Please help 50 points mark brainliest

y=∣x−1∣={

1−x

x−1

x<1

x≥1

}

Similarly y=3−∣x∣={

3+x

3−x

x<0

x≥0

}

The required region is a rectangle with length 2

2

and breadth

2

.

Hence, the area is 4. Hence, all the options are correct

  a) area = 2 -ln(3)

  b) volume = 2π(4/3 -ln(3))

Step-by-step explanation:

The area under the curve is the definite integral of the function between the given limits. The integral is made easier if we rewrite the function first:

  y = ((x +1) -1)/(x +1) = 1 -1/(x +1)

__

The volume of revolution is the integral of the differential volumes. Each such differential volume can be considered a disc of radius y and a thickness of dx.

_____

The numerical values of area and volume are shown in the attachment.

there is no question content here, but you for the free points! : d

my answer got deleted but it's a

step-by-step explanation: