_{-1}^4{t}^{2}dt\\ & =& \frac{2}{3}{\left[\frac{t^3}{3}\right]}_{-1}^4\\ & =& \frac{2}{9}\left(64-1\right)\\ & =& \frac{2}{9}\times 63\\ & =& 14\\ & =& B\end{array}$

$\begin{array}{rcl}\therefore A:B& =& 14:14\\ & =& \mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{1}\end{array}$

Hence, Option ‘A’ is Correct.