   Chapter 5.6, Problem 20E ### 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 ) = 10 x 2 + 5    [ 0 , 2 ]

To determine

To calculate: The area bounded by the function f(x)=10(x2+5) in the interval [0,2] by mid-point rule. Also plot the graph between function f and x-axis.

Explanation

Given Information:

The provided function is f(x)=10(x2+5). Area of this function is calculated on the interval [0,2] 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=2 in the above mid-point rule as;

Δx=204=12

Width Δx of the provided interval will be 12.

So, the interval [0,2] with width 12 will be divided into 4 subintervals as;

[0,12], [12,1], [1,32], [32,2]

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 [0,12] mid-point is ;

[0,12]=0+1223=34

For subinterval [12,1] mid-point is ;

[12,1]=12+12=34

For subinterval [1,32] mid-point is ;

[1,32]=1+322=54

For subinterval [32,2] mid-point is ;

[32,2]=32+22=74

So, the mid points of sub intervals are 14, 34, 54, 74 respectively

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