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

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 \( f(1) \):
\( f(1) = \frac{(3-1)^{5/2}}{5} - (3-1)^{3/2} \)
\( f(1) = \frac{2^{5/2}}{5} - 2^{3/2} \)
\( f(1) = 2^{3/2} \left( \frac{2}{5} - 1 \right) \)
\( f(1) = 2^{3/2} \left( -\frac{3}{5} \right) \)
\( f(1) = -\frac{3}{5} \cdot 2 \sqrt{2} \)
\( f(1) = -\frac{6\sqrt{2}}{5} \)