# Definite integral: why is this wrong?

• tahayassen
In summary, according to Wolfram Alpha, the correct answer for the integral from 0 to 1 of -1 over the cube root of the absolute value of x minus 1 is -3/4. However, there are three valid answers for this integral, and Wolfram Alpha may have chosen the wrong branch for its calculation.
tahayassen
According to wolfram alpha, the correct answer is -3/4.

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Never mind. It is right.

Wolfram isn't getting 3/4, it's getting (-3/2)(-1)^(2/3) which it is computing to be approximately equal to 0.75-1.29904i

The real question of the day is why (-1)^(2/3) doesn't evaluate to 1.

ArcanaNoir said:
The real question of the day is why (-1)^(2/3) doesn't evaluate to 1.
There are three valid answers. Wolfram is calculating this one:
$$(-1)^{2/3} = (e^{i\pi})^{2/3} = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -1/2 + i\sqrt{3}/2$$
These two also work:
$$(-1)^{2/3} = (e^{-i\pi})^{2/3} = e^{-i2\pi/3} = \cos(-2\pi/3) + i\sin(-2\pi/3) = -1/2 - i\sqrt{3}/2$$
$$(-1)^{2/3} = (e^{i3\pi})^{2/3} = e^{i2\pi} = 1$$

jbunniii said:
There are three valid answers. Wolfram is calculating this one:
$$(-1)^{2/3} = (e^{i\pi})^{2/3} = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -1/2 + i\sqrt{3}/2$$
These two also work:
$$(-1)^{2/3} = (e^{-i\pi})^{2/3} = e^{-i2\pi/3} = \cos(-2\pi/3) + i\sin(-2\pi/3) = -1/2 - i\sqrt{3}/2$$
$$(-1)^{2/3} = (e^{i3\pi})^{2/3} = e^{i2\pi} = 1$$

There are other valid answers as well. If we put ## \sqrt[3]{x-1} = -\sqrt[3]{1-x} \; (0 \leq x \leq 1)## we have
$$\text{answer } = -\int_0^1 \frac{1}{\sqrt[3]{1-x}}\, dx = -3/2.$$

Mind has now been totally blown.

tahayassen said:
Mind has now been totally blown.

Oh, I wouldn't blow that hard. If you are doing this as a real integral then you want to pick the branch of cube root which is always real. Your original answer was fine. Wolfram just picked the wrong branch. You could also with equal justification claim that the integral from 0 to 1 of x^(1/3) has three possible answers. Or even more if you get more creative. But that's certainly not the usual answer. Wolfram just misinterpreted the question.

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## 1. Why is the definite integral wrong?

The definite integral is not inherently wrong. It is a mathematical concept used to find the area under a curve or the accumulation of a function over a given interval. However, there can be errors in the calculation or interpretation of a definite integral that can make it incorrect.

## 2. How do I know if a definite integral is wrong?

If you have a specific definite integral in question, you can double-check your work and calculations to see if there were any errors. Additionally, you can use the Fundamental Theorem of Calculus to evaluate the integral and compare it to your answer. If they do not match, there may be a mistake in your work.

## 3. What are common mistakes when working with definite integrals?

Some common mistakes when working with definite integrals include incorrect limits of integration, incorrect use of integration rules, miscalculations, and misinterpretation of the meaning of the integral. It is essential to follow the correct steps and pay attention to detail to avoid these errors.

## 4. Can a definite integral be negative?

Yes, a definite integral can be negative. This means that the area under the curve is below the x-axis, and it has a negative value. It is essential to pay attention to the orientation of the curve and the limits of integration to determine if the integral will be positive or negative.

## 5. Are there any real-world applications of definite integrals?

Definite integrals have various real-world applications, such as calculating the total distance traveled by an object with changing velocity, finding the total amount of fluid flowing through a pipe, and determining the total amount of work done by a force over a distance. They are also used in economics, physics, and engineering to solve various problems.

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