recategorized by
529 views
0 votes
0 votes

The area of the region bounded by the curves $y=\sqrt x,$ $2y+3=x$ and $x$-axis in the first quadrant is

  1. $9$
  2. $\frac{27}{4}$
  3. $36$
  4. $18$
recategorized by

1 Answer

2 votes
2 votes

Here, The curve $y=\sqrt{x}$ and the line $2y+3=x$ will intersect each other. So

$\begin{align} y&=\sqrt{2y+3}\\\Rightarrow y^2-2y-3&=0\\\Rightarrow (y-3)(y+1)&=0\\\therefore y &= -1,~3\end{align}$

From the curve $y=\sqrt{x}$, we get $y$ is always non-negative i.e. $y\ge 0$. Therefore $y\ne -1$ so only $y=3$ is solution of the equations. Now putting $y=3$ to the equation $2y+3=x$, we obtain $x=9$.

Therefore, the curve and the line intersect at the point $(9,3)$. Here is shown the graph below.

 

 

Now, the line $2y+3=x\Rightarrow y = \frac{x-3}{2}$ cuts $x$-axis at the point $(3,0)$. $\therefore$ The required region is bounded the by points $(0,0), (3,0)\text{ and }(9,3)$.

 

$\begin{align} \therefore \mathrm{Area} &= \int_{0}^{9}\sqrt{x}~\mathrm{d}x-\int_{3}^{9}\frac{x-3}{2}~\mathrm{d}x\\&= \left[ \frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{9}-\left[ \frac{(x-3)^2}{4}\right]_{3}^{9}\\&=\left[ \frac{2}{3}(9)^{\frac{3}{2}} -0\right]-\left[ \frac{6^2}{4}-0 \right]\\&=18-9\\&=9  \end{align}$

 

So the correct answer is A.

Related questions

0 votes
0 votes
0 answers
1
gatecse asked Sep 18, 2019
363 views
The shaded region in the following diagram represents the relation$y\:\leq\: x$$\mid \:y\mid \:\leq\: \mid x\:\mid $$y\:\leq\: \mid x\:\mid$$\mid \:y\mid\: \leq\: x$
0 votes
0 votes
1 answer
2
gatecse asked Sep 18, 2019
275 views
The set $\{(x,y)\: :\: \mid x\mid+\mid y\mid\:\leq\:1\}$ is represented by the shaded region in
0 votes
0 votes
1 answer
3
gatecse asked Sep 18, 2019
325 views
The area bounded by $y=x^{2}-4,y=0$ and $x=4$ is$\frac{64}{3}$$6$$\frac{16}{3}$$\frac{32}{3}$
0 votes
0 votes
1 answer
4
Arjun asked Sep 23, 2019
276 views
The area under the curve $x^2+3x-4$ in the positive quadrant and bounded by the line $x=5$ is equal to$59 \frac{1}{6}$$61 \frac{1}{3}$$40 \frac{2}{3}$$72$