Q. 46 - Calculus

Question

Repeat Exercise 45, using

(a) the midpoint sum with n = 4 rectangles;

(b) the trapezoid sum with n = 4 trapezoids.

If the exact distance travelled is just over 2,346 feet, then which of these approximations is the most accurate?

Step-by-Step Solution

Answer

(a) midpoint sum with $n=4$ rectangles is $3520$

(b) midpoint sum with $n=4$ trapezoids $11509.23$

Step 1. Given Information

The velocity of the car is $v\left(t\right)=-0.22t^2+8.8t$ and total distance travelled is equal to the area under the velocity curve on [0, 40],

Step 2. Step 2. Finding ∆ x

$$\begin{gathered} ∆x=\frac{b-a}{n}=\frac{40-0}{4}=10 \\ {x}_k=a+k∆x=0+10k=10k \end{gathered}$$

$x_{k-1}=10\left(k-1\right)$

Part (a) Step 1. Finding Midpoint Sum

$$\begin{gathered} \sum _{k=1}^{4}f\left(10k-10\right)=-\frac{.22{\left(10k-10\right)}^3}{3}+4.4{\left(10k-10\right)}^2 \\ =3520 \end{gathered}$$

Part (b) Step 1. Finding Trapezoid Sum

$\sum _{k=1}^4\frac{f\left(10k-1\right)}{2}=5642.56$

$\sum _{k=1}^4\frac{f\left(10k\right)}{2}=5866.67$

So, $\sum _{k=1}^n\frac{f\left(x_{k-1}\right)+f\left(x_k\right)}{2}=\sum _{k=1}^4\frac{f(10k-1)+f\left(10k\right)}{2}=5642.56+5866.67=11509.23$