Question

If \( \int_{0}^{3} x\left(\sqrt[5]{9-x^{2}}\right) \, dx = k \cdot 3^{1/k} \), then \( k = \)

• A. \( \frac{9}{5} \)
• B. \( \frac{5}{9} \)
• C. \( \frac{5}{12} \)
• D. \( \frac{12}{5} \)

Hint:

Swapping the lower and upper limits of a definite integral introduces a negative sign. This nicely cancels out the negative sign that comes from differentiating the substitution variable \( du = -2x dx \).

Answer 1

The Correct Option is C

Step 1: Choose the substitution.
Let $u=9-x^2$, so $du=-2x\,dx$, which gives $x\,dx=-\frac{du}{2}$.
Step 2: Change the limits.
When $x=0$, $u=9$. When $x=3$, $u=9-9=0$.
Step 3: Rewrite the integral.
\[ I=\int_9^0 u^{1/5}\left(-\frac{du}{2}\right)=\frac{1}{2}\int_0^9 u^{1/5}\,du \]
Step 4: Use the power rule.
\[ I=\frac{1}{2}\left[\frac{u^{6/5}}{6/5}\right]_0^9=\frac{1}{2}\cdot\frac{5}{6}\,9^{6/5} \]
Step 5: Simplify the power of 9.
Since $9=3^2$, $9^{6/5}=3^{12/5}$, so \[ I=\frac{5}{12}\,3^{12/5} \]
Step 6: Match the given form.
The form is $k\cdot3^{1/k}$. Comparing, $k=\frac{5}{12}$ and $\frac{1}{k}=\frac{12}{5}$, which matches the exponent. \[ \boxed{k=\tfrac{5}{12}} \]