   Chapter 5.6, Problem 13E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Using the Midpoint Rule In Exercises 7-14, use the Midpoint Rule with n = 5 to approximate the area of the region bounded by the graph of f and the x-axis over the interval. Sketch the region. See Examples 2 and 3. Function Interval f ( x ) = 8 x 2 + 1                     [ − 5 , 5 ]

To determine

To calculate: the area bounded by the function f(x)=8x2+1 in the interval [5,5] by mid-point rule. Also plot the graph between function f and x-axis.

Explanation

Given Information:

The provided function is f(x)=8x2+1 area of this function is calculated on the interval [5,5] and the provided interval will be divided into 5 sub interval as n=5.

Formula used:

The approximate area of any definite integral baf(x)dx by the use of midpoint rule is calculated by the use of following steps.

Step1: firstly, divide the provided interval of the function into n sub intervals by the use of formula;

Δx=ban where [a,b] are the intervals and n is the subinterval value and Δx is the width.

Step2: by the use of above calculated value of subinterval find the midpoint of each sub interval.

Step3: final step is to obtain approximate area by calculating function f at each mid-point from the use of formula;

abf(x)dx=ban[f(x1)+f(x2)+f(x3)++f(xn)]

Calculation:

Consider the provided function;

Now, put n equal to 5, a=5 and b=5 in the above mid-point rule as;

Δx=5(5)5Δx=2

Width Δx of the provided interval will be 2.

So, the interval [5,5] with width 2.will be divided into 5 subintervals as;

[5,3],[3,1],[1,1],[1,3], and [3,5].

Now, the value of midpoint of each sub interval is calculated as;

[x1,x2]=x1+x22 where x1,x2 are the upper value and lower value of the subintervals

So, for subinterval [5,3] mid-point is ;

[5,3]=532=4

Similarly for subinterval [3,1] mid-point is ;

[3,1]=312=2

For subinterval [1,3] mid-point is ;

[1,3]</

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