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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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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 ) = 1 ( x 2 + 1 ) 2               [ 1 , 3 ]

To determine

To calculate: The area bounded by the function f(x)=1(x2+1)2 in the interval [1,3] by mid-point rule. Also plot the graph between function f and x-axis.

Explanation

Given Information:

The provided function is f(x)=1(x2+1)2. area of this function is calculated on the interval [1,3] 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=1 and b=3 in the above mid-point rule as;

Δx=3(1)4=44Δx=1

Width Δx of the provided interval will be 1.

So, the interval [1,3] with width 1 will be divided into 4 subintervals as;

[1,0], [0,1], [1,2], [2,3]

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

for subinterval [1,0] mid-point is ;

[1,0]=1+02=12

For subinterval [0,1] mid-point is ;

[0,1]=0+12=12

For subinterval [1,2] mid-point is ;

[1,2]=1+22=32

For subinterval [2,3] mid-point is ;

[2,3]=2+32=52

So, the mid points of sub intervals are 12, 12, 32, 52 respectively

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