Question

Only one of the functions \(\sqrt{x} \sqrt[3]{1-x}\) and \(\sqrt{1-x} \sqrt[3]{1-x}\) has an elementary antiderivative. Find the function.

Short answer

Answer: The function \(g(x) = \sqrt{1-x} \sqrt[3]{1-x}\) has an elementary antiderivative given by \(G(x) = -\frac{6}{11}(1-x)^{\frac{11}{6}} + C\).

Step-by-step solution

Simplify the function

In this step, we simplify the function: \(f(x) = x^{\frac{1}{2}}(1-x)^{\frac{1}{3}}\)

Attempt substitution

We try to apply the substitution method to find the elementary antiderivative. Let's substitute \(u = 1-x\). Hence, \(-du = dx\). The function becomes: \(f(u) = (1-u)^{\frac{1}{2}}u^{\frac{1}{3}}\) Since this substitution only created another product of two functions raised to fractional exponents, it does not lead to an elementary antiderivative for the first function. Now let's analyze the second function: Function: \(g(x) = \sqrt{1-x} \sqrt[3]{1-x}\)

Simplify the function

In this step, we simplify the function: \(g(x) = (1-x)^{\frac{1}{2}}(1-x)^{\frac{1}{3}}\)

Rewrite and simplify the function

We rewrite the function by combining the terms: \(g(x) = (1-x)^{\frac{1}{2} + \frac{1}{3}}\) Now simplify the exponent: \(g(x) = (1-x)^{\frac{5}{6}}\)

Apply substitution for antiderivative

Now we try substitution for the second function. Let's substitute \(u = 1-x\). Consequently, \(-du = dx\). The function becomes: \(g(u) = u^{\frac{5}{6}}\) The antiderivative is now much easier to find: \(G(u) = \int u^{\frac{5}{6}}(-du) = -\int u^{\frac{5}{6}}du\) Now we can find the antiderivative using the power rule for integration: \(G(u) = -\frac{6}{11}u^{\frac{11}{6}} + C\) Now substitute back in terms of x: \(G(x) = -\frac{6}{11}(1-x)^{\frac{11}{6}} + C\) We can see that the second function, \(g(x) = \sqrt{1-x} \sqrt[3]{1-x}\), has an elementary antiderivative. Hence, it is the function that has an elementary antiderivative it was asked for.