Question

Let $ f(x) = \int x^3 \sqrt{3-x^2} dx $. If $ 5f(\sqrt{2}) = -4 $, then $ f(1) $ is equal to

• A. \( -\frac{2\sqrt{2}}{5} \)
• B. \( -\frac{8\sqrt{2}}{5} \)
• C. \( -\frac{4\sqrt{2}}{5} \)
• D. \( -\frac{6\sqrt{2}}{5} \)

Hint:

Use substitution to simplify the integral and then use the given condition to find the constant of integration.

Answer 1

The Correct Option is D

To solve for \( f(1) \), we first need to understand the given function \( f(x) = \int x^3 \sqrt{3-x^2} \, dx \).

Given: \( 5f(\sqrt{2}) = -4 \).

Let's try to find the integral of \( x^3 \sqrt{3-x^2} \). Begin by using a substitution method. Set \( u = 3 - x^2 \), then \( du = -2x \, dx \) or \( x \, dx = -\frac{1}{2} \, du \). Also, \( x^2 = 3 - u \), so \( x^3 = x(3-u) \).

Substituting this into the integral:

\[ f(x) = \int x^3 \sqrt{3-x^2} \, dx = \int x(3-u)\sqrt{u} \left(-\frac{1}{2} \right) \, du \]

Simplifying the integral further:

\[ = -\frac{1}{2} \int (3 - u)\sqrt{u} \, du = -\frac{1}{2} \left( \int 3\sqrt{u} \, du - \int u^{3/2} \, du \right) \]

Evaluating these integrals individually:

\[ \text{Integral of } 3\sqrt{u}: \int 3u^{1/2} \, du = 3 \cdot \frac{2}{3} u^{3/2} = 2u^{3/2} \]

\[ \text{Integral of } u^{3/2}: \int u^{3/2} \, du = \frac{2}{5} u^{5/2} \]

Putting these together:

\[ f(x) = -\frac{1}{2} \left( 2u^{3/2} - \frac{2}{5} u^{5/2} \right) = -\frac{1}{2} \left( 2(3-x^2)^{3/2} - \frac{2}{5} (3-x^2)^{5/2} \right) \]

Now, using the condition \( 5f(\sqrt{2}) = -4 \), we substitute \( x = \sqrt{2} \) in the expression:

\[ 5 \left( -\frac{1}{2} \left( 2(3-(\sqrt{2})^2)^{3/2} - \frac{2}{5} (3-(\sqrt{2})^2)^{5/2} \right) \right) = -4 \]

Simplifying within the conditions:

\[ u = 3 - (\sqrt{2})^2 = 1 \]

The expression then reduces to:

\[ 5 \left( -\frac{1}{2} \cdot \left( 2 \cdot 1^{3/2} - \frac{2}{5} \cdot 1^{5/2} \right) \right) = -4 \]

After simplifying and solving this, we find:

\[ 5 \left( -\frac{1}{2} \cdot \left(2 - \frac{2}{5} \right) \right) = -4 \]

This gives us the following calculation:

\[ f(\sqrt{2}) = \frac{4}{5} \]

Then find \( f(1) \).

Substituting \( x = 1 \):

\[ f(1) = -\frac{1}{2} \cdot (2 \cdot (3-1)^{3/2} - \frac{2}{5} \cdot (3-1)^{5/2}) \]

Evaluate:

\[ u = 2 \]

The expression becomes:

\[ -\frac{1}{2} \cdot (2 \cdot 2^{3/2} - \frac{2}{5} \cdot 2^{5/2}) = -\frac{6\sqrt{2}}{5} \]

The value of \( f(1) \) is \( -\frac{6\sqrt{2}}{5} \).

Answer 2

Let \( 3 - x^2 = t^2 \)
\( -2x dx = 2t dt \)
\( x dx = -t dt \)
\( f(x) = \int x^3 \sqrt{3-x^2} dx \)
\( f(x) = \int x^2 \sqrt{3-x^2} x dx \)
\( f(x) = \int (3-t^2) \cdot t \cdot (-t dt) + c \)
\( f(x) = \int (t^4 - 3t^2) dt + c \)
\( f(x) = \frac{t^5}{5} - t^3 + c \)
\( f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} + c \)
Given \( 5f(\sqrt{2}) = -4 \), we have
\( 5 \left( \frac{(3-2)^{5/2}}{5} - (3-2)^{3/2} + c \right) = -4 \)
\( 5 \left( \frac{1}{5} - 1 + c \right) = -4 \)
\( 1 - 5 + 5c = -4 \)
\( -4 + 5c = -4 \)
\( 5c = 0 \)
\( c = 0 \)
Therefore, \( f(x) = \frac{(3-x^2)^{5/2}}{5} - (3-x^2)^{3/2} \)
Now, we need to find \( f(1) \):
\( f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2} \)
\( f(1) = \frac{2^{5/2}}{5} - 2^{3/2} \)
\( f(1) = 2^{3/2} \left( \frac{2}{5} - 1 \right) \)
\( f(1) = 2^{3/2} \left( -\frac{3}{5} \right) \)
\( f(1) = -\frac{3}{5} \cdot 2 \sqrt{2} \)
\( f(1) = -\frac{6\sqrt{2}}{5} \)