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

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Chapter 5.6, Problem 2E
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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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Approximating the Area of a Plane Region In Exercises 1-6, use the rectangles to approximate the area of the region. Compare your result with the exact area obtained using a definite integral. See Example 1.

f ( x ) = f 1 x , [ 1 , 5 ]

Chapter 5.6, Problem 2E, Approximating the Area of a Plane Region In Exercises 1-6, use the rectangles to approximate the

To determine

To calculate: The area bounded by the function f(x)=1x between interval [1,5] by mid-point rule. Also compare this approximate area with exact area using definite integral.

Calculus: An Applied Approach (MindTap Course List), Chapter 5.6, Problem 2E

Explanation

Given Information:

The provided function is f(x)=1x. Area of this function is calculated between the interval [1,5] as shown below:

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: First 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: Use above 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 f(x)=1x. The graph of provided function is shown below:

In provided graph, interval [1,5] is divided into 4 rectangle.

Now, put n=4, a=1 and b=5 to find the width of each rectangle as;

Δx=5−14=1

Width Δx of each rectangle will be 1.

So, the interval [1,5] will be divided into 4 subintervals as;

[1,2],[2,3],[3,4],[4,5].

Now, the value of midpoint of each rectangle is calculated as;

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

So, for first rectangle [1,2] mid-point is;

1+22=1.5

Similarly, for second rectangle [2,3] mid-point is;

2+32=2.5

For third rectangle [3,4] mid-point is;

3+42=3.5

For last rectangle [4,5] mid-point is;

4+52=4

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