# How to evaluate this integral?

Discussion in 'Math' started by heathhosty, Aug 9, 2009.

Aug 8, 2009
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Mar 20, 2007
1,068
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3. ### steveb Senior Member

Jul 3, 2008
2,433
469
Notice that 3x^2 is the derivative of x^3. That should give you a clue to the easy way to solve this one.

4. ### heathhosty Thread Starter New Member

Aug 8, 2009
4
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So the integral is just exp(x^3)?

Aug 8, 2009
4
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6. ### steveb Senior Member

Jul 3, 2008
2,433
469
Not quite. Watch out for the negative signs. You said it right above - "substitution". In particular, use $u=-x^3$ which implies $du=-3x^2 dx$.

$\int 3x^2e^{-x^3}dx=-\int e^u du=-e^u=-e^{-x^3}$

7. ### Mark44 Well-Known Member

Nov 26, 2007
626
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Much better as it is a simpler technique to apply than integration by parts.

8. ### Mark44 Well-Known Member

Nov 26, 2007
626
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Plus a constant...
$\int 3x^2e^{-x^3}dx=-\int e^u du=-e^u + C =-e^{-x^3} + C$

9. ### Ratch New Member

Mar 20, 2007
1,068
4
heathhosty,

Yes, indeed. I did not look at the problem close enough, and gave you bad advice. Substitution is the way to go on this one.

Ratch