   Chapter 5.6, Problem 7E ### 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 ) = x 2   [ 1 , 6 ]

To determine

To calculate: The area bounded by the function f(x)=x2 and x axis over the interval [1,6] by mid-point rule. Also graph the area of the respective region.

Explanation

Given Information:

The provided function is f(x)=x2 area of this function is calculated on the interval [1,6] 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: First 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: 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)dxban[f(x1)+f(x2)+f(x3)+....+f(xn)].

Calculation:

Consider the provided function f(x)=x2.

Now, put n=5, a=1 and b=6 to find the width of subinterval as;

Δx=615=1

Width Δx of the provided interval will be 1.

So, the interval [1,6] with width 1 will be divided into 5 subintervals as;

[1,2],[2,3],[3,4],[4,5],[5,6].

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 sub-interval

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

1+22=32

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

2+32=52

For subinterval [3,4] mid-point is;

3+42=72

for subinterval [4,5] mid-point is;

4+52=92

For subinterval [5,6] mid-point is;

5+62=112

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