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Let a Denote the Area Enclosed by the Graph (x)=1x+1( x ) = \frac { 1 } { x + 1 }

Question 23

Multiple Choice

Let A denote the area enclosed by the graph (x) =1x+1( x ) = \frac { 1 } { x + 1 } , the x-axis, and the lines x = 1 and x = 5. Graphing the region and using plane geometry, we can find that A is


A) limnk=0n(14kn+1) (4n) \lim _ { n \rightarrow \infty } \sum _ { k = 0 } ^ { n } \left( \frac { 1 } { \frac { 4 k } { n } + 1 } \right) \left( \frac { 4 } { n } \right)
B) limnk=0n1(14kn+1) (4n) \lim _ { n \rightarrow \infty } \sum _ { k = 0 } ^ { n - 1 } \left( \frac { 1 } { \frac { 4 k } { n } + 1 } \right) \left( \frac { 4 } { n } \right)
C) limnk=1n(15kn+1) (5n) \lim _ { n \rightarrow \infty } \sum _ { k = 1 } ^ { n } \left( \frac { 1 } { \frac { 5 k } { n } + 1 } \right) \left( \frac { 5 } { n } \right)
D) limnk=0n1(14k+2) (4n) \lim _ { n \rightarrow \infty } \sum _ { k = 0 } ^ { n - 1 } \left( \frac { 1 } { 4 k } + 2 \right) \left( \frac { 4 } { n } \right)
E) limnk=0n1(15kn+1) (4n) \lim _ { n \rightarrow \infty } \sum _ { k = 0 } ^ { n - 1 } \left( \frac { 1 } { \frac { 5 k } { n } + 1 } \right) \left( \frac { 4 } { n } \right)

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