Question

Integrals?
http://prntscr.com/3s6ow2

Answer 1

\[\Large\rm \int\limits_0^{\pi} x^{-1/3}~dx\]Understand that step? :)
From here we can apply the power rule for integration.
Need help with that step?

Answer 2

ok

Answer 3

Before proceeding with the integral, notice that \(\dfrac{1}{x^{1/3}}\) is undefined at \(x=0\), so you must treat the integral as a limit:
\[\large \int_0^\pi x^{-1/3}~dx=\lim_{c\to0^+}\int_c^\pi x^{-1/3}~dx\]

Answer 4

can you help with this (x^2/3)/-2+1

Answer 5

\[\Large\rm \int\limits\limits_0^{\pi} x^{-1/3}~dx =\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}|_0^{\pi}\]Hmm I'm not sure where your -2 is coming from.
It looks like we end up with:\[\Large\rm \frac{x^{2/3}}{2/3}=\frac{3}{2}x^{2/3}\]

Answer 6

why are we not using -2

Answer 7

When we integrate, we `add 1` to the exponent, yes?
Not subtract.

Is that how you were getting a -2 or ... something ? :U
I can't seem to figure out what you did >.<

Answer 8

iwas told the formula (x^n+1)/-2+1

Answer 9

No silly, the formula is:\[\Large\rm \int\limits x^{n}~dx=\frac{x^{n+1}}{n+1}\]

Answer 10

You increase the exponent by 1,
then you divide by the new exponent.

Answer 11

ok

Answer 12

\[\int\limits_{-1}^{1}\frac{ dx }{ x^{-1} }\]

Answer 13

@ash2326

Answer 14

\[\int_{-1}^{1} x dx\]
Can you try now?

Answer 15

can you tell me is its final anwser zero

Answer 16

yes. it's zero. As x is an odd function, integral from -1 to 1 will result zero

Answer 17

can you show me the solution so i call compare my own solution please

Answer 18

@ash2326