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EXAMPLE 2 For the region under ( f(x)=5 x^{2} ) on ( [0,4] ), show that the sum of the areas of the upper approximating rectangle approaches ( frac{320}{3} ), that is [ lim _{n ightarrow infty} R_{n}=frac{320}{3} ] SOLUTION ( R_{n} ) is the sum of the areas of the ( n ) rectangles in the figure. Each rectangle has width ( frac{4}{n} ) and the heights are the values of the function ( f(x)=5 x^{2} ) at the points ( frac{4}{n}, frac{8}{n}, frac{12}{n}, ldots, frac{4 n}{n} ); that is, the heights are ( 5left(frac{4}{n} ight)^{2}, 5left(frac{8}{n} ight)^{2}, 5left(frac{12}{n} ight)^{2}, ldots, 5left(frac{4 n}{n} ight)^{2} ). Thus, [ egin{aligned} R_{n} & =frac{4}{n} cdot 5left(frac{4}{n} ight)^{2}+frac{4}{n} cdot 5left(frac{8}{n} ight)^{2}+frac{4}{n} cdot 5left(frac{12}{n} ight)^{2}+ldots+frac{4}{n} cdot 5left(frac{4 n}{n} ight)^{2} \ & =frac{20}{n} cdot square cdotleft(1^{2}+2^{2}+3^{2}+ldots+n^{2} ight) \ & =square cdot frac{320}{n^{3}}left(1^{2}+2^{2}+3^{2}+ldots+n^{2} ight) . end{aligned} ] Here we need the formula for the sum of the squares of the first ( n ) positive integers: [ 1^{2}+2^{2}+3^{2}+ldots+n^{2}=frac{n(n+1)(2 n+1)}{6} . ] Perhaps you have seen this formula before. Putting this formula into our expression for ( R_{n} ), we get [ R_{n}=square cdot frac{n(n+1)(2 n+1)}{6}=square . ] Thus we have [ egin{aligned} lim _{n ightarrow infty} R_{n} & =lim _{n ightarrow infty} square \ & =lim _{n ightarrow infty} squareleft(frac{n+1}{n} ight)left(frac{2 n+1}{n} ight) \ & =lim _{n ightarrow infty} squareleft(1+frac{1}{n} ight)left(2+frac{1}{n} ight) \ & =square cdot 1 cdot 2=square . end{aligned} ]

          EXAMPLE 2 For the region under ( f(x)=5 x^{2} ) on ( [0,4] ), show that the sum of the areas of the upper approximating rectangle approaches ( frac{320}{3} ), that is
[
lim _{n 
ightarrow infty} R_{n}=frac{320}{3}
]
SOLUTION ( R_{n} ) is the sum of the areas of the ( n ) rectangles in the figure. Each rectangle has width ( frac{4}{n} ) and the heights are the values of the function ( f(x)=5 x^{2} ) at the points ( frac{4}{n}, frac{8}{n}, frac{12}{n}, ldots, frac{4 n}{n} ); that is, the heights are ( 5left(frac{4}{n}
ight)^{2}, 5left(frac{8}{n}
ight)^{2}, 5left(frac{12}{n}
ight)^{2}, ldots, 5left(frac{4 n}{n}
ight)^{2} ). Thus,
[
egin{aligned}
R_{n} & =frac{4}{n} cdot 5left(frac{4}{n}
ight)^{2}+frac{4}{n} cdot 5left(frac{8}{n}
ight)^{2}+frac{4}{n} cdot 5left(frac{12}{n}
ight)^{2}+ldots+frac{4}{n} cdot 5left(frac{4 n}{n}
ight)^{2} \
& =frac{20}{n} cdot square cdotleft(1^{2}+2^{2}+3^{2}+ldots+n^{2}
ight) \
& =square cdot frac{320}{n^{3}}left(1^{2}+2^{2}+3^{2}+ldots+n^{2}
ight) .
end{aligned}
]
Here we need the formula for the sum of the squares of the first ( n ) positive integers:
[
1^{2}+2^{2}+3^{2}+ldots+n^{2}=frac{n(n+1)(2 n+1)}{6} .
]
Perhaps you have seen this formula before. Putting this formula into our expression for ( R_{n} ), we get
[
R_{n}=square cdot frac{n(n+1)(2 n+1)}{6}=square .
]
Thus we have
[
egin{aligned}
lim _{n 
ightarrow infty} R_{n} & =lim _{n 
ightarrow infty} square \
& =lim _{n 
ightarrow infty} squareleft(frac{n+1}{n}
ight)left(frac{2 n+1}{n}
ight) \
& =lim _{n 
ightarrow infty} squareleft(1+frac{1}{n}
ight)left(2+frac{1}{n}
ight) \
& =square cdot 1 cdot 2=square .
end{aligned}
]

$EXAMPLE 2 For the region under ( f(x)=5 x^2 ) on ( [0,4] ), show that the sum of the areas of the upper approximating rectangle approaches ( frac3203 ), that is [ lim n ightarrow infty Rn=frac3203 ] SOLUTION ( Rn ) is the sum of the areas of the ( n ) rectangles in the figure. Each rectangle has width ( frac4n ) and the heights are the values of the function ( f(x)=5 x^2 ) at the points ( frac4n, frac8n, frac12n, ldots, frac4 nn ); that is, the heights are ( 5left(frac4n ight)^2, 5left(frac8n ight)^2, 5left(frac12n ight)^2, ldots, 5left(frac4 nn ight)^2 ). Thus, [ eginaligned Rn =frac4n cdot 5left(frac4n ight)^2+frac4n cdot 5left(frac8n ight)^2+frac4n cdot 5left(frac12n ight)^2+ldots+frac4n cdot 5left(frac4 nn ight)^2 =frac20n cdot square cdotleft(1^2+2^2+3^2+ldots+n^2 ight) =square cdot frac320n^3left(1^2+2^2+3^2+ldots+n^2 ight) . endaligned ] Here we need the formula for the sum of the squares of the first ( n ) positive integers: [ 1^2+2^2+3^2+ldots+n^2=fracn(n+1)(2 n+1)6 . ] Perhaps you have seen this formula before. Putting this formula into our expression for ( Rn ), we get [ Rn=square cdot fracn(n+1)(2 n+1)6=square . ] Thus we have [ eginaligned lim n ightarrow infty Rn =lim n ightarrow infty square =lim n ightarrow infty squareleft(fracn+1n ight)left(frac2 n+1n ight) =lim n ightarrow infty squareleft(1+frac1n ight)left(2+frac1n ight) =square cdot 1 cdot 2=square . endaligned ]$

Added by Cassie T.

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