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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

How do your answers to Problems 27(a)-(e) compare with the corresponding calculations in the discussion (after Example 1) of the area under y =   2 x using right- hand endpoints?

To determine

The comparison in the calculations to compute the area under the curve of the function y=2x from x=0 to x=1 and n equal subintervals where the function is evaluated at left and right end points of each subinterval.

Explanation

Given Information:

The curve is y=2x from x=0 to x=1 and n subintervals and the function is evaluated at left end points of the subintervals.

Explanation:

First consider the measures using left end points.

Consider the curve y=2x from x=0 to x=1 and n subintervals where the function is evaluated at left end points of the subintervals.

Recall that the base of the rectangles to approximate the area is ban where the function is defined on [a,b].

Since, curve is defined from x=0 to x=1.

Thus,

base=10n=1n

Thus, the n subintervals, of length 1n are [0,1n], [1n,2n],.... [n1n,1].

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 left end point of the interval containing the base.

Since, the left end point of the first interval is 0.

Thus, height of the rectangle is

y=2(0)=0

Recall that area of a rectangle is base×height.

Thus, area is

base×height=1n×0=0

Do similar calculation and record these values in a table.

Rectangle Base Left endpoint Height  Area=base×height
1 1n x1=0 y=2(0)=0 1n×0=0
2 1n x2=1n y=2(1n)=2n 1n×2n=2n2
3 1n x3=2n y=2(2n)=4n 1n×4n=4n2
n 1n xn=n1n y=2(n1n)=2n2n 1n×2n2n=2n2n2

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

Thus, area under the curve is 0+2n2+4n2++2n2n2.

Simplify 0+2n2+4n2++2n2n2.

0+2n2+4n2++2n2n2=2n2+4n2++2n2n2=1n(2n+4n++2n2n)=1n×2n(1+2++(n1))=2n2n(n1)2

Further simplify.

2n2n(n1)2=1n(n1)1=(n1)n

Thus, a formula for the sum of the areas of the n rectangles is SL=(n1)n.

Consider the formula SL=(n1)n.

Substitute 10 for n in SL=(n1)n.

S(10)=(101)10=910=0.9

Thus, the value of SL(10) is 0.9.

Now, substitute 100 for n in SL=(n1)n.

S(100)=(1001)100=99100=0.99

Thus, the value of SL(100) is 0.99.

Substitute 1000 for n in SL=(n1)n.

S(1000)=(10001)1000=9991000=0.999

Thus, the value of SL(1000) is 0.999.

Consider the formula SL=(n1)n.

Recall that for functions f(x) and g(x), the value of the limit limn[f(x)+g(x)]=limnf(x)+limng(x).

Consider limnS.

Substitute SL=(n1)n.

limnSL=limn((n1)n)=limnnnlimn1n=limn1limn1n

Use limn1n=0 to simplify further.

limn1limn1n=10=1

Thus, the value of limnSL=1.

Hence, the value of limnSL is 1.

Now consider the measures using right end points.

Consider the curve y=2x from x=0 to x=1 and n subintervals where the function is evaluated at right end points of the subintervals.

Recall that the base of the rectangles to approximate the area is ban where the function is defined on [a,b].

Since, curve is defined from x=0 to x=1

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Chapter 13 Solutions

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