Calc08_3day1 - 8.3 day one Improper Integrals Greg Kelly...

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8.3 day one Improper Integrals Greg Kelly, Hanford High School, Richland, Washington
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Until now we have been finding integrals of continuous functions over closed intervals. Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals .
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Example 1: 1 0 1 1 x dx x + - Since x = 1 is an asymptote, the function has no maximum. Can we find the area under an infinitely high curve? We could define this integral as: 0 1 1 lim 1 b b x dx x - + - (left hand limit) We must approach the limit from inside the interval.
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0 1 1 lim 1 b b x dx x - + - 1 1 1 1 x x x dx x + + + - Rationalize the numerator . 2 1+x 1 dx x - 2 2 1 x 1 1 dx dx x x + - - 2 1 u x = - 2 du x dx = - 1 2 du x dx - = 1 1 2 1 sin 2 x u du - - -
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2 2 1 x 1 1 dx dx x x + - - 2 1 u x = - 2 du x dx = - 1 2 du x dx - = 1 1 2 1 sin 2 x u du - - - 1 1 2 sin x u - - 1 2 1 0 lim sin 1 b b x x - - - - ( 29 ( 29 1 lim 1 1 b - - - - 1 2 π = + 2 0 0 This integral converges because it approaches a solution.
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Example 2: 1 0 dx x 1 0 lim ln b b x + 0 lim ln1 ln b b + - 0 1 lim ln b b + = ∞ This integral diverges . (right hand limit) We approach the limit from inside the interval. 1 0 1 lim b b dx x +
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Example 3: ( 29 3 2 0 3 1 dx x - The function approaches when
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Calc08_3day1 - 8.3 day one Improper Integrals Greg Kelly...

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