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CalculusCalculus: An Applied Approach (MindTap Course List)10th Edition

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Chapter 5.6, Problem 3CP
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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Checkpoint 3 Worked-out solution available at LarsonAppliedCalculus.com

Use the Midpoint Rule with n = 4 to approximate the area of the region bounded by the graph of   f ( x ) = x 2 − 1 , the x-axis, and the lines x = 2 and x = 4.

To determine

To calculate: the area bounded by the function f(x)=x2−1 and the line x=2 and x=4 in by mid-point rule. Also plot the graph between function f and x-axis.

Explanation

Given Information:

The provided function is f(x)=x2−1. area of this function is calculated in the line x=2 and x=4 which means area is calculated in the interval [2,4] and the provided interval will be divided into 4 sub interval as n=4.

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=b−an 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=b−an[f(x1)+f(x2)+f(x3)+⋯+f(xn)]

Calculation:

Consider the provided function;

Now, put n equal to 4, a=2 and b=4 in the above mid-point rule as;

Δx=4−24Δx=12

Width Δx of the provided interval will be 12.

So, the interval [2,4] with width 12 will be divided into 4 subintervals as;

[2,52], [52,3], [3,72] and  [72,4].

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

For subinterval [2,52] mid-point is ;

[2,52]=2+522=94

For subinterval [52,3] mid-point is ;

[52,3]=52+32=1

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