   Chapter 5.6, Problem 16E ### 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 15-20, use the Midpoint Rule with n = 4 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 − x 3               [ 0 , 1 ]

To determine

To calculate: The area bounded by the function f(x)=x2x3 in the interval [0,1] by mid-point rule. Also plot the graph between function f and x-axis.

Explanation

Given Information:

The provided function is f(x)=x2x3. area of this function is calculated on the interval [0,1] 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=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 4, a=0 and b=1 in the above mid-point rule as;

Δx=104Δx=14

Width Δx of the provided interval will be 14.

So, the interval [0,1] with width 14 will be divided into 4 subintervals as;

[0,14], [14,12], [12,34], [34,1].

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 [0,14] mid-point is ;

[0,14]=0+142=18

Similarly for subinterval [14,12] mid-point is ;

[14,12]=14+122=34×2=38

For subinterval [34,1] mid-point is ;

[34,1]=34+12=78

So, the mid points of sub intervals are 18, 38, 58, 78 respectively

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