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Area Under the Curves question

2018 · 15 Apr · Shift 1 · Q42

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Mathematics
Area Under the Curves
2018 · 15 Apr · Shift 1 · Q42

Area Under the Curves question

2018 · 15 Apr · Shift 1 · Q42

JEE MainMathematicsArea Under the CurvesMCQ+4 / −1

The area (in sq. units) of the region {x

\in

R : x

\ge

0, y

\ge

0, y

\ge

x

-

2 and y

\le \sqrt x

}, is :

A
${{13} \over 3}$
B
${{8} \over 3}$
C
${{10} \over 3}$
D
${{5} \over 3}$

View written solutionFree

Correct answer: C

Interpret the region

We need the area of the set satisfying: $x\ge 0,\quad y\ge 0,\quad y\ge x-2,\quad y\le \sqrt{x}.$

So the region lies:

in the first quadrant,
above both $y=0$ and $y=x-2$ ,
below $y=\sqrt{x}$ .

Thus for each $x\ge 0$ , the lower boundary is: $y=\max(0,x-2),$ and the upper boundary is: $y=\sqrt{x}.$

The region exists where $\max(0,x-2)\le \sqrt{x}.$

Find where the curves intersect

We solve: $\sqrt{x}=x-2.$

Since $\sqrt{x}\ge 0$ , we must have $x-2\ge 0\Rightarrow x\ge 2$ .

Now square both sides: $x=(x-2)^2=x^2-4x+4$ $x^2-5x+4=0$ $(x-1)(x-4)=0.$

So $x=1$ or $x=4$ . But because $x\ge 2$ , only $x=4$ is valid.

Hence $y=\sqrt{x}$ and $y=x-2$ meet at $(4,2)$ .

Determine the interval of integration

For $0\le x\le 2$ , we have $x-2\le 0$ , so the condition $y\ge x-2$ is weaker than $y\ge 0$ . Thus lower curve is $y=0$ .
For $2\le x\le 4$ , we have $x-2\ge 0$ , so lower curve is $y=x-2$ .
For $x>4$ , note that $x-2>\sqrt{x}$ , so no region exists.

Therefore area is: $A=\int_0^2 \sqrt{x}\,dx+\int_2^4\big(\sqrt{x}-(x-2)\big)dx.$

Evaluate the integrals

First,

=\frac{2}{3}(2\sqrt{2}).$$ So, $$\int_0^2 \sqrt{x}\,dx=\frac{4\sqrt{2}}{3}.$$ Now second integral: $$\int_2^4\big(\sqrt{x}-(x-2)\big)dx =\int_2^4 \sqrt{x}\,dx-\int_2^4(x-2)dx.$$ Compute each part: $$\int_2^4 \sqrt{x}\,dx=\left[\frac{2}{3}x^{3/2}\right]_2^4 =\frac{2}{3}(8-2\sqrt{2}) =\frac{16-4\sqrt{2}}{3}.$$ And $$\int_2^4(x-2)dx=\left[\frac{x^2}{2}-2x\right]_2^4 =(8-8)-(2-4)=2.$$ Hence, $$\int_2^4\big(\sqrt{x}-(x-2)\big)dx =\frac{16-4\sqrt{2}}{3}-2 =\frac{10-4\sqrt{2}}{3}.$$ Therefore, $$A=\frac{4\sqrt{2}}{3}+\frac{10-4\sqrt{2}}{3}=\frac{10}{3}.$$ --- 5. **Match with options** $$\boxed{\frac{10}{3}}$$ So the correct option is: **C**.

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