# Definite Integ

1. Nov 23, 2003

### 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?

2. Nov 23, 2003

### ShawnD

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.

3. Nov 23, 2003

### Hurkyl

Staff Emeritus
$$\frac{x-x^2}{2x^{1/3}} = (x-x^2) (\frac{1}{2} x^{-1/3})$$

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

4. Nov 23, 2003

### 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?

5. Nov 23, 2003

### Hurkyl

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

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

6. Nov 23, 2003

### VikingStorm

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