Computing with rational exponets

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Homework Help Overview

The discussion revolves around evaluating the integral \(\int_{-1}^{1} (x^{4/3} + 4 x^{1/3}) dx\). Participants express challenges with algebraic computations related to the integral and its antiderivative.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the computation of specific terms in the integral, particularly the evaluation of \(3/7 (1)^{7/3}\) and \(4 (3/4) (1)^{4/3}\). There are questions about the correct interpretation of exponents and the use of notation in both LaTeX and calculators.

Discussion Status

Some participants provide clarifications regarding the evaluation of the terms and the notation used. There is an ongoing exploration of the algebraic steps involved, with no explicit consensus reached on the final outcome.

Contextual Notes

Participants note that the problem may have been misclassified in the forum, suggesting it belongs in a more advanced category than initially posted. There is also mention of discrepancies between personal calculations and those found in a textbook.

Jimmy84
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Homework Statement


Solve [tex]\int_-1^1 (x^{4/3} + 4 x^{1/3}) dx[/tex]

Im having difficulties in algebra when solving this problem.


Homework Equations





The Attempt at a Solution



[tex]\int_-1^1 (x^4/3 + 4 x^1/3) dx[/tex] = 3/7 (1)^7/3 + 4 3/4 (1)^4/3 - 3/7 (-1)^7/3 + 4 3/4 (-1)^4/3

= 3/7 + 3 - (-3/7 +3) = 6/7




Im having difficulties computing 3/7 (1)^7/3 and 4 3/4 (1)^4/3

since 1^ 7/3 = 1/3 and 3/7 1/3 = 1/ 7 and tahts not the answer on the book

and (1)^4/3 = 1/3 and that's not right either.



Thanks in advance
 
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Sorry I meant the integral from -1 to 1 of (x^4/3 + 4 x^1/3)
 
You don't show the antiderivative of x4/3 + 4x1/3, which is an intermediate step for what you're doing. What did you get for that?

With some effort I could do this, but you're the one who should be doing this, not me.

Also, this problem should be in Calculus and Beyond, not Precalculus.
 
The problem is already solved in my book so I already gave the answer here. I just don't understand how to compute some algebra.

I don't know what is the result of 3/7 (1)^7/3 and of 4 3/4 (1)^4/3

since 1^ 7/3 = 1/3 and 3/7 1/3 = 1/ 7 and that's a different answer from the result on my book.

Which says that 3/7 (1)^7/3 = 3/7

and that 4 3/4 (1)^4/3 = 3


Thats all I am looking for, I am not sure what was done to get those results.
 
[tex]\frac{3}{7}\left(1\right)^{7/3}=\frac{3}{7}[/tex]

since

[tex]1^{7/3}=1[/tex] and to right this in text, you would have to say 1^(7/3) and not (1)^7/3

but if the question were shown differently such as how you expressed it (1)^7/3, this is read as:

[tex]\frac{1^7}{3}=\frac{1}{3}[/tex]

Now, can you also figure out why the second one is wrong? It should be written as 4(3/4).1^(4/3)
note: the dot before the 1 just means multiplication.
 
thanks, I am sorry, when trying it with my calculator I got a different result. I used the key ^
 
You probably made the same error on your calculator that you did in LaTex here!

In Latex, to get the entire fraction in the exponent use { } around it. Otherwise Latex interprets "x^1/3" as "(x^1)3". You need "x^{1/3}" (not just parentheses because LaTex treats parentheses as just another symbol).

Similarly, on your calculator, if you enter "x^1/3" your calculator will give you (x^1)/3 or just x divided by 3. Here you do use parentheses: x^(1/3).
 

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