Express
$\int \text{cos}\sqrt{x}dx$ as an infinite series. Evaluate
${\int}_{0}^{1}\text{cos}\sqrt{x}dx$ to within an error of
$0.01.$
$C+{\displaystyle \sum _{n=1}^{\infty}{\left(\mathrm{-1}\right)}^{n+1}\frac{{x}^{n}}{n\left(2n-2\right)\text{!}}}$ The definite integral is approximately
$0.514$ to within an error of
$0.01.$
As mentioned above, the integral
$\int {e}^{\text{\u2212}{x}^{2}}dx$ arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean
$\mu $ and standard deviation
$\sigma ,$ then the probability that a randomly chosen value lies between
$x=a$ and
$x=b$ is given by
To simplify this integral, we typically let
$z=\frac{x-\mu}{\sigma}.$ This quantity
$z$ is known as the
$z$ score of a data value. With this simplification, integral
[link] becomes
In
[link] , we show how we can use this integral in calculating probabilities.
Using maclaurin series to approximate a probability
Suppose a set of standardized test scores are normally distributed with mean
$\mu =100$ and standard deviation
$\sigma =50.$ Use
[link] and the first six terms in the Maclaurin series for
${e}^{\text{\u2212}{x}^{2}\text{/}2}$ to approximate the probability that a randomly selected test score is between
$x=100$ and
$x=200.$ Use the alternating series test to determine how accurate your approximation is.
Since
$\mu =100,\sigma =50,$ and we are trying to determine the area under the curve from
$a=100$ to
$b=200,$ integral
[link] becomes
Using the first five terms, we estimate that the probability is approximately
$0.4922.$ By the alternating series test, we see that this estimate is accurate to within
Use the first five terms of the Maclaurin series for
${e}^{\text{\u2212}{x}^{2}\text{/}2}$ to estimate the probability that a randomly selected test score is between
$100$ and
$150.$ Use the alternating series test to determine the accuracy of this estimate.
The estimate is approximately
$0.3414.$ This estimate is accurate to within
$0.0000094.$
An integral of this form is known as an
elliptic integral of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.
Period of a pendulum
The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length
$L$ that makes a maximum angle
${\theta}_{\text{max}}$ with the vertical, its period
$T$ is given by
where
$g$ is the acceleration due to gravity and
$k=\text{sin}\left(\frac{{\theta}_{\text{max}}}{2}\right)$ (see
[link] ). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and
$\text{sin}\phantom{\rule{0.1em}{0ex}}\theta $ is approximated by
$\theta .)$ Use the binomial series
If
${\theta}_{\text{max}}$ is small, then
$k=\text{sin}\left(\frac{{\theta}_{\text{max}}}{2}\right)$ is small. We claim that when
$k$ is small, this is a good estimate. To justify this claim, consider
For larger values of
${\theta}_{\text{max}},$ we can approximate
$T$ by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate
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Rafiq
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Rafiq
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Anam
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Damian
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?