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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,

$ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { i-1 +2n \over i-1+n } \Big... ...i-1+2n \over i-1+n } { {1 \over n} \over {1 \over n} } \Big({1 \over n}\Big) } $

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { {i \over n}-{1 \over n}+{2n \over n} \over {i \over n}-{1 \over n}+{n \over n} } \Big({1 \over n}\Big) } $

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { {i-1 \over n}+2 \over {i-1 \over n}+1 } \Big({1 \over n}\Big) } $ .

Choose the sampling point $ c_{i} $ to be

$ c_{i} = \displaystyle{ i-1 \over n } $

for $ i=1, 2, 3, ..., n $ . Then $ c_{i} $ represents the left-hand endpoints of $ n $ equal-sized subdivisions of the interval $ [0, 1] $ and

$ \Delta x_{i} = \displaystyle{ 1 \over n } $

for $ i=1, 2, 3, ..., n $ . Thus,

$ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { {i-1 \over n}+2 \over {i-1... ...e{ \lim_{n \to \infty} \sum_{i=1}^{n} { c_{i}+2 \over c_{i}+1 } \Delta x_{i} } $

(Let $ f(x) = \displaystyle{ x+2 \over x+1 } $ .)

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} }$

$ = \displaystyle{ \int^{1}_{0} f(x) \, dx } $

$ = \displaystyle{ \int^{1}_{0} { x+2 \over x+1 } \, dx } $ .

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

$ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( {12 \over n} + {8i \ov... ...o \infty} \sum_{i=1}^{n} \Big( 12 + {8i \over n} \Big) \Big({1 \over n}\Big) } $ .

Choose the sampling point $ c_{i} $ to be

$ c_{i} = \displaystyle{ i \over n } $

for $ i=1, 2, 3, ..., n $ . (Note that other choices for $ c_{i} $ also lead to correct answers. For example, $ c_{i} = \displaystyle{ 8i \over n } $ or $ c_{i} = 12 + \displaystyle{ 8i \over n } $ also works. Each choice determines a different interval and a different function !) Then $ c_{i} $ represents the right-hand endpoints of $ n $ equal-sized subdivisions of the interval $ [0, 1] $ and

$ \Delta x_{i} = \displaystyle{ 1 \over n } $

for $ i=1, 2, 3, ..., n $ . Thus,

$ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( 12 + {8i \over n} \Big... ...tyle{ \lim_{n \to \infty} \sum_{i=1}^{n} \Big( 12 + 8 c_{i} \Big) \Delta_{i} } $

(Let $ f(x) = \displaystyle{ 12 + 8x } $ .)

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} }$

$ = \displaystyle{ \int^{1}_{0} f(x) \, dx } $

$ = \displaystyle{ \int^{1}_{0} ( 12 + 8x ) \, dx } $ .

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

$ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { 9i+3n \over 3in+2n^2} } = ... ...lim_{n \to \infty} \sum_{i=1}^{n} { 9i+3n \over 3i+2n} \Big({1 \over n}\Big) } $

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { 9i+3n \over 3i+2n} { {1 \over n} \over {1 \over n} } \Big({1 \over n}\Big) } $

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { {9i \over n}+{3n \over n} \over {3i \over n}+{2n \over n} } \Big({1 \over n}\Big) } $

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { {9i \over n}+ 3 \over {3i \over n}+ 2 } \Big({1 \over n}\Big) } $ .

Choose the sampling point $ c_{i} $ to be

$ c_{i} = \displaystyle{ i \over n } $

for $ i=1, 2, 3, ..., n $ . (Note that other choices for $ c_{i} $ also lead to correct answers. For example, $ c_{i} = \displaystyle{ 3i \over n } $ , $ c_{i} = \displaystyle{ 9i \over n } $ , $ c_{i} = \displaystyle{ 3i \over n } + 2 $ or $ c_{i} = \displaystyle{ 9i \over n } + 3 $ also works. Each choice determines a different interval and a different function !) Then $ c_{i} $ represents the right-hand endpoints of $ n $ equal-sized subdivisions of the interval $ [0, 1] $ and

$ \Delta x_{i} = \displaystyle{ 1 \over n } $

for $ i=1, 2, 3, ..., n $ . Thus,

$ \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} { {9i \over n}+ 3 \over {3i ... ...m_{n \to \infty} \sum_{i=1}^{n} { 9c_{i} + 3 \over 3 c_{i}+ 2 } \Delta x_{i} } $

(Let $ f(x) = \displaystyle{ 9x+3 \over 3x+2 } $ .)

$ = \displaystyle{ \lim_{n \to \infty} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} }$

$ = \displaystyle{ \int^{1}_{0} f(x) \, dx } $

$ = \displaystyle{ \int^{1}_{0} { 9x+3 \over 3x+2 } \, dx } $ .

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SOLUTION 14 : Use the limit definition of definite integral to evaluate $ \displaystyle{ \int^{b}_{a} K \, dx } $ , where $ K $ is a constant. Use an arbitrary partition $ a=x_{0}, x_{1}, x_{2}, x_{3}, ... , x_{n-2}, x_{n-1}, x_{n}=b $ and arbitrary sampling numbers $ c_{i} $ for $ i=1, 2, 3, ..., n $ . Let

$ f(x) = K $

and recall that

$ \Delta x_{i} = x_{i} - x_{i-1} $

and the mesh of the partition is

$ mesh = \displaystyle{ \max_{1 \le i \le n} \{ x_{i} - x_{i-1} \}} $

for $ i=1, 2, 3, ..., n $ . Thus, the definite integral of $ f $ on the interval $ [a, b] $ is defined to be

$ \displaystyle{ \int^{b}_{a} f(x) \, dx} = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} } $

$ = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} K (x_{i} - x_{i-1}) } $

$ = \displaystyle{ \lim_{mesh \to 0} K \sum_{i=1}^{n} (x_{i} - x_{i-1}) } $

$ = \displaystyle{ \lim_{mesh \to 0} K \{ (x_{1} - x_{0})+(x_{2} - x_{1})+(x_{3} - x_{2})+ \cdots + (x_{n-1} - x_{n-2})+(x_{n} - x_{n-1}) \} } $

(This is a telescoping sum.)

$ = \displaystyle{ \lim_{mesh \to 0} K \{ x_{n} - x_{0} \} } $

$ = \displaystyle{ \lim_{mesh \to 0} K \{ b - a \} } $

$ = K \{ b - a \} $ .

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SOLUTION 15 : Use the limit definition of definite integral to evaluate $ \displaystyle{ \int^{b}_{a} {1 \over x^2} \, dx } $ . Use an arbitrary partition $ a=x_{0}, x_{1}, x_{2}, x_{3}, ... , x_{n-2}, x_{n-1}, x_{n}=b $ and the sampling number $ c_{i} = \sqrt{ x_{i-1}x_{i} } $ for $ i=1, 2, 3, ..., n $ . Begin by showing that $ x_{i-1} < c_{i} < x_{i} $ for $ i=1, 2, 3, ..., n $ . Assume that $ 0 < a < b $ . Note that

$ x_{i-1}^2 < x_{i-1} x_{i} $

since $ x_{i-1} < x_{i} $ , so that

$ \sqrt{ x_{i-1}^2 } < \sqrt{ x_{i-1} x_{i} } $

or

$ x_{i-1} < \sqrt{ x_{i-1} x_{i} } = c_{i} $ .

Similarly,

$ x_{i-1} x_{i} < x_{i}^2 $

since $ x_{i-1} < x_{i} $ , so that

$ \sqrt{ x_{i-1} x_{i} } < \sqrt{ x_{i}^2 } $

or

$ c_{i} = \sqrt{ x_{i-1} x_{i} } < x_{i} $ .

This proves that $ x_{i-1} < c_{i} < x_{i} $ for $ i=1, 2, 3, ..., n $ . Let

$ f(x) = \displaystyle{1 \over x^2} $

and recall that

$ \Delta x_{i} = x_{i} - x_{i-1} $

for $ i=1, 2, 3, ..., n $ and the mesh of the partition is

$ mesh = \displaystyle{ \max_{1 \le i \le n} \{ x_{i} - x_{i-1} \}} $ .

Thus, the definite integral of $ f $ on the interval $ [a, b] $ is defined to be

$ \displaystyle{ \int^{b}_{a} f(x) \, dx} = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} f(c_{i}) \Delta x_{i} } $

$ = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} f(\sqrt{ x_{i-1} x_{i} }) (x_{i} - x_{i-1}) } $

$ = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} {1 \over (\sqrt{ x_{i-1} x_{i} })^2 } (x_{i} - x_{i-1}) } $

$ = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} { 1 \over x_{i-1} x_{i} } (x_{i} - x_{i-1}) } $

$ = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} \Big( { x_{i} \over x_{i-1} x_{i} } - { x_{i-1} \over x_{i-1} x_{i} } \Big) } $

$ = \displaystyle{ \lim_{mesh \to 0} \sum_{i=1}^{n} \Big( { 1 \over x_{i-1} } - { 1 \over x_{i} } \Big) } $

$ = \displaystyle{ \lim_{mesh \to 0} \Big\{ \Big( { 1 \over x_{0} } - { 1 \over ... ...x_{n-1} } \Big) + \Big( { 1 \over x_{n-1} } - { 1 \over x_{n} } \Big) \Big\} } $

(This is a telescoping sum.)

$ = \displaystyle{ \lim_{mesh \to 0} \Big\{ { 1 \over x_{0} } - { 1 \over x_{n} } \Big\} } $

$ = \displaystyle{ \lim_{mesh \to 0} \Big\{ { 1 \over a } - { 1 \over b } \Big\} } $

$ = \displaystyle{ { 1 \over a } - { 1 \over b } } $ .

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Duane Kouba 2000-06-08