Given two polynomial functions and their graph, students set up the integral for the area of a bounded region, compute the volume of a solid whose cross sections perpendicular to the x-axis are rectangles, and set up (without evaluating) the volume integral when a region is revolved about a horizontal line.
Given two polynomial functions and their graph, students set up the integral for the area of a bounded region, compute the volume of a solid whose cross sections perpendicular to the x-axis are rectangles, and set up (without evaluating) the volume integral when a region is revolved about a horizontal line.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
On $0\le x\le 2$ the graph shows $f(x)=x^2+2$ above $g(x)=x^2-2x$ (indeed $f(x)-g(x)=2x+2>0$ there). The area of $R$ is the integral of (top $-$ bottom): $$\text{Area}=\int_0^2\big(f(x)-g(x)\big)\,dx=\int_0^2\big((x^2+2)-(x^2-2x)\big)\,dx=\int_0^2(2x+2)\,dx.$$ (The problem asks only to write the expression, not evaluate it; for reference its value is $8$.)
2 points: 1 for the correct integrand f(x)-g(x) (top minus bottom, with f above g on [0,2]); 1 for the complete definite-integral expression with limits 0 and 2. Reversing the order (g-f) loses the integrand point unless presented as an absolute value; evaluating is not required.
On $2\le x\le 5$, region $S$ lies between $g(x)=x^2-2x$ and the $x$-axis, with $g(x)\ge 0$ there, so the base of each cross section has length $g(x)$. The cross section is a rectangle whose height is half its base, so height $=\tfrac12 g(x)$ and cross-sectional area $$A(x)=\text{base}\times\text{height}=g(x)\cdot\tfrac12 g(x)=\tfrac12\big(g(x)\big)^2=\tfrac12\big(x^2-2x\big)^2.$$ The volume is $$V=\int_2^5 \tfrac12\big(x^2-2x\big)^2\,dx=\tfrac12\int_2^5\big(x^4-4x^3+4x^2\big)\,dx.$$ Antidifferentiate: $$\tfrac12\left[\frac{x^5}{5}-x^4+\frac{4x^3}{3}\right]_2^5.$$ At $x=5$: $\frac{3125}{5}-625+\frac{500}{3}=625-625+\frac{500}{3}=\frac{500}{3}$. At $x=2$: $\frac{32}{5}-16+\frac{32}{3}=\frac{96-240+160}{15}=\frac{16}{15}$. So $$V=\tfrac12\left(\frac{500}{3}-\frac{16}{15}\right)=\tfrac12\cdot\frac{2500-16}{15}=\tfrac12\cdot\frac{2484}{15}=\frac{1242}{15}=\frac{414}{5}.$$ The volume of the solid is $\dfrac{414}{5}=82.8$.
4 points: 1 for the integrand of the form (1/2)(g(x))^2 (base g, height g/2); 1 for the correct limits x=2 and x=5; 1 for a correct antiderivative of (1/2)(x^2-2x)^2; 1 for the answer 414/5 (=82.8). Using f instead of g, or omitting the 1/2 height factor, forfeits the integrand point and cascades; arithmetic slips on the antiderivative cost only the answer point if the setup is correct.
Revolve $S$ about the horizontal line $y=20$. Since $S$ lies between $y=0$ and $y=g(x)$ (with $0\le g(x)\le 20$ on $[2,5]$) and the axis $y=20$ is above the region, each slice generates a **washer**: outer radius reaches from $y=20$ down to $y=0$, giving $R=20$; inner radius reaches from $y=20$ down to $y=g(x)$, giving $r=20-g(x)$. The volume is $$V=\pi\int_2^5\Big(R^2-r^2\Big)\,dx=\pi\int_2^5\Big(20^2-\big(20-g(x)\big)^2\Big)\,dx,$$ where $g(x)=x^2-2x$. The problem asks only for this expression, not its value. (As a numeric check, $V\approx 1741.699$.)
3 points: 1 for the washer-integrand form pi(20^2-(20-g(x))^2) (outer radius 20, inner radius 20-g); 1 for writing g(x)=x^2-2x explicitly in the integrand; 1 for the correct limits x=2 to x=5 and the constant/pi factor. Using radius g(x) (disk about the x-axis) instead of 20-g(x), or revolving f, does not earn the form point; the integral is not evaluated.
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