How do I integrate e^(y^3) in a definite integral without using erf(x)?

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SUMMARY

The discussion focuses on integrating the function \(\int^3_0 \int^1_{\sqrt{\frac{x}{3}}} e^{y^3} \, dy \, dx\) without using the error function, erf(x). The user initially struggled with the inner integral \(\int^1_{\sqrt{\frac{x}{3}}} e^{y^3} \, dy\) and sought alternatives. A successful approach was suggested by changing the order of integration to \(\int_{y=0}^1 \int_{x=0}^{3y^2} e^{y^3} \, dx \, dy\), which simplified the problem significantly.

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The function i need to integrate is

\displaystyle\int^3_0 \int^1_{\sqrt{\frac{x}{3}}} e^y^3,dydx

however my problem is integrating:
\int^1_{\sqrt{\frac{x}{3}}} e^y^3 dy

I have looked on the forum and everyone is mentioning erf(x) which i don't understand yet. Do i need this for this definite integral? If so where can i learn about it and how to use it?
 
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Try changing the order of integration.

For this integral, x ranges from 0 to 3 and, for each x, y from \sqrt{x/3}.

When x= 3, y= \sqrt{x/3}= 1 and when x= 0, y= \sqrt{0/3}= 0 so y ranges from 0 to 1. For each y, since y= \sqrt{x/3} leads to y^2= x/3 or x= 3y^2, x ranges from 0 to 3y^2. Try
\int_{y= 0}^1 \int_{x= 0}^{3y^2} e^{y^3} dx dy
 
that actually worked out quite nicely. Thank you.
 

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