- #1
FancyNut
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I stopped at the last step while calculating this improper integral:
integral of x^3\ ( x^4 - 3)^1\2 with limits from 1 to infinity...
that's x cubed over the square root of x raised to 4 minus 3...
after replacing infinity with b and taking the limits it seems that I have to take limits of two expressions one that goes to infinity which is ( (b^4 - 3)^1\2) / 8 but the other is undefined since I have a negative 2 inside a square root. The second expression is the square root of negative two all over 8.
Thanks for any help XD
integral of x^3\ ( x^4 - 3)^1\2 with limits from 1 to infinity...
that's x cubed over the square root of x raised to 4 minus 3...
after replacing infinity with b and taking the limits it seems that I have to take limits of two expressions one that goes to infinity which is ( (b^4 - 3)^1\2) / 8 but the other is undefined since I have a negative 2 inside a square root. The second expression is the square root of negative two all over 8.
Thanks for any help XD