1 Answer

Joao Luiz Cadore · Answer 1 · 2021-08-02T19:58:49+0000

(1) Using a trapezoidal sum, we approximate the integral by

$\displaystyle\int_0^8R(t)\,\mathrm dt\approx\frac{(R(0)+R(2))(2-0)+(R(3)+R(2))(3-2)+(R(7)+R(3))(7-3)+(R(8)+R(7))(8-7)}2$

$\displaystyle\int_0^8R(t)\,\mathrm dt\approx24.830$

measured in gallons of water.

(2) Notice that the table shows the specific values of
R(t) increasing, meaning
R'(t)>0 . There's not enough evidence to suggest that we have
R'(t)=0 at any time.

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