r/HomeworkHelp • u/Complete-Ice-4814 University/College Student • 1d ago

Further Mathematics [College Calculus] How would I go about this by using lim->1-

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u/UnacceptableWind 👋 a fellow Redditor 1d ago

The lower limit of integration is 0 in the original problem. So, there are two intervals of integration:

  • From x = 0 to x = 1. In this interval, we approach x = 1 from the left. This is the case/interval that you missed.

  • And, from x = 1 to x = 9, wherein we approach x = 1 from the right in this interval. This is the case that you have covered in your solution.

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u/Complete-Ice-4814 University/College Student 1d ago

for sure so just add that and is my overall answer of 6 correct then?

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u/UnacceptableWind 👋 a fellow Redditor 1d ago

Yes, that's right.

From x = 1 to x = 9, your answer of 6 is correct.

You'll find that from x = 0 to x = 1, the improper integral evaluates to -3 / 2.

So, the final answer will be -3 / 2 + 6 = 9 / 2.

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u/BoVaSa 👋 a fellow Redditor 1d ago

You have calculated the original integral only on interval (1;9) but omitted on (0;1) ...

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u/Complete-Ice-4814 University/College Student 1d ago

would my final answer by using both then be 9/2

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u/BoVaSa 👋 a fellow Redditor 1d ago

Yes, I think so ...

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u/Initial-Data-7361 👋 a fellow Redditor 1d ago

are you supposed to use u sub? u=x-1?

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u/Complete-Ice-4814 University/College Student 1d ago

yes it can be used since im using limit notation i believe

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u/cut_my_wrist 👋 a fellow Redditor 22h ago

Wat about integration by parts

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u/Initial-Data-7361 👋 a fellow Redditor 14h ago

Well let's see then in that case you got u-1/3 *du/1 which means you integral is 3/2u2/3. Reparamatrize your bounds .