How do you integrate int xroot3(4+x^2) from [0,2]?

2 Answers
Nov 11, 2016

The answer is =6-3/2^(1/3)

Explanation:

Use the substitution, u=4+x^2
so, du=2xdx
:. intx(4+x^2)^(1/3)dx=intu^(1/3)(du)/2
=1/2u^(4/3)/(4/3)=3/8(4+x^2)^(4/3)+C

So int_0^2x(4+x^2)^(1/3)dx=[3/8(4+x^2)^(4/3)] _0^2

=3/8*(4+4)^(4/3)-3/8*4^(4/3)
=3/8*16-3/8*4*4^(1/3)
=6-3/2^(1/3)

See answer below:
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