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

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Q. The base of a solid is the region enclosed by the graphs of

y=\ln (x)

and

y=0.1 x^{3}-0.5 x^{2}

, between

x=2

and

x=5

\newline

Cross sections of the solid perpendicular to the

x

-axis are rectangles whose height is

6-x

\newline

Which one of the definite integrals gives the volume of the solid?

\newline

Choose

1

answer:

\newline

(A)

\int_{2}^{5}\left[0.1 x^{3}-0.5 x^{2}-\ln (x)\right](6-x) d x

\newline

(B)

\int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right](6-x) d x

\newline

(C)

\int_{2}^{5}\left[\ln ^{2}(x)+\left(0.1 x^{3}+0.5 x^{2}\right)^{2}\right] d x

\newline

(D)

\int_{2}^{5}\left[\ln (x)-0.1 x^{3}+0.5 x^{2}\right]^{2} d x

Understand the Problem: Understand the problem. $\newline$ 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 $y=0.1x^{3}-0.5x^{2}$ 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. $\newline$ The area $A$ of a rectangle is given by $A = \text{length} \times \text{width}$ . In this case, the length of the rectangle is the difference between the two functions $0.1x^{3}-0.5x^{2} - \ln(x)$ and the width is the height of the rectangle $6-x$ .
Set Up Integral: Set up the integral to calculate the volume. $\newline$ The volume $V$ of the solid is the integral of the cross-sectional area $A$ from $x=2$ to $x=5$ . Therefore, $V = \int_{2}^{5} A \, dx$ , where $A = (0.1x^{3}-0.5x^{2} - \ln(x))(6-x)$ .
Identify Correct Integral: Identify the correct integral from the given options. $\newline$ 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. $\newline$ The correct integral is $(A) \int_{2}^{5}(0.1x^{3}-0.5x^{2}-\ln(x))(6-x)dx$ because it represents the area of the cross section $(0.1x^{3}-0.5x^{2} - \ln(x))$ multiplied by the height $(6-x)$ and integrated from $x=2$ to $x=5$ .