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The base of a solid is the region enclosed by the graphs of 
y=2+(x)/(2) and 
y=2^((x)/(2)), between the 
y-axis and 
x=4.
Cross sections of the solid perpendicular to the 
x-axis are rectangles whose height is 
4-x.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) 
int_(0)^(4)[2^((x)/(2))-(x)/(2)-2](4-x)dx
(B) 
int_(0)^(4)[2^((x)/(2))+(x)/(2)+2](4-x)dx
(C) 
int_(0)^(4)[2+(x)/(2)-2^((x)/(2))](4-x)dx
(D) 
int_(0)^(4)[2^((x)/(2))-(x)/(2)+2](4-x)dx

The base of a solid is the region enclosed by the graphs of y=2+x2 y=2+\frac{x}{2} and y=2x2 y=2^{\frac{x}{2}} , between the y y -axis and x=4 x=4 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 4x 4-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) 04[2x2x22](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-2\right](4-x) d x \newline(B) 04[2x2+x2+2](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}+\frac{x}{2}+2\right](4-x) d x \newline(C) 04[2+x22x2](4x)dx \int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{2}}\right](4-x) d x \newline(D) 04[2x2x2+2](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}+2\right](4-x) d x

Full solution

Q. The base of a solid is the region enclosed by the graphs of y=2+x2 y=2+\frac{x}{2} and y=2x2 y=2^{\frac{x}{2}} , between the y y -axis and x=4 x=4 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 4x 4-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) 04[2x2x22](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-2\right](4-x) d x \newline(B) 04[2x2+x2+2](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}+\frac{x}{2}+2\right](4-x) d x \newline(C) 04[2+x22x2](4x)dx \int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{2}}\right](4-x) d x \newline(D) 04[2x2x2+2](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}+2\right](4-x) d x
  1. Understand the Problem: Understand the problem.\newlineWe need to find the volume of a solid with a given base and cross-sectional area. The base is bounded by the curves y=2+(x/2)y = 2 + (x/2) and y=2(x/2)y = 2^{(x/2)}, and the limits of integration are from the y-axis (x=0x=0) to x=4x=4. The height of the cross sections is given by 4x4-x.
  2. Determine Cross Section Area: Determine the area of a cross section.\newlineThe area A(x)A(x) of a cross section at a given xx is the difference between the upper function and the lower function, multiplied by the height (4x)(4-x). The upper function is y=2(x/2)y = 2^{(x/2)}, and the lower function is y=2+(x/2)y = 2 + (x/2).\newlineA(x)=[2(x/2)(2+x/2)](4x)A(x) = [2^{(x/2)} - (2 + x/2)](4-x)
  3. Set Up Integral for Volume: Set up the integral for the volume.\newlineThe volume VV of the solid is the integral of the cross-sectional area A(x)A(x) from x=0x=0 to x=4x=4.\newlineV=04A(x)dxV = \int_{0}^{4} A(x) \, dx\newlineV=04[2(x/2)(2+x/2)](4x)dxV = \int_{0}^{4} [2^{(x/2)} - (2 + x/2)](4-x) \, dx
  4. Match Integral with Choices: Match the integral with the given choices.\newlineWe need to find which of the given choices matches the integral we have set up. The correct integral should have the form of the area function we found in Step 22.\newlineThe correct choice is:\newline(C) 04[2+(x/2)2(x/2)](4x)dx\int_{0}^{4} [2 + (x/2) - 2^{(x/2)}](4-x) \, dx

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