To evaluate the integral

$I = \int_0^1 x^3 (1 - x)^{4/3} \, dx,$

we can solve it using the Beta function.

The Beta function $B(x, y)$ is defined as:

$B(x, y) = \int_0^1 t^{x-1} (1 - t)^{y-1} \, dt.$

We can express the given integral in a form that matches the Beta function.

Step 1: Comparison with the Beta function form

The integral is:

$I = \int_0^1 x^3 (1 - x)^{4/3} \, dx.$

First, rewrite it to match the form of the Beta function:

$I = \int_0^1 x^{3} (1 - x)^{4/3} \, dx = \int_0^1 x^{(4) - 1} (1 - x)^{(7/3) - 1} \, dx.$

Now we can recognize that the integral is in the Beta function form:

$I = B(4, \frac{7}{3}).$

Step 2: Using the Beta function identity

The Beta function has the following relationship with the Gamma function:

$B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)}.$

So, applying this identity:

$I = B(4, \frac{7}{3}) = \frac{\Gamma(4) \Gamma\left(\frac{7}{3}\right)}{\Gamma\left(4 + \frac{7}{3}\right)}.$

Step 3: Simplifying the Gamma functions

We know that $\Gamma(4) = 3! = 6$. Now, simplify the other Gamma functions:

$4 + \frac{7}{3} = \frac{12}{3} + \frac{7}{3} = \frac{19}{3}.$

Thus, the expression becomes:

$I = \frac{6 \Gamma\left(\frac{7}{3}\right)}{\Gamma\left(\frac{19}{3}\right)}.$

Final Answer

The integral evaluates to:

$I = \frac{6 \Gamma\left(\frac{7}{3}\right)}{\Gamma\left(\frac{19}{3}\right)}.$

If you need more details about the Gamma functions or further evaluation, feel free to ask!

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Related Questions:

1. How do you compute $\Gamma\left(\frac{7}{3}\right)$ numerically?
2. What is the relationship between the Gamma function and factorials?
3. Can we simplify the Beta function for other types of integrals?
4. How does the Beta function apply to integrals involving polynomial terms?
5. What are the properties of the Gamma function that help in evaluating integrals?

Tip:

The Beta function is a powerful tool for evaluating integrals involving powers of variables and their complements. Understanding its connection to the Gamma function can greatly simplify many complex integrals.