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CalculusCalculus: An Applied Approach (MindTap Course List)10th Edition

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Chapter 5.6, Problem 10E
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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 7-14, use the Midpoint Rule with n = 5 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 ) = 9 − x 2    [ 1 , 3 ]

To determine

To calculate: The area bounded by the function f(x)=9−x2 and x axis over the interval [1,3]. by mid-point rule. Also graph the area of the respective region.

Explanation

Given Information:

The provided function is f(x)=9−x2 area of this function is calculated on the interval [1,3] and the provided interval will be divided into 5 sub interval as n=5.

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: First divide the provided interval of the function into n sub intervals by the use of formula;

Δx=b−an where [a,b] are the intervals and n is the subinterval value. and Δx is the width.

Step2: Use above 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≈b−an[f(x1)+f(x2)+f(x3)+....+f(xn)].

Calculation:

Consider the provided function f(x)=9−x2.

Now, put n=5, a=1 and b=3 to find the width of subinterval as;

Δx=3−15=25

Width Δx of the provided interval will be 25.

So, the interval [1,3] with width 25 will be divided into 5 subintervals as;

[1,75],[75,95],[95,115],[115,135],[135,155].

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 sub-intervals

So, for subinterval [1,75] mid-point is;

1+752=1210

Similarly, for subinterval [75,95] mid-point is;

75+952=1610

For subinterval [95,115] mid-point is;

95+1152=2010

for subinterval [115,135] mid-point is;

115+1352=2410

For subinterval [135,155] mid-point is;

135+1552=2810

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