   Chapter 5.6, Problem 9E ### 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 x + 4   [ 0 , 1 ]

To determine

To calculate: The area bounded by the function f(x)=xx+4 and x axis over the interval [0,1] by mid-point rule. Also graph the area of the respective region.

Explanation

Given Information:

The provided function is f(x)=xx+4 area of this function is calculated on the interval [0,1] 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)=xx+4.

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

Δx=105=15

Width Δx of the provided interval will be 15.

So, the interval [0,1] with width 15 will be divided into 5 subintervals as;

[0,15],[15,25],[25,35],[35,45],[45,55].

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-intervals

So, for subinterval [0,15] mid-point is;

[0+15]2=110

Similarly, for subinterval [15,25] mid-point is;

[15+25]2=310

For subinterval [25,35] mid-point is;

[25+35]2=510

for subinterval [35,45] mid-point is;

[35+45]2=710

For subinterval [45,55] mid-point is;

[45+55]2=910

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