Definite Integ (1 Viewer)

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VikingStorm

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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

Science Advisor
638
1
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

Staff Emeritus
Science Advisor
Gold Member
14,829
14
[tex]
\frac{x-x^2}{2x^{1/3}}
= (x-x^2) (\frac{1}{2} x^{-1/3})
[/tex]

[tex]
(-1)^{8/3} = 1
[/tex]
 
V

VikingStorm

Guest
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

Staff Emeritus
Science Advisor
Gold Member
14,829
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[tex](-1)^{8/3} = 1[/tex]

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

VikingStorm

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

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