How can I calculate #int_-1^0(x^2*sqrt(1+x^3))dx# ?

Question

How can I calculate #int_-1^0(x^2*sqrt(1+x^3))dx#

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Answer 1 · 2018-05-13T23:06:37

#int_(-1)^0 x^2sqrt(1+x^3) = 2/9#

Note that:

#d/dx (1+x^3) = 3x^2#

so:

#int_(-1)^0 x^2sqrt(1+x^3) = 1/3 int_(-1)^0 sqrt(1+x^3) d(1+x^3)#

#int_(-1)^0 x^2sqrt(1+x^3) = 1/3 [(1+x^3)^(3/2)/(3/2)]_(-1)^0#

#int_(-1)^0 x^2sqrt(1+x^3) = 2/9 [(1+0)^(3/2)-(1+(-1)^3)^(3/2)] =2/9#