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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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Approximating the Area of a Plane Region In Exercises 1-6, use the rectangles to approximate the area of the region. Compare your result with the exact area obtained using a definite integral. See Example 1.

f ( x ) = e x / 2 , [ 0 , 3 ]

Chapter 5.6, Problem 6E, Approximating the Area of a Plane Region In Exercises 1-6, use the rectangles to approximate the

To determine

To calculate: The area bounded by the function f(x)=ex2 between interval [0,3] by mid-point rule. Also compare this approximate area with exact area using definite integral.

Calculus: An Applied Approach (MindTap Course List), Chapter 5.6, Problem 6E

Explanation

Given Information:

The provided function is f(x)=ex2. Area of this function is calculated between the interval [0,3] as shown below:

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)=ex2. The graph of provided function is shown below:

In provided graph, interval [0,3] is divided into 6 rectangles.

Now, put n=6, a=0 and b=3 to find the width of each rectangle as;

Δx=306=0.5

Width Δx of each rectangle will be 0.5.

So, the interval [0,3] will be divided into 6 subintervals as;

[0,0.5][0.5,1.0],[1,1.5],[1.5,2],[2,2.5],[2.5,3].

Now, the value of midpoint of each rectangle is calculated as;

[x1,x2]=x1+x22 where x1,x2 are the upper value and lower value of the sub-interval

So, for first rectangle [0,0.5] mid-point is;

0+0.52=0.25

So, for second rectangle [0.5,1.0] mid-point is;

0.5+12=0.75

So, for third rectangle [1,1.5] mid-point is;

1+1.52=1.25

For fourth rectangle [1.5,2] mid-point is;

1.5+22=1.75

For fifth rectangle [2,2.5] mid-point is;

2+2.52=2.25

For last rectangle [2.5,3] mid-point is;

2.5+32=2.75

So, the mid points of all four rectangles are 0.25,0.75,1.25,1.75,2.25,2.75 respectively.

Now, obtain the value of provided function at each mid-point value calculated above

Put x=0

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