heathhosty,

How to evaluate this integral?

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Try integration by parts. http://www.slideshare.net/empoweringminds/integration-by-parts-tutorial-example-calculus-2

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

 

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

So the integral is just exp(x^3)?

 

heathhosty,



Try integration by parts. http://www.slideshare.net/empoweringminds/integration-by-parts-tutorial-example-calculus-2

Ratch

Would integration by substitution be better?

 

So the integral is just exp(x^3)?

Not quite. Watch out for the negative signs. You said it right above - "substitution". In particular, use \(u=-x^3\) which implies \( du=-3x^2 dx\).

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

 

Would integration by substitution be better?

Much better as it is a simpler technique to apply than integration by parts.

 

Not quite. Watch out for the negative signs. You said it right above - "substitution". In particular, use \(u=-x^3\) which implies \( du=-3x^2 dx\).

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

Plus a constant...
\( \int 3x^2e^{-x^3}dx=-\int e^u du=-e^u + C =-e^{-x^3} + C \)

 

heathhosty,

Would integration by substitution be better?

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