The area of the region { ( x , y ) : x 2 y | x 2 - 4

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

Q.
The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

1 $\frac{4}{3} (4 \sqrt{2} + 1)$

2 $\frac{3}{4} (4 \sqrt{2} - 1)$

3 $\frac{3}{4} (4 \sqrt{2} + 1)$

4 $\frac{4}{3} (4 \sqrt{2} - 1)$

Ans.
(4)

$Required area = 2 (\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y)$

$= 2 ({[\frac{y^{3 / 2}}{3 / 2}]}_{1}^{2} - {[\frac{{(4 - y)}^{3 / 2}}{3 / 2}]}_{2}^{4})$

$= \frac{4}{3} [(2 \sqrt{2} - 1) - (0 - 2 \sqrt{2})] = \frac{4}{3} (4 \sqrt{2} - 1)$

Solution

Q.
The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

1 $\frac{4}{3} (4 \sqrt{2} + 1)$

2 $\frac{3}{4} (4 \sqrt{2} - 1)$

3 $\frac{3}{4} (4 \sqrt{2} + 1)$

4 $\frac{4}{3} (4 \sqrt{2} - 1)$

Ans.
(4)

$Required area = 2 (\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y)$

$= 2 ({[\frac{y^{3 / 2}}{3 / 2}]}_{1}^{2} - {[\frac{{(4 - y)}^{3 / 2}}{3 / 2}]}_{2}^{4})$

$= \frac{4}{3} [(2 \sqrt{2} - 1) - (0 - 2 \sqrt{2})] = \frac{4}{3} (4 \sqrt{2} - 1)$

Want Us to Email you About Special Offers & Updates?

Solution

Q. The area of the region {(x,y):x2≤y≤|x2-4|,y≥1} is [2023]

1 43(42+1)

2 34(42-1)

3 34(42+1)

4 43(42-1)

Ans. (4) Required area=2(∫12ydy+∫244-ydy) =2([y3/23/2]12-[(4-y)3/23/2]24) =43[(22-1)-(0-22)]=43(42-1)

Want Us to Email you About Special Offers & Updates?

Q.
The area of the region ${(x, y) : x^{2} \leq y \leq | x^{2} - 4 |, y \geq 1}$ is [2023]

1 $\frac{4}{3} (4 \sqrt{2} + 1)$

2 $\frac{3}{4} (4 \sqrt{2} - 1)$

3 $\frac{3}{4} (4 \sqrt{2} + 1)$

4 $\frac{4}{3} (4 \sqrt{2} - 1)$

Ans.
(4)

$Required area = 2 (\int_{1}^{2} \sqrt{y} d y + \int_{2}^{4} \sqrt{4 - y} d y)$

$= 2 ({[\frac{y^{3 / 2}}{3 / 2}]}_{1}^{2} - {[\frac{{(4 - y)}^{3 / 2}}{3 / 2}]}_{2}^{4})$

$= \frac{4}{3} [(2 \sqrt{2} - 1) - (0 - 2 \sqrt{2})] = \frac{4}{3} (4 \sqrt{2} - 1)$