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Express cos x d x as an infinite series. Evaluate 0 1 cos x d x to within an error of 0.01 .

C + n = 1 ( −1 ) n + 1 x n n ( 2 n 2 ) ! The definite integral is approximately 0.514 to within an error of 0.01 .

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As mentioned above, the integral e x 2 d x 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 μ and standard deviation σ , then the probability that a randomly chosen value lies between x = a and x = b is given by

1 σ 2 π a b e ( x μ ) 2 / ( 2 σ 2 ) d x .

(See [link] .)

This graph is the normal distribution. It is a bell-shaped curve with the highest point above mu on the x-axis. Also, there is a shaded area under the curve above the x-axis. The shaded area is bounded by alpha on the left and b on the right.
If data values are normally distributed with mean μ and standard deviation σ , the probability that a randomly selected data value is between a and b is the area under the curve y = 1 σ 2 π e ( x μ ) 2 / ( 2 σ 2 ) between x = a and x = b .

To simplify this integral, we typically let z = x μ σ . This quantity z is known as the z score of a data value. With this simplification, integral [link] becomes

1 2 π ( a μ ) / σ ( b μ ) / σ e z 2 / 2 d z .

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 μ = 100 and standard deviation σ = 50 . Use [link] and the first six terms in the Maclaurin series for e x 2 / 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 μ = 100 , σ = 50 , and we are trying to determine the area under the curve from a = 100 to b = 200 , integral [link] becomes

1 2 π 0 2 e z 2 / 2 d z .

The Maclaurin series for e x 2 / 2 is given by

e x 2 / 2 = n = 0 ( x 2 2 ) n n ! = 1 x 2 2 1 · 1 ! + x 4 2 2 · 2 ! x 6 2 3 · 3 ! + + ( −1 ) n x 2 n 2 n · n ! + = n = 0 ( −1 ) n x 2 n 2 n · n ! .

Therefore,

1 2 π e z 2 / 2 d z = 1 2 π ( 1 z 2 2 1 · 1 ! + z 4 2 2 · 2 ! z 6 2 3 · 3 ! + + ( −1 ) n z 2 n 2 n · n ! + ) d z = 1 2 π ( C + z z 3 3 · 2 1 · 1 ! + z 5 5 · 2 2 · 2 ! z 7 7 · 2 3 · 3 ! + + ( −1 ) n z 2 n + 1 ( 2 n + 1 ) 2 n · n ! + ) 1 2 π 0 2 e z 2 / 2 d z = 1 2 π ( 2 8 6 + 32 40 128 336 + 512 3456 2 11 11 · 2 5 · 5 ! + ) .

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

1 2 π 2 13 13 · 2 6 · 6 ! 0.00546 .
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Use the first five terms of the Maclaurin series for e x 2 / 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 .

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Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is

0 π / 2 d θ 1 k 2 sin 2 θ .

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 θ max with the vertical, its period T is given by

T = 4 L g 0 π / 2 d θ 1 k 2 sin 2 θ

where g is the acceleration due to gravity and k = sin ( θ max 2 ) (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 sin θ is approximated by θ . ) Use the binomial series

1 1 + x = 1 + n = 1 ( −1 ) n n ! 1 · 3 · 5 ( 2 n 1 ) 2 n x n

to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if

  1. you use only the first term in the binomial series, and
  2. you use the first two terms in the binomial series.
    This figure is a pendulum. There are three positions of the pendulum shown. When the pendulum is to the far left, it is labeled negative theta max. When the pendulum is in the middle and vertical, it is labeled equilibrium position. When the pendulum is to the far right it is labeled theta max. Also, theta is the angle from equilibrium to the far right position. The length of the pendulum is labeled L.
    This pendulum has length L and makes a maximum angle θ max with the vertical.

We use the binomial series, replacing x with k 2 sin 2 θ . Then we can write the period as

T = 4 L g 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ + 1 · 3 2 ! 2 2 k 4 sin 4 θ + ) d θ .
  1. Using just the first term in the integrand, the first-order estimate is
    T 4 L g 0 π / 2 d θ = 2 π L g .

    If θ max is small, then k = sin ( θ max 2 ) is small. We claim that when k is small, this is a good estimate. To justify this claim, consider
    0 π / 2 ( 1 + 1 2 k 2 sin 2 θ + 1 · 3 2 ! 2 2 k 4 sin 4 θ + ) d θ .

    Since | sin x | 1 , this integral is bounded by
    0 π / 2 ( 1 2 k 2 + 1.3 2 ! 2 2 k 4 + ) d θ < π 2 ( 1 2 k 2 + 1 · 3 2 ! 2 2 k 4 + ) .

    Furthermore, it can be shown that each coefficient on the right-hand side is less than 1 and, therefore, that this expression is bounded by
    π k 2 2 ( 1 + k 2 + k 4 + ) = π k 2 2 · 1 1 k 2 ,

    which is small for k small.
  2. For larger values of θ 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
    T 4 L g 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ ) d θ = 2 π L g ( 1 + k 2 4 ) .
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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