Question

Please help 50 points mark brainliest

Answer 1

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

Answer 2

Answer:

  a) area = 2 -ln(3)

  b) volume = 2π(4/3 -ln(3))

Step-by-step explanation:

a)

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)

  $\displaystyle A=\int_0^2{y}\,dx=\int_0^2{1}\,dx-\int_0^2{\dfrac{1}{x+1}}\,dx\\\\=\left.(x-\ln{(x+1)})\right|_{x=0}^{x=2}=\boxed{2-\ln(3)}$

__

b)

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.

  $\displaystyle V=\int_0^2{\pi y^2}\,dx=\pi\int_0^2{\left(1-\dfrac{2}{x+1}+\dfrac{1}{(x+1)^2}\right)}\,dx\\\\=\left.\pi\left(x-2\ln(x+1)-\dfrac{1}{x+1}\right)\right|_{x=0}^{x=2}=\pi\left(2-2\ln{(3)}+\dfrac{2}{3}\right)\\\\=\boxed{2\pi\left(\dfrac{4}{3}-\ln(3)\right)}$

_____

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