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What is #int_(0)^(1) (x^2)*e^(-x^2) dx #?

Answer 1

Charles Arocha

#= -1/(2e) + (sqrt pi)/4 erf (1)#

It's this!!

#int_(0)^(1) (x^2)*e^(-x^2) dx#

#= int_(0)^(1) -1/2x ( - 2 xe^(-x^2)) dx#

#= int_(0)^(1) -1/2x ( e^(-x^2))^prime dx#

Which is Integration by Parts!

#= [-1/2x e^(-x^2)]\_(0)^(1) - int_(0)^(1) (-1/2x)^prime e^(-x^2)dx#

#= -1/(2e) + 1/2 color(blue)( int_(0)^(1) e^(-x^2)dx) qquad triangle#

For that blue bit, I would call on this definition of the error function:

# erf (x) =2/(sqrt (pi )) int _{0}^{x} e^{-t^{2}}\ dt#

We can pattern match this into:

#color(blue)( int_(0)^(1) e^(-x^2)dx) = (sqrt pi)/2 erf (1)#

And so #triangle # becomes

#= -1/(2e) + (sqrt pi)/4 erf (1)#

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Answer 2

Emily Ammerman

The definite integral of ( x^2 \cdot e^{-x^2} ) from 0 to 1 is approximately 0.385872.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.