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[0.1x3−0.5x2−ln(x)](6−x)dx(B) ∫25[ln(x)−0.1x3+0.5x2](6−x)dx(C) ∫25[ln2(x)+(0.1x3+0.5x2)2]dx(D) ∫25[ln(x)−0.1x3+0.5x2]2dx
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[0.1x3−0.5x2−ln(x)](6−x)dx(B) ∫25[ln(x)−0.1x3+0.5x2](6−x)dx(C) ∫25[ln2(x)+(0.1x3+0.5x2)2]dx(D) ∫25[ln(x)−0.1x3+0.5x2]2dx
Understand the Problem: Understand the problem.We 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) and y=0.1x3−0.5x2 between x=2 and x=5. The cross sections are rectangles perpendicular to the x-axis with a height of 6−x.
Determine Cross Section Area: Determine the area of a typical cross section.The area A of a rectangle is given by A=length×width. In this case, the length of the rectangle is the difference between the two functions 0.1x3−0.5x2−ln(x) and the width is the height of the rectangle 6−x.
Set Up Integral: Set up the integral to calculate the volume.The volume V of the solid is the integral of the cross-sectional area A from x=2 to x=5. Therefore, V=∫25Adx, where A=(0.1x3−0.5x2−ln(x))(6−x).
Identify Correct Integral: Identify the correct integral from the given options.We need to find the integral that matches the expression for A we found in Step 3. The correct integral is the one that has the difference of the functions multiplied by (6−x) and is integrated from x=2 to x=5.
Match Integral: Match the integral with the given options.The correct integral is (A)∫25(0.1x3−0.5x2−ln(x))(6−x)dx because it represents the area of the cross section (0.1x3−0.5x2−ln(x)) multiplied by the height (6−x) and integrated from x=2 to x=5.
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