Question

_n=1^_0^1x^3(1-x^2)^ndx=

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. 1
• D. 2

Hint:

Summing an infinite geometric series inside an integral first is almost always much faster than integrating each individual term and summing the results.

Answer 1

The Correct Option is A

Step 1: Concept
By switching the order of summation and integration (justified by uniform convergence on the interval), the expression can be rewritten as $\int_{0}^{1} x^{3} \left( \sum_{n=1}^{\infty} (1-x^{2})^{n} \right) dx$.

Step 2: Meaning
The inner summation $\sum_{n=1}^{\infty} (1-x^{2})^{n}$ is an infinite geometric series with first term $a = (1-x^2)$ and common ratio $r = (1-x^2)$. For $x \in (0,1)$, $|r| < 1$, so the sum is $\frac{a}{1-r} = \frac{1-x^2}{1-(1-x^2)} = \frac{1-x^2}{x^2}$.

Step 3: Analysis
Substitute this simplified series sum back into our definite integral: $\int_{0}^{1} x^3 \left(\frac{1-x^2}{x^2}\right) dx = \int_{0}^{1} x(1-x^2) dx = \int_{0}^{1} (x - x^3) dx$. Now compute the simple power rule integral: $\left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = \left(\frac{1}{2} - \frac{1}{4}\right) - (0 - 0) = \frac{1}{4}$.

Step 4: Conclusion
The resulting value is $\frac{1}{4}$, which perfectly matches option (A).

Final Answer: (A)