anonymous
  • anonymous
whats the anti derivative of 1/(x^3)?? i have an answer but shows its wrong//// help?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
my answer is x^-2/2
anonymous
  • anonymous
When doing integrals with a power on the bottom, its often helpful to rewrite it like so anti derivative of x^-3 So lets use the general rule for integration, -3 + 1 = -2 So the anti derivative would be x^(-2) / -2 or in proper form 1 / 2*x^2
anonymous
  • anonymous
I'm sorry, I mistyped that. The answer is -2 / x^2

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amistre64
  • amistre64
x^-3 -> x^-4/4: 1/(4x^4)
anonymous
  • anonymous
Wait I keep messing that up.... 1 / -2*x^2
amistre64
  • amistre64
hah!!... i missed it too lol
amistre64
  • amistre64
-1/2x^2 is right ;)
anonymous
  • anonymous
thanks you haha ... so how did you get that>?
amistre64
  • amistre64
its the reverse of the power rule of a derivative.
anonymous
  • anonymous
got it .. thanks
amistre64
  • amistre64
\[D(X^n) = n * X^{n-1} \rightarrow D(x^5) = 5x^4\]
dumbcow
  • dumbcow
you do it correctly you just forgot to include the negative when you divided by the exponent
amistre64
  • amistre64
\[\int\limits_{} X^n dx \rightarrow \frac{X^{n+1}}{n+1}\]
anonymous
  • anonymous
ahh i got it ,.,.,.
anonymous
  • anonymous
thanks

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