Question:
If \( \int_{0}^{3} x\left(\sqrt[5]{9-x^{2}}\right) \, dx = k \cdot 3^{1/k} \), then \( k = \)
If \( \int_{0}^{3} x\left(\sqrt[5]{9-x^{2}}\right) \, dx = k \cdot 3^{1/k} \), then \( k = \)
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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 \( du = -2x dx \).
Updated On: Jun 8, 2026
- \( \frac{9}{5} \)
- \( \frac{5}{9} \)
- \( \frac{5}{12} \)
- \( \frac{12}{5} \)
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The Correct Option is C
Solution and Explanation
Concept:
We solve this definite integral using a direct linear substitution for the expression inside the fifth root radical.
Step 1: Performing the substitution and shifting integration limits.
Let \( u = 9 - x^2 \implies du = -2x \, dx \implies x \, dx = -\frac{du}{2} \). Now find the new integration limits:
• Lower limit: When \( x = 0 \implies u = 9 - 0 = 9 \)
• Upper limit: When \( x = 3 \implies u = 9 - 3^2 = 0 \)
Step 2: Evaluating the transformed integral.
\[ I = \int_{9}^{0} u^{1/5} \left(-\frac{du}{2}\right) = \frac{1}{2} \int_{0}^{9} u^{1/5} \, du \] Using the standard power rule for integration: \[ I = \frac{1}{2} \left[ \frac{u^{6/5}}{6/5} \right]_{0}^{9} = \frac{1}{2} \times \frac{5}{6} \left[ 9^{6/5} - 0 \right] = \frac{5}{12} \left(3^2\right)^{6/5} = \frac{5}{12} 3^{12/5} \]
Step 3: Matching with the given format to identify \( k \).
The problem statement writes the answer as \( k \cdot 3^{1/k} \). Comparing this structure with our evaluated result: \[ k = \frac{5}{12} \] Let us check the exponent match: \( 1/k = \frac{1}{5/12} = \frac{12}{5} \), which matches the power of 3 perfectly. Thus, \( k = \frac{5}{12} \) is correct.
Step 1: Performing the substitution and shifting integration limits.
Let \( u = 9 - x^2 \implies du = -2x \, dx \implies x \, dx = -\frac{du}{2} \). Now find the new integration limits:
• Lower limit: When \( x = 0 \implies u = 9 - 0 = 9 \)
• Upper limit: When \( x = 3 \implies u = 9 - 3^2 = 0 \)
Step 2: Evaluating the transformed integral.
\[ I = \int_{9}^{0} u^{1/5} \left(-\frac{du}{2}\right) = \frac{1}{2} \int_{0}^{9} u^{1/5} \, du \] Using the standard power rule for integration: \[ I = \frac{1}{2} \left[ \frac{u^{6/5}}{6/5} \right]_{0}^{9} = \frac{1}{2} \times \frac{5}{6} \left[ 9^{6/5} - 0 \right] = \frac{5}{12} \left(3^2\right)^{6/5} = \frac{5}{12} 3^{12/5} \]
Step 3: Matching with the given format to identify \( k \).
The problem statement writes the answer as \( k \cdot 3^{1/k} \). Comparing this structure with our evaluated result: \[ k = \frac{5}{12} \] Let us check the exponent match: \( 1/k = \frac{1}{5/12} = \frac{12}{5} \), which matches the power of 3 perfectly. Thus, \( k = \frac{5}{12} \) is correct.
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