(a) Use the Midpoint Rule and the given data to estimate the value of the integral ∫1

(a) Use the Midpoint Rule and the given data to...

$\begin{array}{l}{\text { (a) Use the Midpoint Rule and the given data to estimate }} \\ {\text { the value of the integral } \int_{1}^{5} f(x) d x}\end{array}$ $\begin{array}{l}{\text { (b) If it is known that }-2 \leqslant f^{\prime \prime}(x) \leqslant 3 \text { for all } x, \text { estimate }} \\ {\text { the error involved in the approximation in part (a). }}\end{array}$

$\begin{array}{l}{\text { (a) Use the Midpoint Rule and the given data to estimate }} \\ {\text { the value of the integral } \int_{1}^{5} f(x) d x}\end{array}$
$\begin{array}{l}{\text { (b) If it is known that }-2 \leqslant f^{\prime \prime}(x) \leqslant 3 \text { for all } x, \text { estimate }} \\ {\text { the error involved in the approximation in part (a). }}\end{array}$

Key Concepts

Midpoint Rule

The Midpoint Rule is a numerical integration technique that estimates the value of a definite integral by dividing the integration interval into subintervals, computing the function value at the midpoint of each subinterval, and then multiplying by the width of each subinterval. This method simplifies the integration process by approximating the area under the curve using rectangles whose heights are determined by the function values at the midpoint, leading to an overall approximation of the integral.

Error Analysis for the Midpoint Rule

Error analysis in the context of the Midpoint Rule involves estimating the difference between the true value of the integral and the approximation obtained by the rule. This analysis typically makes use of the bound on the second derivative of the function, as the error is linked to the curvature of the function. A known error bound formula states that the error is proportional to the maximum absolute value of the second derivative over the interval and inversely proportional to the square of the number of subintervals, allowing one to estimate and control the accuracy of the numerical approximation.

Transcript

00:01 In part a, we are going to estimate the integrals from 1 to 5 of the function f of x differential of x by using the midpoint rule, and we are given the data corresponding to the images of some points in the domain of this function.

00:20 In part b, if we know the fact that the second derivative of this function f is greater than or equal to negative 2 and less than or equal to 3 for all x, we want to estimate the error in the approximation to the integral given by the mean point rule calculated in bar a.

00:47 So we start with part a.

00:48 Here we have the data.

00:50 We can see that the values where we have data given are 1 .5, 2, 2 .53, 3 .5, 4 .5 and 5 .5 and 5.

01:03 And 5 and then looking at this data then we can apply midpoint rule by taking delta x equal 1 and this implies that the nodes are 1 1 2 3 4 and 5 that is the nodes are these values here and that is because we need the midpoint of these intervals to apply midpoint rule.

01:54 Then if these are the nodes, that is the sub -intervals are from 1 to 2, from 2 to 3 to 4 and from 4 to 5, then the midpoints are in the table, that is the midpoint between 1 and 2 is 1 .5, the midpoint between 2 and 3 is 2 .5, between 3 and 4 is 3 .5, and between 4.

02:21 And 5 is 4 .5 so that's the reason why looking at the graph we decide that this is the step size these are the nodes and the midpoints are 1 .5 2 .5 3 .5 and 4 .5 and with this selection n equal 4 because we have 4 sub interval from one to two to three three four four to five and having said that we can say that the midpoint rule using four sub intervals will be delta x times f at the first midpoint 1 .5 plus f at the second midpoint 1 .5 plus f at the third midpoint 3 .5 plus f at the last midpoint 4 .5 that is and 4 is one times the image of 1 .5 is 2 .9 so we get 2 .9 plus the image of 2 .5 is 3 .6 plus the image of 3 .5 is 4 plus the image of 4 .5 is 3 .9 .5 is 3 .9.

04:29 So it's the sum of these values because here we have factor 1 and the result is 14 .4.

04:38 That is the midpoint rule is in force of intervals and the data in this table give us 14 .4.

04:52 And we know this is an approximation to the interval.

04:56 That is the interval is approximately equal to 14...

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