Integrating Definite Integral of (x - x^2)*(2x^(-1/3)) from -8 to -1

[VikingStorm]

INT[-8 to -1] x - x^2 / 2*x^(1/3) dx

(x - x^2)*(2x^(-1/3))

Distributed:
2x^(2/3) - 2x^(5/3)

[6x^(5/3) / 5] -[ 3x^(8/3) / 4]

Plug in -1, and -8

-1.2 - -.75 = -.45

-38.4 - - 192 = 153.6

-.45 - 153.6 = -154.05

When I put this into the calculator straight to check my work, I get 57.112 as the answer. What did I do wrong here?

 

[ShawnD]

Originally posted by VikingStorm
x - x^2 / 2*x^(1/3) dx

(x - x^2)*(2x^(-1/3))

There is the error. In your original equation, the 2 is in the denominator. In your second equation, you brought the 2 to the top without putting a negative exponent on it; you only put a negative exponent on the x.

 

[Hurkyl]

$$\frac{x-x^2}{2x^{1/3}}<br /> = (x-x^2) (\frac{1}{2} x^{-1/3})$$

$$(-1)^{8/3} = 1$$

 

[VikingStorm]

Ah, I thought it was just a constant multiplier that stuck with the x.

Hmm...

So that would make it:
[ 2^-1 * x^(2/3)] - [2^-1 * x^(5/3)]

5x^(5/3)/6 - 3x^(8/3)/16

For -8, I get -26.7 - - 48 = 21.3

-1, -.833 - -.1875 = -.6455

Ay... I must have done something else wrong?

 

[Hurkyl]

$$(-1)^{8/3} = 1$$

(and this time you multiplied by five-thirds instead of divided)

 

[VikingStorm]

Urgh, hopefully I won't make these simple mistakes on the test.