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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Evaluate lim n n 2 i = 1 n j = 1 n 2 1 n 2 + n i + j .

To determine

To evaluate: The value of limnn2i=1nj=1n21n2+ni+j.

Explanation

Formula used:

The Riemann sum can be expressed as, limm,ni=1mj=1nf(xi,yj)ΔA=Df(x,y)dA,

Here, ΔA=lb, where l,b is the length and breadth of the rectangle.

The sample points of the upper right corner of each square is denoted by (xi,yj).

The image value of the sample points under the function f is denoted by f(xi,yj).

The Riemann sum constants are denoted by m, n.

Calculation:

Rewrite the given summation as follows.

n2i=1nj=1n21n2+ni+j=i=1nj=1n2(1n2)1n2+ni+j=i=1nj=1n2(1n3)11nn2+ni+j=i=1nj=1n2(1n3)1n2n1+in+jn2=i=1nj=1n2(1n3)11+in+jn2

Since, the denominator root contains in and jn2, all the values of this are contain inside 0 and 1. Thus, this equation can also be expressed in terms of a Riemann sum for two variables in the region R={(x,y)|0x1,0y1} as 11+x+y.

Since, x is denoted by in, the rectangular region R is divided into n subintervals on the x-axis. Similarly, since, y is denoted by jn2, the rectangular region R is divided into n2 subintervals on the y-axis. Take the sample points anywhere from the sub rectangle. The area of each sub rectangle is given by,

Area=1n(1n2)=1n3

Thus when n the above equation becomes,

limnn2i=1nj=1n21n2+ni+j=limn,n2i=1nj=1n2(1n3)11+in+jn2=limn,n2i=1nj=1n2f(xij*,yij*)ΔA

Use the formula mentioned above to integrate it

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Chapter 15 Solutions

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