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:

\[ = 39. \]

Answer 2

The problem asks to evaluate the definite integral \( 18 \int_{0}^{\sqrt{2}\sqrt[4]{5}} x^2 g(x) \, dx \), where the function \( f(x) \) is defined as \( f(x) = \frac{x}{(1 + x^4)^{1/4}} \) and \( g(x) \) is the composition of \( f \) with itself four times, i.e., \( g(x) = f(f(f(f(x)))) \).

Concept Used:

To solve this problem, we first need to find a simplified expression for the composite function \( g(x) \). We can find a pattern by analyzing the structure of \( f(x) \). Let's manipulate the expression for \( f(x) \):

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

Raising both sides to the power of 4, we get:

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

This reveals a recursive relationship for repeated compositions. If we let \( f_n(x) \) denote the n-th composition of \( f \), we have \( \frac{1}{(f_n(x))^4} = \frac{1}{(f_{n-1}(x))^4} + 1 \). This implies \( \frac{1}{(f_n(x))^4} = \frac{1}{x^4} + n \). After finding \(g(x)\), the integral is evaluated using the method of substitution.

Step-by-Step Solution:

Step 1: Determine the explicit form of \( g(x) \).

The function \( g(x) \) is the result of applying \( f \) four times, so \( g(x) = f_4(x) \). Using the relationship derived above, with \( n=4 \):

\[ \frac{1}{(g(x))^4} = \frac{1}{x^4} + 4 \]

Now, we solve for \( g(x) \):

\[ \frac{1}{(g(x))^4} = \frac{1 + 4x^4}{x^4} \] \[ (g(x))^4 = \frac{x^4}{1 + 4x^4} \]

Taking the fourth root of both sides gives the expression for \( g(x) \):

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

Step 2: Set up the integral with the expression for \( g(x) \).

The integral to be evaluated is \( I = 18 \int_{0}^{\sqrt{2}\sqrt[4]{5}} x^2 g(x) \, dx \). Substituting the expression for \( g(x) \) we found:

\[ I = 18 \int_{0}^{\sqrt{2}\sqrt[4]{5}} x^2 \left( \frac{x}{(1 + 4x^4)^{1/4}} \right) \, dx \] \[ I = 18 \int_{0}^{\sqrt{2}\sqrt[4]{5}} \frac{x^3}{(1 + 4x^4)^{1/4}} \, dx \]

Step 3: Perform a substitution to evaluate the integral.

Let \( u = 1 + 4x^4 \). The differential is \( du = 16x^3 \, dx \), which implies \( x^3 \, dx = \frac{du}{16} \).

Next, we change the limits of integration from \( x \) to \( u \):

• When \( x = 0 \), \( u = 1 + 4(0)^4 = 1 \).
• When \( x = \sqrt{2}\sqrt[4]{5} \), we first compute \( x^4 \): \( x^4 = (\sqrt{2}\sqrt[4]{5})^4 = (\sqrt{2})^4 (\sqrt[4]{5})^4 = 4 \cdot 5 = 20 \). Thus, \( u = 1 + 4(20) = 81 \).

Substituting \( u \) and \( du \) into the integral:

\[ I = 18 \int_{1}^{81} \frac{1}{u^{1/4}} \left( \frac{du}{16} \right) = \frac{18}{16} \int_{1}^{81} u^{-1/4} \, du = \frac{9}{8} \int_{1}^{81} u^{-1/4} \, du \]

Step 4: Compute the definite integral.

First, find the antiderivative of \( u^{-1/4} \):

\[ \int u^{-1/4} \, du = \frac{u^{-1/4 + 1}}{-1/4 + 1} = \frac{u^{3/4}}{3/4} = \frac{4}{3} u^{3/4} \]

Now, apply the limits of integration:

\[ I = \frac{9}{8} \left[ \frac{4}{3} u^{3/4} \right]_{1}^{81} = \frac{9}{8} \cdot \frac{4}{3} \left[ u^{3/4} \right]_{1}^{81} = \frac{3}{2} \left[ (81)^{3/4} - (1)^{3/4} \right] \]

Final Computation & Result:

We evaluate the expression at the limits:

\[ (81)^{3/4} = (3^4)^{3/4} = 3^3 = 27 \] \[ (1)^{3/4} = 1 \]

Substituting these values back into the expression for \( I \):

\[ I = \frac{3}{2} (27 - 1) = \frac{3}{2} (26) = 3 \times 13 = 39 \]

The value of the integral is 39.