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

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

1 Answer
May 13, 2018

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

Explanation:

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#