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FP1 Calculus URGENT HELP please!

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Q3 is from FP1 Calculus...have no idea how to do it since I got two 'problem values' here :frown:

Need help please!!!
Reply 1
I've only just finished learning FP1, and this is how I would go about it :smile: Replace the infinity with 'a', then integrate as normal. As 'a' tends towards infinity, the integral also tends towards infinity! As I said I'm also just a student so let me know if anyone disagrees :smile:
Original post by Oliviazh
image.jpeg
Q3 is from FP1 Calculus...have no idea how to do it since I got two 'problem values' here :frown:

Need help please!!!


Integrate as normal. When it comes to substituting limits, you cannot substitute \infty because it is not a number. Instead, you assign a different variable, such as aa, and take the limit of the integral as aa\rightarrow \infty
Reply 3
Original post by RDKGames
Integrate as normal. When it comes to substituting limits, you cannot substitute \infty because it is not a number. Instead, you assign a different variable, such as aa, and take the limit of the integral as aa\rightarrow \infty


But how about 0? Cuz my teacher also said that we need to replace 0 with a different variable.... that's why I'm confused cuz it then will be two variables😂
Or is it in this case we don't need to consider 0?
Original post by Oliviazh
But how about 0? Cuz my teacher also said that we need to replace 0 with a different variable.... that's why I'm confused cuz it then will be two variables😂
Or is it in this case we don't need to consider 0?


Well technically speaking the function 1x\frac{1}{\sqrt{x}} is undefined at x=0 so you would need to replace it as well, but personally I would skip this because the integrated function will be defined at x=0.

If it was 0x3/2\displaystyle \int_0^{\infty} x^{-3/2} or something then you would indeed need to replace the 0.
Reply 5
Original post by RDKGames
Well technically speaking the function 1x\frac{1}{\sqrt{x}} is undefined at x=0 so you would need to replace it as well, but personally I would skip this because the integrated function will be defined at x=0.

If it was 0x3/2\displaystyle \int_0^{\infty} x^{-3/2} or something then you would indeed need to replace the 0.


Thanks a lot!:smile:
Reply 6
Original post by spoiler-s
I've only just finished learning FP1, and this is how I would go about it :smile: Replace the infinity with 'a', then integrate as normal. As 'a' tends towards infinity, the integral also tends towards infinity! As I said I'm also just a student so let me know if anyone disagrees :smile:


Thank you!:smile:

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