integrate t*(t^2 - 1)^(1/3) dt over (0,3)
I substitute u = t^2 - 1
du = 2t dt
which leads to
integrate (1/2) u^(1/3) du over(-1,8)
= (3/8) * u^(4/3) over (-1/8)
= 3/8 * [8^(4/3) - (-1)^(4/3)]
I would guess that (-1)^(4/3) is +1, since the cube root of -1 is -1 and taking that to the fourth power is +1, so that would lead to a final answer of 45/8.
The book says 51/8 which would make sense if (-1)^(4/3) is -1.
Computer calculators give me complex (real + imaginary) answers.
Which is correct?
If you put (-1)^(4/3) in a calculator you get = -0.5 - 0.866025404 i (I used google calculator)
If you put in "cube root (-1)"
you get -1. -1 * 4 = -1
Using,
(-1)^(4/3) = -1
3/8 (8)^(4/3) - 3/8 (-1)^(4/3)
3/8 (16) + 3/8
6 + 3/8 = 51/8
So, the book answer of 51/8 is correct?
Also, depending on which integration calculator you use online, most give the value of this integral as,
6.1875 + 0.32476 i
Ask your teacher about the different calculator values for (-1)^(4/3).