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I just need these solutions checked.

Thank you in advance!

**Consider the region bounded by the following curves ##y=x-3, y=5-x, \text{and}\ y=3##:**

1.) set up an integral expression that would give the area of the region of y as a function of x:

##y = x-3 = 5-x##

##x + x - 3 - 5 = 0##

##2x-8=0##

##x = 4##

##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##

2.) set up an integral expression that would give the area of the region of x as a function of y:

##\displaystyle \int_0^5 |\left(5-y\right)|##

3.) compute the area of the region by evaluating one of the expressions in (1.) or (2.):

##\displaystyle \int_0^3 |\left(2x-8\right)|\ dx##

##\left(x-8\right)+\ \text{C}##

##\displaystyle = \lim_{x \to 0^{+}} ((x-8)x)=0##

##\displaystyle = \lim_{x \to 3^{-}} ((x-8)x)=-15##

##\displaystyle |-15-0| = 15##

4.) set up an integral expression that would give the volume of the solid created by rotating the region about y = 1, using the disk/washer method:

##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##

5.) find the volume of the solid by evaluating the integral/integrals in (4.):

##\displaystyle \pi \int_0^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##

##=\ -\dfrac{29}{6}\pi##

##\approx 15\ \text{cubic units}##

6.) set up an integral expression that would give the volume of the solid created by rotating the region about y = -2, using the disk/washer method:

##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##

7.) compute the volume of the solid by evaluating the integral or integrals in (6.):

##\displaystyle \pi \int_{-2}^1 \left(5-x\right)^2-\left(x-3)^2\right)\ dx##

##=\ -\dfrac{27}{2}\pi##

##\approx 42\ \text{cubic units}##