Question

integral 0 to 1 dx/x^3
is it the same thing as doing integral 0 to 1 dx/x^2 right?

Answer 1

what do you mean? Integrating x^3 over 0 to 1 is not the same as integrating x^2 over that same interval.

But I'm not sure I understand your question...

Answer 2

how can it be different?

Answer 3

the two functions are different, right? Why wouldn't the integral be different?

Am I confused about the question? I am still not sure I understand... don't want to make you confused too!! :)

Answer 4

let me just type the problem XD
integral 0 to 1 dx/x^3
its those integrals where limits r involved

Answer 5

Like this? \[\int\limits_{0}^{1} x ^{3} dx\] and \[\int\limits_{0}^{1}x ^{2}dx\]

Answer 6

or like this:\[\int\limits_{0}^{1}\frac{ dx }{ x ^{3} }\]

Answer 7

its like this
\[\int\limits_{0}^{1} \frac{ dx }{ x^3 } \]
and
\[\int\limits_{0}^{1} \frac{ dx }{ x^2 }\]

Answer 8

I think you can integrate both of these by treating them like x^(-3) and x^(-2), respectively. There might be an easier way or a formula but this should work.

Integrate: x^(-3) to get -(1/2)x^(-2)

Answer 9

\[\int\limits_{0}^{1}\frac{ dx }{ x ^{3} } = \int\limits_{0}^{1}x ^{-3}dx = -\frac{ 1 }{ 2 }x ^{-2}\]

Answer 10

how did u get the 1/2?

Answer 11

oh nvm

Answer 12

I'm struggling to recall... integration is the opposite of differentiation... so d/dx of (-1/2) (x^(-2)) becomes (-1/2)(-2)x^(-3) or just x^(-3)

Answer 13

As I said, there may be a "rule" for integrating an expression with a negative exponent, i.e. something with x raised to whatever in the denominator.

Answer 14

then after getting -1/2(-2)x^-3, then use the 0 to 1 limit to get my answer?
sorry for asking so much. just trying to get understanding >.<

Answer 15

no, I'm sorry for being confusing...

you need to integrate x^-3 I think that gives you the expression, -(1/2)x^-2

Then you evaluate that over the 0 to 1 limit.

But I am struggling to remember and also explain... sorry!

You might want to consider posting the question again without all this discussion and see if you get a better explanation :)

Answer 16

hahaha maybe i should just type the problem instead XD. thank u :3

Answer 17

I'm not sure you can calculate either integral over that 0 to 1 interval... both expressions have x in the denominator, and so when you evaluate at the lower limit x = 0, it puts zero in the denominator. That's not good :)