Integral 1/x^(2/3)dx from -1 to 1: Solution

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In summary, the conversation discusses finding the integral of 1/x^(2/3) from -1 to 1. It is split into two integrals and the limits are taken as they approach 0. The correct answer is found to be 6, but Wolfram Alpha gives the wrong answer due to plugging in the antiderivative. The conversation also briefly touches on finding the integral of 2dx/(x^2-1) from -infinity to -2, with the conclusion being that there is no need to break it at -1.
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miglo
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Homework Statement


integral 1/x^(2/3)dx from -1 to 1


Homework Equations





The Attempt at a Solution


so i split it up into two integrals, one with limits going from -1 to b and the other with limits going from c to 1, and taking the limits as b and c go to 0
i know my antiderivative is 3x^(1/3) and i plugged in my limits of integration and then took the limit as both b and c went to 0 and i got an answer of 3-3(-1)^(1/3) i know this is wrong simply because the cube root of -1 will be give me a complex number but i tried it on wolfram alpha and it gave me the same answer. i looked at the back of my book and it says the answer is 6, i don't understand how the book got an answer of 6
 
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  • #2
Wolfram Alpha is wrong and so is plugging into the antiderivative (which it's likely WA did) for an improper integral. 1/x^(2/3) is actually an even function if you take it to be (1/x^(1/3))^2, isn't it? That should mean you can take the improper integral over (0,1] and double it. What's that integral? BTW (-1)^(1/3) is not necessarily a complex number.
 
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  • #3
ohh thanks! ill watch out for even and odd functions

one more quick question though
i have another improper integral, its 2dx/(x^2-1) from -infinity to -2
so i broke it up into two integrals, the first one has limits from a to -1 with the limit as a approaches -infinity and the second is from -1 to -2
am i doing this right?
 
  • #4
miglo said:
ohh thanks! ill watch out for even and odd functions

one more quick question though
i have another improper integral, its 2dx/(x^2-1) from -infinity to -2
so i broke it up into two integrals, the first one has limits from a to -1 with the limit as a approaches -infinity and the second is from -1 to -2
am i doing this right?

Why do you need the integral from -1 to -2? -1 isn't in (-infinity,-2). There's no need to break it there. Just integrate from a to -2.
 
  • #5
ohh alright
thanks again
 

1. What is the integral of 1/x2/3dx from -1 to 1?

The integral of 1/x2/3dx from -1 to 1 is equal to 0. This can be solved using the power rule of integration, which states that the integral of xndx is equal to xn+1/(n+1) + C. In this case, n = -1/3, so the integral is 3x-1/3. Evaluating this at the limits of -1 and 1 gives us 0.

2. How do you solve the integral of 1/x2/3dx from -1 to 1?

The integral of 1/x2/3dx from -1 to 1 can be solved using the power rule of integration, which states that the integral of xndx is equal to xn+1/(n+1) + C. In this case, n = -1/3, so the integral is 3x-1/3. Evaluating this at the limits of -1 and 1 gives us 0.

3. Is the integral of 1/x2/3dx from -1 to 1 convergent or divergent?

The integral of 1/x2/3dx from -1 to 1 is convergent. This can be determined by calculating the improper integral at each limit separately. At x = -1, the integral becomes 3/(-1)1/3 = -3, and at x = 1, the integral becomes 3/(1)1/3 = 3. The difference between these two values is 6, which is a finite value, indicating convergence.

4. Can the integral of 1/x2/3dx from -1 to 1 be solved without using the power rule?

Yes, the integral of 1/x2/3dx from -1 to 1 can also be solved using substitution. Let u = x1/3, then du = (1/3)x-2/3dx. Substituting these values into the integral, we get ∫1/(u3) * (3/u) du = 3∫u-4 du = -3/u3 + C = -3/x + C. Evaluating this at the limits of -1 and 1 gives us 0, confirming the same solution as using the power rule.

5. What is the graph of the integral of 1/x2/3dx from -1 to 1?

The graph of the integral of 1/x2/3dx from -1 to 1 is a horizontal line at y = 0. This is because the integral of 1/x2/3dx from -1 to 1 is equal to 0, as determined by evaluating the integral at the limits of -1 and 1. This can also be seen by looking at the graph of 1/x2/3, which has a vertical asymptote at x = 0, indicating that the integral from -1 to 1 would not have any area above or below the x-axis, resulting in a zero value.

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