How do you find the definite integral of #3x^2 sqrt(x^3+1)dx# from #[-1, 1]#?

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Answer 1 · 2016-06-24T09:11:50

# (4 sqrt(2))/3 #

there's a very nice pattern here - notice the red and green

#3x^\color{red}{2} sqrt(x^color{green}{3}+1)dx#

so a good trial solution is #a (x^3 + 1)^{3/2} # where a = const.

testing that:

# d/dx [a (x^3 + 1)^{3/2}] = (3/2) a (x^3 + 1)^{1/2} (3x^2)#

# = ((9a)/2) x^2 (x^3 + 1)^{1/2} #

#\implies (9a)/2 = 3, a = 2/3#

personally think that's better than going trough the motions of a substitution...

so we have:

#int_{-1}^{1} \ 3x^2 sqrt(x^3+1) \ dx = [2/3 (x^3 + 1)^{3/2}]_{-1}^{1} #

#= [2/3 (1 + 1)^{3/2}] - [2/3 (-1 + 1)^{3/2}]#

#= (2 sqrt(8))/3 = (4 sqrt(2))/3 #