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: Swap the sum and integral.
The expression is $\displaystyle\sum_{n=1}^{\infty}\int_0^1 x^3(1-x^2)^n\,dx=\int_0^1 x^3\left(\sum_{n=1}^{\infty}(1-x^2)^n\right)dx$.
Step 2: Sum the geometric series inside.
With first term $r=1-x^2$ and ratio $r$ (and $|r|<1$ for $0<x<1$), \[ \sum_{n=1}^{\infty}(1-x^2)^n=\frac{1-x^2}{1-(1-x^2)}=\frac{1-x^2}{x^2}. \]
Step 3: Put it back in.
\[ \int_0^1 x^3\cdot\frac{1-x^2}{x^2}\,dx=\int_0^1 x(1-x^2)\,dx. \]
Step 4: Expand.
\[ \int_0^1 (x-x^3)\,dx. \]
Step 5: Integrate.
\[ \left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1=\frac{1}{2}-\frac{1}{4}. \]
Step 6: Simplify.
\[ \frac{1}{2}-\frac{1}{4}=\frac{1}{4}. \] \[ \boxed{\dfrac{1}{4}} \]