Computing with rational exponets

[Jimmy84]

1. The problem statement, all variables and given/known data
Solve \int_-1^1 (x^{4/3} + 4 x^{1/3}) dx

Im having difficulties in algebra when solving this problem.


2. Relevant equations



3. The attempt at a solution

\int_-1^1 (x^4/3 + 4 x^1/3) dx = 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 thats not right either.



Thanks in advance

[Jimmy84]

Sorry I meant the integral from -1 to 1 of (x^4/3 + 4 x^1/3)

[Mark44]

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.

[Jimmy84]

The problem is already solved in my book so I already gave the answer here. I just dont understand how to compute some algebra.

I dont 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 thats 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 im looking for, Im not sure what was done to get those results.

[Mentallic]

\frac{3}{7}\left(1\right)^{7/3}=\frac{3}{7}

since

1^{7/3}=1 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:

\frac{1^7}{3}=\frac{1}{3}

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.

[Jimmy84]

thanks, im sorry, when trying it with my calculator I got a different result. I used the key ^

[HallsofIvy]

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).