Home
Class 12
MATHS
Draw a rough sketch of the given curve y...

Draw a rough sketch of the given curve `y=1+abs(x+1),x=-3, x=3, y=0` and find the area of the region bounded by them, using integration.

Text Solution

Verified by Experts

The correct Answer is:
N/a

We have `y=1+|x+1|,`,x=-3,x =3 ,x=3 and y =0

`therefore y={:{(-x, if x lt-1),(x+2, ifx ge -1):}`
`therefore "Area of shaded refion" ,A =int _(-3)^(-1) -x dx + int _(-1)^(3)(x+2)dx`
`=-[(x^(2))/(2)]^(-1)+[(x^(2))/(2)+2x]_(-1)^(3)`
`-=[1/2-9/2]+[9/2+6-1/2+]`
`=- [-4]+[8+4]`
=12+4 = 16 sq units
Promotional Banner

Topper's Solved these Questions

  • APPLICATION OF INTEGRALS

    NCERT EXEMPLAR ENGLISH|Exercise Objective Type Questions|22 Videos
  • APPLICATION OF INTEGRALS

    NCERT EXEMPLAR ENGLISH|Exercise Objective Type Questions|22 Videos
  • APPLICATION OF DERIVATIVES

    NCERT EXEMPLAR ENGLISH|Exercise Fillers|5 Videos
  • CONTINUITY AND DIFFERENTIABILITY

    NCERT EXEMPLAR ENGLISH|Exercise True/False|10 Videos

Similar Questions

Explore conceptually related problems

Draw a rough sketch of the curves y = (x-1)^2 and y = |x-1| . Hence, find the area of the region bounded by these curves.

Draw a rough sketch of the curve y= (x-1)^2(x-2)(x-3)^3

Draw the rough sketch of the curve y=x^(4)-x^(2) .

Draw the rough sketch of the curve y=(x-1)^(2)(x-3)^(3) .

Draw a rough sketch of the curves y =sin x varies from 0 to pi/2 and find the area of the region enclosed by them and x-axis

The area of the region bounded by the curves y=|x-1|andy=3-|x| is

Find the area of the region bounded by y=-1,y=2, x=y^3 and x=0.

Find the area of the region bounded by the curve y^(2)=9x" and " y=3x .

Find the area of the region bounded by the curves x=|y^(2)-1| and y=x-5 .

Draw a rough sketch of the curves y^(2)=xandy^(2)=4-3x and find the area enclosed between them.