Given the graphs of two functions (one explicit, one only constrained by supplied conditions), students find the area of a shaded region between them, evaluate an improper integral or show that it diverges, and evaluate an integral involving x times a derivative using integration by parts and the given conditions.
Given the graphs of two functions (one explicit, one only constrained by supplied conditions), students find the area of a shaded region between them, evaluate an improper integral or show that it diverges, and evaluate an integral involving x times a derivative using integration by parts and the given conditions.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
On $0\le x\le 3$ the graph of $f$ lies above the graph of $g$, so the shaded area is $$\text{Area}=\int_0^3\bigl(f(x)-g(x)\bigr)\,dx=\int_0^3 f(x)\,dx-\int_0^3 g(x)\,dx.$$ The first integral is given: $\int_0^3 f(x)\,dx=10$. For the second, antidifferentiate $g(x)=\dfrac{12}{3+x}$: $$\int_0^3\frac{12}{3+x}\,dx=12\Bigl[\ln(3+x)\Bigr]_0^3=12\bigl(\ln 6-\ln 3\bigr)=12\ln 2.$$ Therefore $$\text{Area}=10-12\ln 2\approx 1.682.$$
3 points: P1 for an integrand of f(x)-g(x) (or g(x)-f(x), or absolute value) in a definite integral; an implied f-integrand with explicit g, e.g. 10 - int_0^3 g dx, also earns it. P2 for the antiderivative of g, namely a*ln(3+x) form, here 12 ln(3+x). P3 (only after P1, P2) for the correct answer 10 - 12 ln 2. The answer need not be simplified, but any attempted simplification must be correct. Incorrect u-sub limits forfeit P3.
Since $g(x)=\dfrac{12}{3+x}$, we have $\bigl(g(x)\bigr)^2=\dfrac{144}{(3+x)^2}$. Write the improper integral as a limit: $$\int_0^{\infty}\bigl(g(x)\bigr)^2\,dx=\lim_{b\to\infty}\int_0^{b}\frac{144}{(3+x)^2}\,dx.$$ An antiderivative of $\dfrac{144}{(3+x)^2}$ is $-\dfrac{144}{3+x}$, so $$\int_0^{b}\frac{144}{(3+x)^2}\,dx=\left[-\frac{144}{3+x}\right]_0^{b}=-\frac{144}{3+b}+\frac{144}{3}.$$ Taking the limit, $$\lim_{b\to\infty}\left(-\frac{144}{3+b}+48\right)=0+48=48.$$ The improper integral **converges** to $48$.
3 points: P1 for correct limit notation throughout (rewriting the improper integral as lim_{b->inf} of int_0^b), with no arithmetic on infinity. P2 for an antiderivative of the form -a/(3+x), here -144/(3+x). P3 for the answer 48. If the antiderivative constant a != 144, P3 is lost. Incorrect u-sub limits forfeit P3.
With $h(x)=x\cdot f'(x)$, evaluate $\int_0^3 x\,f'(x)\,dx$ by integration by parts. Let $$u=x,\quad dv=f'(x)\,dx\;\Longrightarrow\; du=dx,\quad v=f(x).$$ Then $$\int_0^3 x\,f'(x)\,dx=\Bigl[x\,f(x)\Bigr]_0^3-\int_0^3 f(x)\,dx.$$ Evaluate the boundary term using $f(3)=2$ (the $x=0$ term vanishes because of the factor $x$), and use the given $\int_0^3 f(x)\,dx=10$: $$=\bigl(3\cdot f(3)-0\cdot f(0)\bigr)-10=3\cdot 2-0-10=-4.$$ So $\displaystyle\int_0^3 h(x)\,dx=-4.$
3 points: P1 for the integration-by-parts choice u = x and dv = f'(x) dx (may be implied by the next line). P2 for the antiderivative structure x*f(x) - int f(x) dx (tabular method acceptable). P3 (only if P1, P2 earned) for the correct value -4, obtained from 3*f(3) - 0 - 10 = 3*2 - 10.
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