### SOLUTIONS TO THE LIMIT DEFINITION OF A DEFINITE INTEGRAL

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

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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,

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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

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Similarly,

since , so that

or

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This proves that for . Let

and recall that

for and the mesh of the partition is

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

(This is a telescoping sum.)

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