Question

The function for the standard normal distribution is

$$

f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}

$$

Its graph is that of the bell curve. Probability and statistics books often have tables like the one following, which lists some approximate areas under the bell curve:

Use the information given in the table, properties of definite integrals, and symmetry to find

(a) \(\frac{1}{\sqrt{2 \pi}} \int_{-0.5}^{1.5} e^{-x^{2} / 2} d x\)

(b) \(\frac{1}{\sqrt{2 \pi}} \int_{1.5}^{2} e^{-x^{2} / 2} d x\)

Short answer

(a).

$$

\text { the definite integral is } \int_{0}^{40}\left(0.22 t^{2}+8.8 t\right) d t \text {. }

$$

(b). $$

\text { The exact distance is } 2346.67 \text { feet }

$$

Step-by-step solution

Part (a) Step 1: Given information

The expression is

$\frac{1}{\sqrt{2 \pi}} \int_{-0.5}^{1.5} e^{-x^{2} / 2} d x$

Part (a) Step 2: Simplification

The velocity of a car is,

$$

v(t)=-0.22 t^{2}+8.8 t

$$

The car is driven for 40 seconds.

The total distance travelled is equal to the area under the velocity curve on $[0,40]$.

a.

The objective is to find the definite integral to compute the exact distance travelled by the car over 40 seconds.

The definite integral is,

$$

\int_{0}^{40}\left(0.22 t^{2}+8.8 t\right) d t

$$

Therefore, the definite integral is $\int_{0}^{40}\left(0.22 t^{2}+8.8 t\right) d t$.

Part (b) Step 1: Given information

The expression is

$\frac{1}{\sqrt{2 \pi}} \int_{1.5}^{2} e^{-x^{2} / 2} d x$

Part (b) Step 2: Simplification

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,

$\Delta x$

$$

\begin{aligned}

&=\frac{40-0}{n} \\

&=\frac{40}{n}

\end{aligned}

$$

And.

$$

\begin{aligned}

&x_{k} \\

&=0+k\left(\frac{40}{n}\right) \\

&=\frac{40 k}{n}

\end{aligned}

$$

The right sum is,

$$

\begin{aligned}

&\sum_{k=1}^{n}\left(-0.22\left(\frac{40 k}{n}\right)^{2}+8.8\left(\frac{40 k}{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) \sum_{k=1}^{n}(k) \\

&=\left[\left(\frac{40}{n}\right)^{3}(-0.22)\left(\frac{n(n+1)(2 n+1)}{6}\right)\right]+\left[\left(\frac{40}{n}\right)^{2}(8.8) \frac{n(n+1)}{2}\right]

\end{aligned}

$$

The exact value is,

$$

\begin{aligned}

&\lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(x_{k}{ }^{*}\right) \Delta x \\

&=\lim _{n \rightarrow \infty}\left(\left[\left(\frac{40}{n}\right)^{3}(-0.22)\left(\frac{n(n+1)(2 n+1)}{6}\right)\right]+\left[\left(\frac{40}{n}\right)^{2}(8.8) \frac{n(n+1)}{2}\right]\right) \\

&=2346.67

\end{aligned}

$$

Therefore, the exact distance is $2346.67$ feet.