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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)-2^((x)/(2))](4-x)dx
(B) 
int_(0)^(4)[2^((x)/(2))+(x)/(2)+2](4-x)dx
(C) 
int_(0)^(4)[2^((x)/(2))-(x)/(2)-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[2+x22x2](4x)dx \int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{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[2x2x22](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-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[2+x22x2](4x)dx \int_{0}^{4}\left[2+\frac{x}{2}-2^{\frac{x}{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[2x2x22](4x)dx \int_{0}^{4}\left[2^{\frac{x}{2}}-\frac{x}{2}-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. Identify Volume Calculation: To find the volume of the solid, we need to integrate the area of the cross sections along the xx-axis from x=0x=0 to x=4x=4. The area of each cross section is given by the difference in the yy-values of the two functions times the height of the rectangle (4x)(4-x).
  2. Find Difference in Y-values: First, we need to find the difference in the yy-values of the two functions y=2+(x/2)y=2+(x/2) and y=2(x/2)y=2^{(x/2)}. This difference will be the length of the base of the rectangle for the cross section at each point xx.
    Difference in yy-values: (2+(x/2))2(x/2)(2 + (x/2)) - 2^{(x/2)}
  3. Calculate Area of Cross Section: Now, we multiply the difference in yy-values by the height of the rectangle (4x)(4-x) to get the area of the cross section at each point xx.\newlineArea of cross section: [2+(x2)2(x2)]×(4x)[2 + (\frac{x}{2}) - 2^{(\frac{x}{2})}] \times (4-x)
  4. Integrate Area for Volume: To find the volume of the solid, we integrate the area of the cross sections from x=0x=0 to x=4x=4.
    Volume of solid: 04[(2+(x/2))2(x/2)](4x)dx\int_{0}^{4}[(2 + (x/2)) - 2^{(x/2)}] \cdot (4-x) \, dx
  5. Match Answer Choices: Looking at the answer choices, we see that option (A) matches our expression for the volume of the solid.\newlineFinal answer: (A)04[2+(x/2)2(x/2)](4x)dx(A) \int_{0}^{4}[2 + (x/2) - 2^{(x/2)}] * (4-x) \, dx

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