* SOLUTION 11 :* First algebraically manipulate the expression in order to make a choice for the sampling points. Thus,

.

Choose the sampling point to be

for . Then represents the left-hand endpoints of equal-sized subdivisions of the interval and

for . Thus,

(Let .)

.

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* SOLUTION 12 :* First algebraically manipulate the expression in order to make a choice for the sampling points. Thus,

.

Choose the sampling point to be

for . (Note that other choices for also lead to correct answers. For example, or also works. Each choice determines a different interval and a different function !) Then represents the right-hand endpoints of equal-sized subdivisions of the interval and

for . Thus,

(Let .)

.

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* SOLUTION 13 :* First algebraically manipulate the expression in order to make a choice for the sampling points. Thus,

.

Choose the sampling point to be

for . (Note that other choices for also lead to correct answers. For example, , , or also works. Each choice determines a different interval and a different function !) Then represents the right-hand endpoints of equal-sized subdivisions of the interval and

for . Thus,

(Let .)

.

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* SOLUTION 14 :* Use the limit definition of definite integral to evaluate
, where is a constant. Use an arbitrary partition
and arbitrary sampling numbers for
. Let

and recall that

and the mesh of the partition is

for . Thus, the definite integral of on the interval is defined to be

(This is a telescoping sum.)

.

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* SOLUTION 15 :* Use the limit definition of definite integral to evaluate
. Use an arbitrary partition
and the sampling number
for
. Begin by showing that
for
. Assume that . Note that

since , so that

or

.

Similarly,

since , so that

or

.

This proves that for . Let

and recall that

for and the mesh of the partition is

.

Thus, the definite integral of on the interval is defined to be

(This is a telescoping sum.)

.

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