   Chapter 5.6, Problem 12E ### 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 3 + 4 ) 3/4                 [ − 1 , 1 ]

To determine

To calculate: the area bounded by the function f(x)=((x)3+4)34 in the interval [1,1] by mid-point rule. Also plot the graph between function f and x-axis.

Explanation

Given Information:

The provided function is f(x)=((x)3+4)34 area of this function is calculated on the interval [1,1] 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: 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=1 and b=1 in the above mid-point rule as;

Δx=1(1)5Δx=25

Width Δx of the provided interval will be 25.

So, the interval [1,1] with width 25. will be divided into 5 subintervals as;

[1,0.6],[0.6,0.2],[0.2,0.2],[0.2,0.6] and [0.6,1]

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 [1,0.6] mid-point is ;

[1,0.6]=10.62=0.8

Similarly for subinterval [0.6,0.2] mid-point is ;

[0.6,0.2]=0.60.22=0.4

For subinterval [0.2,0.2] mid-point is ;

[0.2,0.2]=0

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