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

Full solution

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

y=\ln (x)

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

\newline

(D)

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

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. $\newline$ For $y=\ln(x)$ , at $x=3$ , $y=\ln(3)$ . $\newline$ For $y=0.1x^{3}-0.5x^{2}$ , at $x=3$ , $y=0.1(3)^3-0.5(3)^2$ . $\newline$ We calculate these values to see which is greater.
Calculate Values: Calculating the values: $\newline$ $y=\ln(3) \approx 1.0986$ $\newline$ $y=0.1(3)^3-0.5(3)^2 = 0.1(27)-0.5(9) = 2.7 - 4.5 = -1.8$ $\newline$ Since $\ln(3)$ 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 $(\ln(x) - (0.1x^{3} - 0.5x^{2})) \times (6-x)$ .
Correct Integral: The correct integral for the volume is therefore: $\int_{x=2}^{x=5} [\ln(x) - (0.1x^{3} - 0.5x^{2})] \cdot (6-x) \, dx$ . This matches choice (B) from the given options.