Question

Help calculus module 8 DBQ

please show work

Answer 1

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
`R(t)` at the endpoints of each subinterval. The sum is then

`\frac{R(0)+R(2)}2(2-0)+\frac{R(2)+R(3)}2(3-2)+\frac{R(3)+R(7)}2(7-3)+\frac{R(7)+R(8)}2(7-8)=\boxed{24.83}`

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since
`R` is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval
`[a,b]`, it guarantees the existence of some
`c\in(a,b)` such that

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
`c` for which
`R'(c)=0`. I would chalk this up to not having enough information.

3.
`R(t)` gives the rate of water flow, and
`R(t)\approx W(t)`, so that the average rate of water flow over [0, 8] is the average value of
`W(t)`, given by the integral

`R_(\rm avg)=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt`

If doing this by hand, you can integrate by parts, setting

`u=\ln(t^2+7)\implies\mathrm du=(2t)/(t^2+7)\,\mathrm dt`

`\mathrm dv=\mathrm dt\implies v=t`

`R_(\rm avg)=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_(t=0)^(t=8)-\int_0^8(2t^2)/(t^2+7)\,\mathrm dt\right)`

For the remaining integral, consider the trigonometric substitution
`t=\sqrt 7\tan s`, so that
`\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds`. Then

`R_(\rm avg)=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^(-1)(8/\sqrt7)}(7\tan^2s)/(7\tan^2s+7)\sec^2s\,\mathrm ds`

`R_(\rm avg)=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^(-1)(8/\sqrt7)}\tan^2s\,\mathrm ds`

`R_(\rm avg)=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^(-1)(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds`

`R_(\rm avg)=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_(s=0)^{s=\tan^(-1)(8/\sqrt7)}`

`R_(\rm avg)=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^(-1)\frac8{\sqrt7}\right)-\tan^(-1)\frac8{\sqrt7}\right)`

`\boxed{R_(\rm avg)=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^(-1)\frac8{\sqrt7}}`

or approximately 3.0904, measured in gallons per hour (because this is the average value of
`R`).

4. By the fundamental theorem of calculus,

`g'(x)=f(x)`

and
`g(x)` is increasing whenever
`g'(x)=f(x)>0`. This happens over the interval (-2, 3), since
`f(x)=3` on [-2, 0), and
`-x+3>0` on [0, 3).

5. First, by additivity of the definite integral,

`\displaystyle\int_(-2)^xf(t)\,\mathrm dt=\int_(-2)^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt`

Over the interval [-2, 0), we have
`f(x)=3`, and over the interval [0, 6],
`f(x)=-x+3`. So the integral above is

`\displaystyle\int_(-2)^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_(t=-2)^(t=0)+\left(-\frac{t^2}2+3t\right)\bigg|_(t=0)^(t=x)=\boxed{6+3x-\frac{x^2}2}`