   Chapter 5, Problem 22TYS ### 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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# In Exercises 21 and 22, 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. Compare your result with the exact area obtained using a definite integral. Sketch the region. f ( x ) = x 2 + 2 ,     [ − 1 , 1 ]

To determine

To calculate: The area of the region bounded by the graph of the function f(x)=x2+2 with x-axis over the interval [1,1] by the midpoint rule with n as 4 and also calculate the exact area by the use of a definite integral and then sketch the region.

Explanation

Given Information:

The provided function is f(x)=x2+2 and the intervals is intervals are [1,1]

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,

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)]

Steps to solve a definite integral abf(x)dx with the help of basic integration rule is,

abkf(x)dx=kabf(x)dx

Where k is a constant.

Calculation:

Consider the function,

f(x)=x2+2

The intervals are [1,1] with n=4

Now divide the provided interval into 4 subparts as shown below,

Δx=1(1)4=24=12

Therefore, the 4 subintervals are, [1,12],[12,0],[0,12] and [12,1].

Now find the mid points of these intervals because each subinterval has a width of 12.

Therefore, the mid points of these interval are 34,14,14 and 34.

Here the mid points 34,14,14 and 34 lies in the middle of

[1,12],[12,0],[0,12] and [12,1] intervals respectively

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