   Chapter 5, Problem 114RE ### 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 111-114, use the Midpoint Rule with n = 6 to approximate the area of the region bounded by the graph of f and the x-axis over the interval. Sketch the region. Function Interval f ( x ) = 5 ( x + 2 ) 2               [ 0 , 1 ]

To determine

To calculate: The area of the region bounded by the graph of the function f(x)=5(x+2)2 where the intervals are [0,1] use midpoint rule with n as 6 and sketch the region.

Explanation

Given Information:

The provided function is f(x)=5(x+2)2 and the intervals is intervals are [0,1].

Formula used:

Steps to solve a definite integral abf(x)dx with the help of midpoint rule.

Step 1: For a given interval [a,b] divide it into n subintervals with having width of,

Δx=ban

Step 2: Evaluate the midpoint for the given subinterval. Midpoints={x1,x2,x3,xn}

Step 3: Find the value of f at each midpoint and make the sum as shown below:

abf(x)dxban[f(x1)+f(x2)+f(x3)++f(xn)]

Calculation:

Consider the function,

f(x)=5(x+2)2

The intervals are [0,1] with n=6.

Now divide the provided interval into 6 subparts as shown below:

Δx=106=16

Therefore the 6 sub intervals are:

[0,16],[16,13],[13,12],[12,23],[23,56],[56,1]

Now find the mid points of these intervals because each subinterval has a width of 16. Therefore, the mid points of these interval are shown below,

112,14,512,712,34 and1112

Here, the mid points 112,14,512,712,34 and1112  lies in the middle of [0,16],[16,13],[13,12],[12,23],[23,56],[56,1] intervals respectively

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