Q. 53 - Calculus

Question

Suppose that once again you drive in a car for $40$ seconds with velocity $v\left(t\right)=-0.22t^2+8.8t$ feet per second,as shown in the graph that follows. Suppose also that your total distance travelled is equal to the area under the velocity curve on $[0, 40]$.

1. What definite integral would you have to compute in order to find your exact distance travelled over the $40$ seconds of your trip?
2. Find the exact value of that definite integral by taking a limit of Riemann sums.

Step-by-Step Solution

Answer

Part(a) The definite integral is ${\int }_0^{40}\left(0.22t^2+8.8t\right)dt$.

Part(b) The exact value of definite integral is $2346.67\text{ feet}$.

Part(a) Step 1. Given Information

We are given a function and a graph,

$v\left(t\right)=-0.22t^2+8.8t$

Part(a) Step 1. Finding the definite integral

The definite integral that represents the distance travelled over the $40$seconds of your trip is given by,

${\int }_0^{40}\left(0.22t^2+8.8t\right)dt$

Part(b) Step 1. Finding the value of definite integral

The right sum defined for n rectangles on $[a, b]$ is,

 $\sum _{k=1}^{n}f\left(x_k\right)\Delta x$.

Where, $\Delta x=\frac{b-a}{n},x_k=a+k\Delta x$.

The interval is $[0,40]$.

Now,

$$\begin{gathered} \Delta x=\frac{40-0}{n} \\ =\frac{40}{n} \end{gathered}$$

And,

$$\begin{gathered} x_k=0+k\left(\frac{40}{n}\right) \\ =\frac{40k}{n} \end{gathered}$$

Part(b) Step 2. Finding the value of definite integral

The right sum will be,

$$\begin{gathered} \sum _{k=1}^n\left(-0.22{\left(\frac{40k}{n}\right)}^2+8.8\left(\frac{40k}{n}\right)\right)\left(\frac{40}{n}\right)={\left(\frac{40}{n}\right)}^3(-0.22)\sum _{k=1}^n(k{)}^2+{\left(\frac{40}{n}\right)}^2(8.8\left)\sum _{k=1}^n\right(k) \\ =\left[{\left(\frac{40}{n}\right)}^3(-0.22)\left(\frac{n(n+1)(2n+1)}{6}\right)\right]+\left[{\left(\frac{40}{n}\right)}^2(8.8)\frac{n(n+1)}{2}\right] \end{gathered}$$

The exact value will be,

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(x_k^{}\right)\Delta x=\underset{n\to \infty }{\lim }\left(\left[{\left(\frac{40}{n}\right)}^3(-0.22)\left(\frac{n(n+1)(2n+1)}{6}\right)\right]+\left[{\left(\frac{40}{n}\right)}^2(8.8)\frac{n(n+1)}{2}\right]\right) \\ =2346.67 \end{gathered}$$

Hence, the exact distance is $2346.67\text{ feet}$.