The area of the region (x,y):0 y 6-x,y^2 4x-3,x 0 is :

Question

The area of the region ${(x,y):0\le y\le 6-x,{y}^{2}\ge 4x-3,x\ge 0}$ is :

Accepted Answer

We need to find the area of the region

${(x,y):0\le y\le 6-x,\, {y}^{2}\ge 4x-3,\, x\ge 0}.$

Let us understand each condition carefully.

1. From $0\le y\le 6-x$
This means:

$y\ge 0$, so the region lies above the $x$-axis.

$y\le 6-x$, or

$x+y\le 6,$

so the region lies below the line

$y=6-x.$

2. From ${y}^{2}\ge 4x-3$
Rewrite it in terms of $x$:

$4x-3\le {y}^{2}$

$4x\le {y}^{2}+3$

$x\le \frac{{y}^{2}+3}{4}.$

So the region lies to the left of the parabola

$x=\frac{{y}^{2}+3}{4}.$

Also, we are given

$x\ge 0.$

So for a fixed $y$, $x$ lies between $0$ and whichever is smaller:

$x=\frac{{y}^{2}+3}{4}\, \, x=6-y.$

So the horizontal width is

$0\le x\le min(\frac{{y}^{2}+3}{4},\, 6-y).$

Now we find where these two curves intersect.

3. Point of intersection
Set

$\frac{{y}^{2}+3}{4}=6-y.$

Multiply by $4$:

${y}^{2}+3=24-4y$

${y}^{2}+4y-21=0$

$(y+7)(y-3)=0.$

So,

$y=3\, \, y=-7.$

Since $y\ge 0$, we take

$y=3.$

Now compare the two right boundaries:

For $0\le y\le 3$, we have

$\frac{{y}^{2}+3}{4}\le 6-y,$

so width is

$\frac{{y}^{2}+3}{4}.$

For $3\le y\le 6$, we have

$6-y\le \frac{{y}^{2}+3}{4},$

so width is

$6-y.$

Hence area is

$A={\int }_{3}^{0}\frac{{y}^{2}+3}{4}\, dy+{\int }_{6}^{3}(6-y)\, dy.$

4. Evaluate the integrals
First integral:

${\int }_{3}^{0}\frac{{y}^{2}+3}{4}\, dy=\frac{1}{4}{\int }_{3}^{0}({y}^{2}+3)\, dy$

$=\frac{1}{4}{[\frac{{y}^{3}}{3}+3y]}_{3}^{0}=\frac{1}{4}(\frac{27}{3}+9)=\frac{1}{4}(9+9)=\frac{18}{4}=\frac{9}{2}.$

Second integral:

${\int }_{6}^{3}(6-y)\, dy={[6y-\frac{{y}^{2}}{2}]}_{6}^{3}$

$=(36-18)-(18-\frac{9}{2})=18-\frac{27}{2}=\frac{9}{2}.$

So total area is

$A=\frac{9}{2}+\frac{9}{2}=9.$

Therefore, the correct answer is:

$9$

So, Option B is correct.

Solutions · 1