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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

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BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Use rectangles to find the area between

y =   x 2 6 x +   8 and the x-axis from x =   0  to  x   =   2 . Divide the interval [0, 2] into n equal subintervals so that each subinterval has length 2/n.

To determine

To calculate: The area between the curve y=x26x+8 and the x-axis from x=0 to x=2 when the interval [0,2] is divided into n equal subintervals so that the length of each subinterval is 2n.

Explanation

Given Information:

The provided curve is y=x26x+8 from x=0 to x=2 and the interval [0,2] is divided into n equal subintervals so that the length of each subinterval is 2n.

Formula used:

The base of the rectangles to approximate the area is ban where the interval [a,b] on which function is defined is divided into n subintervals.

The height of the rectangles is the value of the function calculated at the right-hand end point of the interval containing the base.

The area of a rectangle is base×height.

The approximated area under the curve is sum of the areas of each rectangle.

The value of the sum k=1nk2=n(n+1)(2n+1)6.

The value of the sum k=1nk=n(n+1)2.

For functions f(x) and g(x),

limn[f(x)+g(x)]=limnf(x)+limng(x).

The value of 1n when n,

limn1n=0

Calculation:

Consider the provided curve y=x26x+8 from x=0 to x=2 and divide the interval [0,2] into n equal subintervals so that the length of each subinterval is 2n.

The base of the rectangles to approximate the area is ban where the interval [a,b] on which function is defined is divided into n subintervals.

Since, the function is defined from x=0 to x=2. So, a=0 and b=2.

Thus,

Base of each rectangle=20n=2n

Thus, the n subintervals, each of length 2n, are [0,2n], [2n,4n],,[2n2n,2].

Since, there are n subintervals, the number of rectangles is n.

Recall that the height of the rectangles is the value of the function calculated at the right-hand end point of the interval containing the base.

Since, the right-hand end point of the first interval [0,2n] is 2n.

Thus, the height of the first rectangle is,

y=x26x+8=(2n)26(2n)+8=4n212n+8

Recall that the area of a rectangle is base×height.

Thus, the area of the first rectangle is,

Area=base×height=2n×(4n212n+8)=2n(4n212n+8)

Do similar calculation to find the area of all n rectangles and record these values in a table.

Rectangle Base Right-endpoint Height Area=base×height
1 2n x1=2n y=(2n)26(2n)+8 2n((2n)26(2n)+8)
2 2n x2=4n y=(4n)26(4n)+8 2n((4n)26(4n)+8)
3 2n x3=6n y=(6n)26(6n)+8 2n((6n)26(6n)+8)
n 2n xn=2nn=2 y=(2nn)26(2nn)+8 2n((2nn)26(2nn)+8)

Recall that the approximated area under the curve is sum of the areas of each rectangle.

Thus, the area under the curve is

2n((2n)26(2n)+8)+2n((4n)26(4n)+8)+2n((6n)26(6n)+8)++2n((2nn)26(2nn)+8).

Simplify 2n((2n)26(2n)+8)+2n((4n)26(4n)+8)+2n((6n)26(6n)+8)++2n((2nn)26(2nn)+8)

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Chapter 13 Solutions

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