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

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In summary, the conversation discusses an error in the second equation where the 2 is brought to the top without putting a negative exponent on it, leading to incorrect calculations. The correct equation is [(1/2)*x^(2/3)] - [(1/2)*x^(5/3)], and when plugged in for -1 and -8, the correct answers are -0.6455 and 21.3 respectively. The conversation also mentions the importance of correctly handling negative exponents, such as (-1)^{8/3} = 1.
  • #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?
 
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  • #2
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.
 
  • #3
[tex]
\frac{x-x^2}{2x^{1/3}}
= (x-x^2) (\frac{1}{2} x^{-1/3})
[/tex]

[tex]
(-1)^{8/3} = 1
[/tex]
 
  • #4
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
[tex](-1)^{8/3} = 1[/tex]

(and this time you multiplied by five-thirds instead of divided)
 
  • #6
Urgh, hopefully I won't make these simple mistakes on the test.
 

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

What does it mean to "integrate" a definite integral?

Integrating a definite integral means finding the area under a curve within a specific range of values. In this case, we are finding the area under the curve of the function (x - x^2)*(2x^(-1/3)) from -8 to -1.

How do you solve a definite integral?

To solve a definite integral, we use the Fundamental Theorem of Calculus, which states that the definite integral of a function can be evaluated by finding the antiderivative of the function and plugging in the upper and lower limits of integration. In this case, we would find the antiderivative of (x - x^2)*(2x^(-1/3)) and evaluate it at -8 and -1, then find the difference between the two values.

What is the purpose of integrating a definite integral?

The purpose of integrating a definite integral is to find the area under a curve, which has many real-world applications in fields such as physics, engineering, and economics. It also helps us to find the total value of a function within a specific range.

What are some techniques for solving a definite integral?

Some common techniques for solving a definite integral include substitution, integration by parts, and partial fraction decomposition. It is also helpful to have a good understanding of basic integration rules and properties.

Can a definite integral have a negative value?

Yes, a definite integral can have a negative value if the function being integrated has negative values within the specified range. This indicates that the area under the curve is below the x-axis. However, we can always take the absolute value of the definite integral to find the total value of the function within the given range.

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