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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)[ln(x)-0.1x^(3)+0.5x^(2)]^(2)dx
(B) 
int_(2)^(5)[ln(x)-0.1x^(3)+0.5x^(2)](6-x)dx
(C) 
int_(2)^(5)[0.1x^(3)-0.5x^(2)-ln(x)](6-x)dx
(D) 
int_(2)^(5)[ln^(2)(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[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 \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[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(D) 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

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[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 \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[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(D) 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
  1. Set Up Integral: To find the volume of the solid, we need to set up an integral that calculates the area of each cross section and then integrates these areas along the xx-axis from x=2x=2 to x=5x=5. The area of each cross section is given by the difference in the yy-values of the two functions (the top function minus the bottom function) times the height of the rectangle (6x6-x).
  2. Determine Top Function: First, we need to determine which function is the top function and which is the bottom function. We can do this by evaluating both functions at a point between x=2x=2 and x=5x=5, say x=3x=3, and comparing their values.\newlineFor y=ln(x)y=\ln(x), at x=3x=3, y=ln(3)y=\ln(3).\newlineFor y=0.1x30.5x2y=0.1x^{3}-0.5x^{2}, at x=3x=3, y=0.1(3)30.5(3)2y=0.1(3)^3-0.5(3)^2.\newlineWe calculate these values to see which is greater.
  3. Calculate Values: Calculating the values:\newliney=ln(3)1.0986y=\ln(3) \approx 1.0986\newliney=0.1(3)30.5(3)2=0.1(27)0.5(9)=2.74.5=1.8y=0.1(3)^3-0.5(3)^2 = 0.1(27)-0.5(9) = 2.7 - 4.5 = -1.8\newlineSince ln(3)\ln(3) is greater than 0.1(3)30.5(3)20.1(3)^3-0.5(3)^2, the function y=ln(x)y=\ln(x) is the top function between x=2x=2 and x=5x=5.
  4. Set Up Integral for Volume: Now we can set up the integral for the volume. The area of each cross section is (ln(x)(0.1x30.5x2))×(6x)(\ln(x) - (0.1x^{3} - 0.5x^{2})) \times (6-x).
  5. Correct Integral: The correct integral for the volume is therefore: x=2x=5[ln(x)(0.1x30.5x2)](6x)dx\int_{x=2}^{x=5} [\ln(x) - (0.1x^{3} - 0.5x^{2})] \cdot (6-x) \, dx. This matches choice (B) from the given options.

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