Accumulation and rates with volume of concrete

Question

$f(x)$ is the volume of concrete, in cubic meters, that has been added to a floor of a skyscraper after $x$ seconds of a $45$-second pour. Values of $r_f$ defined as $r_f(x)=2e^{-\frac{x}{15}}$ give the rate of change of $f$ in m$^3$/sec for all $0 \le x \le 45$ seconds.

First, I was asked to find the rate the concrete was being poured at $x=5.2$. I gave $1.4108$ and was marked correct. Then, I was asked to find the amount of concrete poured at $x=5.4$ given $f(5.2)=8.79$ m$^3$ of concrete. I took that value plus the slope of $f(5.2)$ times the change in $x$, giving me $0.282816$ which was also correct. Then I was asked to find the amount of concrete poured between $5.2$ and $6$ seconds. My answer was marked correct using $1.41408(0.8)$

But then I was given one last problem, "Revise your previous answer by estimating how much concrete is poured every $0.2$ seconds from $5.2$ to $6$ seconds"

I tried $$1.41408(0.2)+1.41408(0.4)+1.41408(0.6)+1.41408(0.8)=2.82816$$

and was marked wrong. I realize I should not be using the same slope $y=1.41408$, and I also noticed I should not be multiplying by successive terms of $0.2$. I instead tried to say that variation is $m(0.2)$ where $m$ is $r_f(x)$. Basically $0.2r_f(5.4)+0.2r_f(5.6)+...+0.2r_f(6)$ After summing this up I got 1.37712 which seemed to make a lot more sense. But I was still marked wrong. What should I try at this point?

Answer 1

I decided to make an answer of all that was discussed in the comments

First, OP hasn't provide the source of the problem, so there will be a few assumption I'll make. OP attempts suggest that it is a problem of linear approximation. Since the rate of change function is integrable, I'll compare every answer with the "true" value obtain by integration.

By integration

The rate of change of $f$ is given by $r_f(x)=2e^{-\frac x{15}}$, $0\leq x\leq 45$. We could find the amount of concrete poured after $t$ second by $$f(t)=\int_0^t r_f(x) dx = \int_0^t 2e^{-\frac x{15}} dx = 30\left(1-e^{-\frac t{15}}\right)$$ Note that $f(5.2)=8.788770921$ which is the value given by the problem.

Find the amount on concrete poured after $5.4$ seconds.

$$f(5.4)=9.069710218\tag{1}$$ The amount that was poured between $5.2$ seconds and $5.4$ seconds is $$f(5.4)-f(5.2)=0.280939297$$

Find the amount of concrete poured between $5.2$ seconds and $6$ seconds.

$$f(6)-f(5.2) = 1.101627698\tag{2}$$

By linear approximation

The rate of change at $5.2$ is $$r_f(5.2)=2e^{-\frac{5.2}{15}}=1.414081939$$

Find the amount on concrete poured after $5.4$ seconds knowing that $f(5.2)=8.79\text{ m}^3$

$$f(5.4) = f(5.2)+r_f(5.2)(5.4-5.2)=8.79+0.282816387\approx 9.072$$ which is less than $1\%$ away from the value computed in $(1)$.

Find the amount of concrete poured between $5.2$ seconds and $6$ seconds.

$$f(6)-f(5.2) = r_f(5.2)(6-5.2) = 1.131265551$$ which is about $3\%$ over the value of $(2)$.

Revise the amount of concrete poured between $5.2$ seconds and $6$ seconds by estimating how much concrete is poured every $0.2$ seconds.

this is where I wanted to compare the different possibilities that were discussed in the comments

OP suggest the following $$0.2*r_f(5.4)+0.2*r_f(5.6)+0.2*r_f(5.8)+0.2*r_f(6)=1.094299834$$ Usualy we use the beginning of the interval for the approximation, and not the end. This will give us $$0.2*r_f(5.2)+0.2*r_f(5.4)+0.2*r_f(5.6)+0.2*r_f(5.8)=1.108988203$$ @YNK suggested two other approachs. first by splitting the interval differently. $$0.1*r_f(5.2)+0.2*r_f(5.4)+0.2*r_f(5.6)+0.2*r_f(5.8)+0.2*r_f(6)=1.101644018$$ Then by the middle of the interval $$0.2*r_f(5.3)+0.2*r_f(5.5)+0.2*r_f(5.7)+0.2*r_f(5.9)=1.101619538$$