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What is the arc length of #f(x) = x^2e^(3-x^2) # on #x in [ 2,3] #?

1 Answer

Jun 3, 2018

#approx 1.8995#

Explanation:

We have
#f(x)=x^2*e^(3-x^2)#
so
#f'(x)=2xe^(3-x^2)+x^2e^(3-x^2)(-2x)#
so
#f'(x)=e^(3-x^2)*(2x-2x^3)#
We have to integrate

#int_2^3sqrt(1+(e^(3-x^2)*(2x-2x^3))^2)dx#
With a numerical method we get
#approx 1.8995#

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