
#1
Jun608, 12:10 PM

P: 13

hi everyone
the integral is : [tex]\[ I = \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx} } } \][/tex] I'm not sure about the answer , but i think it'll be [tex]\[ \frac{{x^3 }}{3} \][/tex] am i right ?????? thanks 



#2
Jun608, 12:18 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

Go back and learn the basics again. Since there is an integral with respect to dx, the result cannot possibly be a function of x. The result here must be a number. Did you forget to do the final integral?




#3
Jun608, 02:03 PM

P: 443

As halls said, pay very close attention to what variable you are integrating with respect to. If you have a different variable within the integrand treat it as a constant both while integrating and evaluating.




#4
Jun608, 02:05 PM

P: 13

Triple Integral
I knew that i was wrong




#5
Jun608, 02:05 PM

P: 443

I am not so sure about that. I got a different answer. Perhaps you want to show your steps?




#6
Jun608, 02:14 PM

P: 13

Sorry , It'll be 1/12 (won't it ??)
[tex]\[ \begin{array}{l} \int\limits_0^1 {\int\limits_0^x {\int\limits_0^y {ydzdydx = \int\limits_0^1 {\int\limits_0^x {\left( {\int\limits_0^y {ydz} } \right)} } } } } dydx = \int\limits_0^1 {\int\limits_0^x {y^2 } dydx} \\ = \int\limits_0^1 {\left( {\int\limits_0^x {y^2 dy} } \right)} dx = \int\limits_0^1 {\frac{{x^3 }}{3}} dx = \left( {\frac{{x^4 }}{{12}}} \right)_0^1 = \frac{1}{{12}} \\ \end{array} \][/tex] Thanks 



#7
Jun608, 02:29 PM

P: 443

There you go.



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