1. Let [tex]a,b,c[/tex] be the three points of intersection, i.e. the solutions to [tex]f(x)=g(x)[/tex]. They are approximately
[tex]a\approx-3.638[/tex]
[tex]b\approx-1.862[/tex]
[tex]c\approx0.889[/tex]
Then the area [tex]R+S[/tex] is
[tex]\displaystyle\int_a^c|f(x)-g(x)|\,\mathrm dx=\int_a^b(g(x)-f(x))\,\mathrm dx+\int_b^c(f(x)-g(x))\,\mathrm dx[/tex]
since over the interval [tex][a,b][/tex] we have [tex]g(x)\ge f(x)[/tex], and over the interval [tex][b,c][/tex] we have [tex]g(x)\le f(x)[/tex].
[tex]\displaystyle\int_a^b\left(\dfrac{x+1}3-\cos x\right)\,\mathrm dx+\int_b^c\left(\cos x-\dfrac{x+1}3\right)\,\mathrm dx\approx\boxed{1.662}[/tex]
2. Using the washer method, we generate washers with inner radius [tex]r_{\rm in}(x)=2-\max\{f(x),g(x)\}[/tex] and outer radius [tex]r_{\rm out}(x)=2-\min\{f(x),g(x)\}[/tex]. Each washer has volume [tex]\pi({r_{\rm out}(x)}^2-{r_{\rm in}(x)}^2)[/tex], so that the volume is given by the integral
[tex]\displaystyle\pi\int_a^b\left((2-\cos x)^2-\left(2-\frac{x+1}3\right)^2\right)\,\mathrm dx+\pi\int_b^c\left(\left(2-\frac{x+1}3\right)^2-(2-\cos x)^2\right)\,\mathrm dx\approx\boxed{18.900}[/tex]
3. Each semicircular cross section has diameter [tex]g(x)-f(x)[/tex]. The area of a semicircle with diameter [tex]d[/tex] is [tex]\dfrac{\pi d^2}8[/tex], so the volume is
[tex]\displaystyle\frac\pi8\int_a^b\left(\frac{x+1}3-\cos x\right)^2\,\mathrm dx\approx\boxed{0.043}[/tex]
4. [tex]f(x)=\cos x[/tex] is continuous and differentiable everywhere, so the the mean value theorem applies. We have
[tex]f'(x)=-\sin x[/tex]
and by the MVT there is at least one [tex]c\in(0,\pi)[/tex] such that
[tex]-\sin c=\dfrac{\cos\pi-\cos0}{\pi-0}[/tex]
[tex]\implies\sin c=\dfrac2\pi[/tex]
[tex]\implies c=\sin^{-1}\dfrac2\pi+2n\pi[/tex]
for integers [tex]n[/tex], but only one solution falls in the interval [tex][0,\pi][/tex] when [tex]n=0[/tex], giving [tex]c=\sin^{-1}\dfrac2\pi\approx\boxed{0.690}[/tex]
5. Take the derivative of the velocity function:
[tex]v'(t)=2t-9[/tex]
We have [tex]v'(t)=0[/tex] when [tex]t=\dfrac92=4.5[/tex]. For [tex]0\le t<4.5[/tex], we see that [tex]v'(t)<0[/tex], while for [tex]4.5<t\le8[/tex], we see that [tex]v'(t)>0[/tex]. So the particle is speeding up on the interval [tex]\boxed{\dfrac92<t\le8}[/tex] and slowing down on the interval [tex]\boxed{0\le t<\dfrac92}[/tex].
