A region is bounded by a quadratic and a sine-plus-line curve. Students find the area of the region, the volume of a solid built from rectangular cross sections on it, set up (without evaluating) a revolution-volume integral about a horizontal line, and find where the two curves have parallel tangent lines.
A region is bounded by a quadratic and a sine-plus-line curve. Students find the area of the region, the volume of a solid built from rectangular cross sections on it, set up (without evaluating) a revolution-volume integral about a horizontal line, and find where the two curves have parallel tangent lines.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
On $R$ the curve $g(x)=x+\sin(\pi x)$ lies above $f(x)=x^2-2x$, and the two graphs meet at $x=0$ and $x=3$ (since $g-f=-x^2+3x+\sin(\pi x)$ vanishes at both). The area is $$\int_0^3\big(g(x)-f(x)\big)\,dx = \int_0^3\big(x+\sin(\pi x)-(x^2-2x)\big)\,dx = 5.136620\approx 5.137.$$ (The exact value is $\frac{4+9\pi}{2\pi}$.) The area of $R$ is about $5.137$.
2 points. P1: an integrand of g(x)-f(x) (or f(x)-g(x), or the absolute value of either) inside a definite integral on [0,3], with or without dx; a difference of two definite integrals also qualifies. P2: the numeric answer 5.137 (or 5.136). Incorrect linkage between integral and answer is treated as scratch work and not penalized.
Each cross section perpendicular to the $x$-axis is a rectangle whose base lies in $R$ (length $g(x)-f(x)$) and whose height is $x$. The cross-sectional area is $x\big(g(x)-f(x)\big)$, so the volume is $$\int_0^3 x\big(g(x)-f(x)\big)\,dx = \int_0^3 x\big(x+\sin(\pi x)-(x^2-2x)\big)\,dx = 7.704930\approx 7.705.$$ (The exact value is $\frac{12+27\pi}{4\pi}$.) The volume of the solid is about $7.705$.
2 points. P3: a definite integral whose integrand is a product of two nonconstant factors, one of which is x and the other g(x)-f(x) (or f(x)-g(x)). P4: the numeric answer 7.705 (or 7.704). The differential dx is not required for either point, and incorrect linkage between integral and answer is ignored.
Rotating $R$ about the horizontal line $y=-2$ produces washers. At each $x$, the outer radius reaches from $y=-2$ up to the higher curve $g$, giving $R(x)=g(x)-(-2)=g(x)+2$, and the inner radius reaches to the lower curve $f$, giving $r(x)=f(x)-(-2)=f(x)+2$. Therefore (set up, not evaluated) $$V=\pi\int_0^3\Big[(g(x)+2)^2-(f(x)+2)^2\Big]\,dx.$$
3 points (setup only; do not evaluate). P5: a definite integral with an R^2 - r^2 form where one radius is correct (or a difference between the proper curve and a nonzero constant). P6: the fully correct integrand (g(x)+2)^2 - (f(x)+2)^2 (or an equivalent form). P7: correct limits 0 to 3, the leading constant pi, and the differential dx; eligibility for P7 requires having earned P5. Omitting pi keeps P5/P6 eligible but loses P7.
The tangent lines are parallel where the slopes match: $f'(x)=g'(x)$. With $f'(x)=2x-2$ and $g'(x)=1+\pi\cos(\pi x)$, $$2x-2 = 1+\pi\cos(\pi x).$$ Solving on $0<x<1$ gives $$x = 0.675819 \approx 0.676.$$ The tangent to $f$ is parallel to the tangent to $g$ at $x\approx 0.676$.
2 points. P8: the general condition f'(x)=g'(x), or any correct equation formed by substituting 2x-2 for f'(x), 1+pi*cos(pi*x) for g'(x), or both. P9: the numeric answer 0.676 (or 0.675), with or without supporting work.
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