The base of a solid is the region enclosed by the graphs of y=ln(x) and y=0.1x3−0.5x2, 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) ∫25[ln(x)−0.1x3+0.5x2]2dx(B) ∫25[ln(x)−0.1x3+0.5x2](6−x)dx(C) ∫25[0.1x3−0.5x2−ln(x)](6−x)dx(D) ∫25[ln2(x)+(0.1x3+0.5x2)2]dx
Q. The base of a solid is the region enclosed by the graphs of y=ln(x) and y=0.1x3−0.5x2, 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) ∫25[ln(x)−0.1x3+0.5x2]2dx(B) ∫25[ln(x)−0.1x3+0.5x2](6−x)dx(C) ∫25[0.1x3−0.5x2−ln(x)](6−x)dx(D) ∫25[ln2(x)+(0.1x3+0.5x2)2]dx
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 x-axis from x=2 to x=5. The area of each cross section is given by the difference in the y-values of the two functions (the top function minus the bottom function) times the height of the rectangle (6−x).
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=2 and x=5, say x=3, and comparing their values.For y=ln(x), at x=3, y=ln(3).For y=0.1x3−0.5x2, at x=3, y=0.1(3)3−0.5(3)2.We calculate these values to see which is greater.
Calculate Values: Calculating the values:y=ln(3)≈1.0986y=0.1(3)3−0.5(3)2=0.1(27)−0.5(9)=2.7−4.5=−1.8Since ln(3) is greater than 0.1(3)3−0.5(3)2, the function y=ln(x) is the top function between x=2 and x=5.
Set Up Integral for Volume: Now we can set up the integral for the volume. The area of each cross section is (ln(x)−(0.1x3−0.5x2))×(6−x).
Correct Integral: The correct integral for the volume is therefore: ∫x=2x=5[ln(x)−(0.1x3−0.5x2)]⋅(6−x)dx. This matches choice (B) from the given options.
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