   Chapter 5.6, Problem 1CP ### 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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# Checkpoint 1 Worked-out solution available at LarsonAppliedCalculus.comUse four rectangles to approximate the area of the region bounded by the graph of f ( x ) = x 2 +   1 ,  the  x − axis , x   = 0 ,  and  x   =   2 .

To determine

To calculate: The area bounded by the function f(x)=x2+1, x axis over the interval [0,2] by mid-point rule by the use of four rectangles.

Explanation

Given Information:

The provided function is f(x)=x2+1 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: First 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: 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)dxban[f(x1)+f(x2)+f(x3)+....+f(xn)].

Calculation:

Consider the provided function f(x)=x2+1.

Now, put 4, a=0 and b=2 to find the width of subinterval as;

Δx=2(0)4=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 sub-interval

So, for subinterval [0,12] mid-point is;

[0+12]2=0.25

Similarly, for subinterval [12,1] mid-point is;

[12+1]2=34=

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