Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The parabolas $y^{2}=4x\; and\; x^{2}=4y$ divide the square region bounded by the lines $x=4,y=4$ and the coordinate axes. If $S_{1},S_{2},S_{3}$ are respectively the areas of these parts numbered from top to bottom then $S_{1}:S_{2}:S_{3}$ is

Option 1)
$1:2:3$

Option 2)
$1:2:1$

Option 3)
$1:1:1$

Option 4)
$2:1:2$

Answers (1)

best_answer

As we learnt in

Area along x axis -

Let $y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x)$ be two curve then area bounded between the curves and the lines

x = a and x = b is

$\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |$

- wherein

Where $\Delta y= f_{2}\left ( x \right )-f_{1}(x)$

$S_2=\int_{0}^{4}\left (\sqrt{4x}-\frac{x^2}{4} \right )dx$

$= \left [ \frac{2x^{3/2}}{3/2} \right ]_{0}^{4}- \left [ \frac{x^3}{12} \right ]_{0}^{4}$

$= \frac{4}{3}\times (8-0)- \frac{4^3}{12}$

$= \frac{32}{3}- \frac{64}{12}$

$= \frac{16}{3}$

Total area = S₁+ S₂+ S₃= 16

Also S₁= S₃

Thus

$S_1=S_2=S_3=\frac{16}{3}$

Option 1)

$1:2:3$

This is incorrect option

Option 2)

$1:2:1$

This is incorrect option

Option 3)

$1:1:1$

This is correct option

Option 4)

$2:1:2$

This is incorrect option

Posted by

divya.saini

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks

anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?

anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question