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Can someone help me understand this?
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@jabez177 @Jaynator495 @AloneS @Awolflover1
When one or both limits are infinity, it is of type 1
when the function has discontinuity at either of the end points, it is of type 2 improper integral
1st problem . 1/(x-7) has discontinuity at x=7
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and it is the lower limit of integration
so can you answer the 1st question now?
Yea B
correct
i guess you can answer all the 4
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What about #4?
cot 0 in undefined
infinite discontinuity. so??
Oh B. But uh how is cot 0 undefined? D:
cot 0= cos 0 / sin 0 sin 0=0 , so denominator is 0
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actually to compute the integral compute
by changing the 0 to epsilon and then after evaluating the integral, put epsilon=0
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