Numerical Integration

Numerical Integration

Review of Left and Right Sums

 We first encourage you to review left and right sums in class.  For more information click here
 

The Midpoint Approximation

To get a better approximation of , for example, 

       \( \int_2^5 x^2 dx \)

we could instead use the y-value of the midpoint of each interval as the height.  We will use six rectangles.  

Notice that the first midpoint is at 2.25 and each rectangle width

                     5 - 2
        Dx   =             =  .5
                       6

The new numbers are as follows:

        2.25 + 0(.5), 2.25 + 1(.5), 2.25 + 2(.5), 2.25 + 3(.5), 2.25 + 4(.5), 2.25 + 5(.5) 

so that the y coordinates are

        (2.25 + 0(.5))², (2.25 + 1(.5))², (2.25 + 2(.5))², (2.25 + 3(.5))², (2.25 + 4(.5))², (2.25 + 5(.5))²

We see that the i^th rectangle has y-coordinate:

        height = (2.25 + i(.5))²

To get the area of the i^th rectangle we multiply the height by the base:

        (2.25 + i(.5))²(.5)

Finally to get the total area we add the terms up:

        S(2.25 + i(.5))²(.5)  =  38.9975

The true solution is 37.66... .  The left endpoint approximation would have yielded 33.875 and the right endpoint approximation would have yielded 44.375.  We see that for this case, the midpoint approximation yields a closer approximation.

This approximation is called the midpoint approximation and is given in general by

The Trapezoidal Approximation

A fourth method involves the trapezoidal rule which geometrically calculates the area of the trapezoid with base on the x-axis and heights f(x_i) and f(x_i+1) 

The area of the trapezoid is 

        Dx
                  ( f(x_i) + f(x_i+1) )
          2


or the base times the average of the heights.  Adding up all the trapeziods gives

Example

Use the trapezoidal approximation with three trapezoids to approximate the integral

       \( \int_0^{12} \frac{x^2}{10+x^2} dx \)

Solution

The picture is shown below.

The x values of interest are 

        x₀  =  0,    x₁  =  4,    x₂  =  8,    x₃  =  12

Plugging in these values into the function gives

        f(0)  =  0,    f(4)  =  0.6154,    f(8)  =  0.8649,    f(12)  =  0.9351

The trapezoid approximation formula gives

                12 - 0
T(n)   =                [0 + 2(0.6154) + 2(0.8649) + 0.9351]
                  2(3)

        =  7.7914

We can compare this with the true answer of 7.8475.

Exercise

Use the Trapezoidal Approximation with 5 trap to approximate the integral

       \( \int_3^{13} \sqrt{1 + x^3}dx \)
 

Error

The error in approximating an integral can be found by subtracting the true value from the estimated value.  The graphs show that the error is directly inked to the concavity of the integrand.  Without proof, bounds for the errors using the midpoint and trapezoid approximations are:

                        B(b - a)³ 
        |E_M|  <                         
                            24n²

                        B(b - a)³ 
        |E_T|  <                         
                            12n²

Where 

        B = max |f ''(x)|

Example:  

If you want to approximate 

       \( \int_0^2 e^{x^2} dx \)


using the midpoint rule with an error of less than .001, we compute 

       \(f''(x) = (4 + 4x^2)e^{x^2} \)

which in an increasing function on [0,2], hence has its maximum at x = 2.  So 

        B = (2 + 4(4))e⁴  < 983

so we find 

            983(2³)
                              < .001 
              24n² 

or

        7864  <  .001(24n²)

        n²  <  327667

Taking square roots of both sides gives 

        n > 572

Hence if we let

        n = 573 

we are guaranteed to have an error less than .001. 
 

Simpson's Estimate

We saw that the Trapezoidal and Midpoint estimates provided better accuracy than the Left and Right endpoint estimates.  It turns out that a certain combination of the Trapezoid and Midpoint estimates is even better. 

Geometrically, if n is an even number then Simpson's Estimate gives the area under the parabolas defined by connecting three adjacent points.  

Let n be even then using the even subscripted x values for the trapezoidal estimate and the midpoint estimate, gives

       \( S_n = \frac{1}{3}[ \frac{b-a}{2n} (f(x_0) + 2f(x_2) + 2f(x_4) + ... + 2f(x_{2n-2}) + f(x_{2n}))] \)

       \( + \frac{2}{3}[ \frac{b-a}{n} (f(x_1) + f(x_3) + f(x_5) + ... + f(x_{2n-1}) )] \)

       \( = \frac{b-a}{3n} (f(x_0) + 4f(x_1) + 2f(x_2)+ 4f(x_3) + 2f(x_4) + 4f(x_5) + ... + 2f(x_{n-2} + 4f(x_{n-1}) + f(x_n)) \)

Notice the 

        1 2 4 2 4 ... 2 4 2 4 1 

pattern.
 

Example

Use Simpson's Approximation with n = 6 to approximate 

       \( \int_1^4 \frac{1}{1 + x^3} dx \)
 

Solution

The key values of x are

        x₀  = 1,     x₁  =  1.5,     x₂  =  2,     x₃  =  2.5,     x₄  =  3,     x₅  =  3.5,     x₆  =  4

and the function values are 

        f(x₀)  =  .5,     f(x₁)  =  .2286,      f(x₂)  =  .1111,    

         f(x₃)  =  .0602,      f(x₄)  =  .0357,      f(x₅)  =  .0228,      f(x₆)  =  .0154

Now we can put these numbers into the Simpson's approximation formula.

             4 - 1
                        (.5 + 4(.2286) + 2(.1111) + 4(.0602) + 2(.0357) + 4(.0228) + .0154)
              3(6)

        =  .3426

Exercise

Use Simpson's approximation  with n  =  4 to approximate 

       \( \int_2^{18} \frac{1}{x} dx \)

Error in Simpson's Estimate

Without proof, we state

Let 

        M = max |f''''(x)| 

and let E_S be the error in using Simpson's estimate then

                        M(b - a)⁵ 
        |E_S|  <                         
                         180n⁴

Example


Determine the value of n that will approximate 

       \( \int_1^3 \frac{1}{x} dx \)

within two decimal places of accuracy.


Solution

We need to have an error less than .005:

        |E_S| < .005

We have 

        b - a  =  3 - 1  =  2

We take derivatives to compute M:

        f(x) = 1/x

        f '(x)  = -1/x²

        f ''(x)  =  2/x³

        f '''(x)  = -6/x⁴

        f ''''(x)  = 24/x⁵

We see that the maximum value of this function between 1 and 3 occurs when x = 1:

        M = |24/1⁵| = 24

We put this together to get

                  (24)(2⁵)
                                    <  .005
                   180n⁴

Multiplying by the denominator gives

        768 < 0.9 n⁴

or

        853.33 < n⁴

Taking fourth roots gives

        n > 5.4

Hence, if we choose n = 6 we are guaranteed two decimals of accuracy. 

Exercise

If you want to approximate 

        \( \int_0^1 cos(x) dx \)

with n  =  5, determine the maximum errors that occur using the midpoint, the trapezoidal, and Simpson's approximation.

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