Calc08_3day1 - 8.3 day one Improper Integrals Greg Kelly Hanford High School Richland Washington Until now we have been finding integrals of continuous

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
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 .
Example 1: 1 0 1 1 x dx x + - The function is undefined at 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.
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 - - -
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 2 1 1 lim sin 1 sin 0 1 b b b - - - - - - - 1 2 π = + 2 π 0 0 This integral converges because it approaches a solution.

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