Question

Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) be a function defined by
$$ f(x) = \frac{x}{(1 + x^4)^{1/4}} $$
and \( g(x) = f(f(f(x))) \). Then
$$ 18 \int_{\sqrt[3]{\frac{8}{3}}}^{\sqrt[3]{\frac{4}{3}}} x^3 g(x) \, dx $$
equals: 

• A. 33
• B. 36
• C. 42
• D. 39

Answer 1

The Correct Option is D

First, let’s calculate \( g(x) = f(f(f(f(x)))) \).

Starting with \( f(x) \):

\[ f(x) = \frac{x}{(1 + x^4)^{1/4}}. \]

Now, let’s calculate \( f(f(x)) \):

\[ f(f(x)) = \frac{f(x)}{\left(1 + f(x)^4\right)^{1/4}} = \frac{\frac{x}{(1 + x^4)^{1/4}}}{\left(1 + \left(\frac{x}{(1 + x^4)^{1/4}}\right)^4\right)^{1/4}} = \frac{x}{(1 + 2x^4)^{1/4}}. \]

Next, we calculate \( f(f(f(x))) \):

\[ f(f(f(x))) = \frac{x}{(1 + 3x^4)^{1/4}}. \]

Finally, we calculate \( f(f(f(f(x)))) \):

\[ f(f(f(f(x)))) = \frac{x}{(1 + 4x^4)^{1/4}}. \]

Thus, \( g(x) = f(f(f(f(x)))) = \frac{x}{(1 + 4x^4)^{1/4}} \).

Now, we need to evaluate the integral:

\[ \int_{0}^{\sqrt[4]{18}} x^3 g(x) \, dx = \int_{0}^{\sqrt[4]{18}} \frac{x^4}{(1 + 4x^4)^{1/4}} \, dx. \]

Let \( u = 1 + 4x^4 \). Then \( du = 16x^3 dx \), which gives \( x^3 dx = \frac{du}{16} \).

When \( x = 0 \), \( u = 1 \); and when \( x = \sqrt[4]{18} \), \( u = 9 \).

Substituting into the integral, we have:

\[ \int_{0}^{\sqrt[4]{18}} \frac{x^4}{(1 + 4x^4)^{1/4}} \, dx = \int_{1}^{9} \frac{(u - 1)/4}{u^{1/4}} \cdot \frac{du}{16}. \]

Simplifying:

\[ = \frac{1}{64} \int_{1}^{9} \left( u^{3/4} - u^{-1/4} \right) du. \]

Now, integrate term by term:

\[ = \frac{1}{64} \left[ \frac{u^{7/4}}{7/4} - \frac{u^{3/4}}{3/4} \right]_{1}^{9}. \]

Simplifying further:

\[ = \frac{1}{64} \left( \frac{4}{7} \cdot 9^{7/4} - \frac{4}{3} \cdot 9^{3/4} - \left( \frac{4}{7} \cdot 1^{7/4} - \frac{4}{3} \cdot 1^{3/4} \right) \right). \]

Calculating each term:

\[ = \frac{1}{64} \left( \frac{4}{7} \cdot 81 - \frac{4}{3} \cdot 9 - \left( \frac{4}{7} - \frac{4}{3} \right) \right). \]

After evaluating all terms, we find:

\[ = 39. \]