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The base of a solid is the region enclosed by the graphs of 
y=ln(x) and 
y=0.1x^(3)-0.5x^(2), between 
x=2 and 
x=5.
Cross sections of the solid perpendicular to the 
x-axis are rectangles whose height is 
6-x.
Which one of the definite integrals gives the volume of the solid?
Choose 1 answer:
(A) 
int_(2)^(5)[0.1x^(3)-0.5x^(2)-ln(x)](6-x)dx
(B) 
int_(2)^(5)[ln(x)-0.1x^(3)+0.5x^(2)](6-x)dx
(C) 
int_(2)^(5)[ln^(2)(x)+(0.1x^(3)+0.5x^(2))^(2)]dx
(D) 
int_(2)^(5)[ln(x)-0.1x^(3)+0.5x^(2)]^(2)dx

The base of a solid is the region enclosed by the graphs of y=ln(x) y=\ln (x) and y=0.1x30.5x2 y=0.1 x^{3}-0.5 x^{2} , between x=2 x=2 and x=5 x=5 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 6x 6-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) 25[0.1x30.5x2ln(x)](6x)dx \int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x \newline(B) 25[ln(x)0.1x3+0.5x2](6x)dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x \newline(C) 25[ln2(x)+(0.1x3+0.5x2)2]dx \int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x \newline(D) 25[ln(x)0.1x3+0.5x2]2dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x

Full solution

Q. The base of a solid is the region enclosed by the graphs of y=ln(x) y=\ln (x) and y=0.1x30.5x2 y=0.1 x^{3}-0.5 x^{2} , between x=2 x=2 and x=5 x=5 .\newlineCross sections of the solid perpendicular to the x x -axis are rectangles whose height is 6x 6-x .\newlineWhich one of the definite integrals gives the volume of the solid?\newlineChoose 11 answer:\newline(A) 25[0.1x30.5x2ln(x)](6x)dx \int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x \newline(B) 25[ln(x)0.1x3+0.5x2](6x)dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x \newline(C) 25[ln2(x)+(0.1x3+0.5x2)2]dx \int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x \newline(D) 25[ln(x)0.1x3+0.5x2]2dx \int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x
  1. Understand the Problem: Understand the problem.\newlineWe need to find the volume of a solid with a known base and a specific cross-sectional area. The base is bounded by the curves y=ln(x)y=\ln(x) and y=0.1x30.5x2y=0.1x^{3}-0.5x^{2} between x=2x=2 and x=5x=5. The cross sections are rectangles perpendicular to the xx-axis with a height of 6x6-x.
  2. Determine Cross Section Area: Determine the area of a typical cross section.\newlineThe area AA of a rectangle is given by A=length×widthA = \text{length} \times \text{width}. In this case, the length of the rectangle is the difference between the two functions 0.1x30.5x2ln(x)0.1x^{3}-0.5x^{2} - \ln(x) and the width is the height of the rectangle 6x6-x.
  3. Set Up Integral: Set up the integral to calculate the volume.\newlineThe volume VV of the solid is the integral of the cross-sectional area AA from x=2x=2 to x=5x=5. Therefore, V=25AdxV = \int_{2}^{5} A \, dx, where A=(0.1x30.5x2ln(x))(6x)A = (0.1x^{3}-0.5x^{2} - \ln(x))(6-x).
  4. Identify Correct Integral: Identify the correct integral from the given options.\newlineWe need to find the integral that matches the expression for AA we found in Step 33. The correct integral is the one that has the difference of the functions multiplied by (6x)(6-x) and is integrated from x=2x=2 to x=5x=5.
  5. Match Integral: Match the integral with the given options.\newlineThe correct integral is (A)25(0.1x30.5x2ln(x))(6x)dx(A) \int_{2}^{5}(0.1x^{3}-0.5x^{2}-\ln(x))(6-x)dx because it represents the area of the cross section (0.1x30.5x2ln(x))(0.1x^{3}-0.5x^{2} - \ln(x)) multiplied by the height (6x)(6-x) and integrated from x=2x=2 to x=5x=5.

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