Help calculus module 8 DBQplease show work

Accepted Solution

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of [tex]R(t)[/tex] at the endpoints of each subinterval. The sum is then[tex]\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}[/tex]which is measured in units of gallons, hence representing the amount of water that flows into the tank.2. Since [tex]R[/tex] is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval [tex][a,b][/tex], it guarantees the existence of some [tex]c\in(a,b)[/tex] such that[tex]\dfrac{R(b)-R(a)}{b-a)=R'(c)[/tex]Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such [tex]c[/tex] for which [tex]R'(c)=0[/tex]. I would chalk this up to not having enough information.3. [tex]R(t)[/tex] gives the rate of water flow, and [tex]R(t)\approx W(t)[/tex], so that the average rate of water flow over [0, 8] is the average value of [tex]W(t)[/tex], given by the integral[tex]R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt[/tex]If doing this by hand, you can integrate by parts, setting[tex]u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt[/tex][tex]\mathrm dv=\mathrm dt\implies v=t[/tex][tex]R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)[/tex]For the remaining integral, consider the trigonometric substitution [tex]t=\sqrt 7\tan s[/tex], so that [tex]\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds[/tex]. Then[tex]R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds[/tex][tex]R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds[/tex][tex]R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds[/tex][tex]R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}[/tex][tex]R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)[/tex][tex]\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}[/tex]or approximately 3.0904, measured in gallons per hour (because this is the average value of [tex]R[/tex]).4. By the fundamental theorem of calculus,[tex]g'(x)=f(x)[/tex]and [tex]g(x)[/tex] is increasing whenever [tex]g'(x)=f(x)>0[/tex]. This happens over the interval (-2, 3), since [tex]f(x)=3[/tex] on [-2, 0), and [tex]-x+3>0[/tex] on [0, 3).5. First, by additivity of the definite integral, [tex]\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt[/tex]Over the interval [-2, 0), we have [tex]f(x)=3[/tex], and over the interval [0, 6], [tex]f(x)=-x+3[/tex]. So the integral above is[tex]\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}[/tex]