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

Answer: (D)

39

Answer 2

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:

$= 39.$