Q. 5 - Calculus

Question

Limits of Riemann sums: In the reading we saw that the

area between the graph of $f\left(x\right)=x^2-2x+2$ and the

$x-$axis on $\left[1,3\right]$could be approximated with the right sum

$\sum _{k=1}^n\left(1+\frac{4k^2}{n^2}\right)\left(\frac{2}{n}\right)$. Let $A\left(n\right)$ be equal to this $n-$rectangle right-sum approximation. The following table describes various values of:

Describe the meaning of the entries in this table, and verify that the entry for $A\left(10\right)$is correct.

Step-by-Step Solution

Answer

Area under the graph of the velocity is $3.4$ square feet and total distance traveled is $50.6$ feet.

Step. 1 Given Information

$A\left(n\right)$be equal to the $n-$rectangle right-sum approximation.
The table with various values of $A\left(n\right)$:

Step 2. Area under the graph

The relationship is $S=A+54$

The distance from the stop sign (in feet) $t$ seconds after stepping on the breaks is given by the function,
$s\left(t\right)=3t^3-12t^2-9t+54$

The area under the graph of the velocity on $\left[0,4.6\right]$ (The velocity was found by differentiation of$s\left(t\right)=3t^3-12t^2-9t+54$).
$v\left(t\right)=9t^2-24t-9$
$A\left(t\right)={\int }_0^{4.6}\left(9t^2-24t-9\right)dt$
$A\left(t\right)={\left[9\frac{t^3}{3}\right]}_0^{4.6}-{\left[24\frac{t^2}{2}\right]}_0^{4.6}-{\left[9\frac{t}{1}\right]}_0^{4.6}$

$A\left(t\right)=292-254-41.4$

$A\left(t\right)=-3.4$

Step 3. Total distance traveled

$s\left(t\right)=3t^3-12t^2-9t+54$
$s(4.6)=3{{\left(}_{\right)}}^{4.6}-12{{\left(}_{\right)}}^{4.6}-9(4.6)+54$
$s(4.6)=292-254-41.4+54$

$s(4.6)=50.6$
The relationship between the two is $S=A+54$.