Question

$\int_0^1 x^{12} (1-x)^{10} dx$ is equal to

• A. $\frac{1}{132}$
• B. $\frac{1}{156}$
• C. $\frac{1}{182}$
• D. None of the above

Hint:

$\int_0^1 x^{m-1}(1-x)^{n-1} dx = B(m,n) = \frac{(m-1)!(n-1)!}{(m+n-1)!}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Beta function: $\int_0^1 x^{m-1} (1-x)^{n-1} dx = B(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$.
Step 2: Detailed Explanation:
Here $m-1 = 12 \Rightarrow m = 13$, $n-1 = 10 \Rightarrow n = 11$.
So $\int_0^1 x^{12} (1-x)^{10} dx = B(13,11) = \frac{12! \cdot 10!}{23!} = \frac{1}{23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13} \cdot (12! \cdot 10!)$
Better: $B(13,11) = \frac{12! \cdot 10!}{23!} = \frac{12! \cdot 10!}{23 \cdot 22 \cdots 13 \cdot 12!} = \frac{10!}{23 \cdot 22 \cdots 13}$
$\frac{10!}{23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13} = \frac{3628800}{23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}$
Calculate denominator: $23 \cdot 22 = 506$, $\cdot 21 = 10626$, $\cdot 20 = 212520$, $\cdot 19 = 4037880$, $\cdot 18 = 72681840$, $\cdot 17 = 1235591280$, $\cdot 16 = 19769460480$, $\cdot 15 = 296541907200$, $\cdot 14 = 4151586700800$, $\cdot 13 = 53970627110400$
So $\frac{3628800}{53970627110400} = \frac{1}{14873536}$? That's not matching. Actually, $B(13,11) = \frac{12!10!}{23!} = \frac{1}{23 \binom{22}{10}} = \frac{1}{23 \cdot 646646} = \frac{1}{14872858}$ approx. But options are fractions like $1/132$, etc. So maybe the integral is $\int_0^1 x^{12}(1-x)^{10} dx = \frac{12!10!}{23!} = \frac{1}{23 \cdot \binom{22}{10}}$. $\binom{22}{10} = 646646$,
so $23 \times 646646 = 14872858$, so $1/14872858$, which is none of the options. Perhaps the integral is $\int_0^1 x^{11}(1-x)^{10} dx$? Then $m=12, n=11$, $B(12,11) = \frac{11!10!}{22!} = \frac{1}{22 \cdot \binom{21}{10}} = \frac{1}{22 \cdot 352716} = \frac{1}{7759752}$.
Still not matching. Let's compute $B(13,11)$ numerically:
$\frac{12!10!}{23!} = \frac{479001600 \cdot 3628800}{25852016738884976640000}$? Actually, $\int_0^1 x^{12}(1-x)^{10} dx = \frac{12!10!}{23!} = \frac{1}{23 \cdot \binom{22}{10}} = \frac{1}{23 \cdot 646646} = \frac{1}{14872858}$.
None match. But option (C) is $1/182 \approx 0.0055$, while our value is $6.72 \times 10^{-8}$.
So clearly different. Perhaps the integral is $\int_0^1 x^{12}(1-x)^{10} dx$ is a beta function and the answer is a small fraction.
Given options, $1/182$ is the largest among them, but still much larger. Actually $1/182 \approx 0.0055$
while the integral should be very small because $x^{12}$ is very small near 0. Wait, at $x=1$, $(1-x)^{10}=0$, so the function peaks somewhere in between. The maximum of $x^{12}(1-x)^{10}$ occurs at $x=12/22=6/11 \approx 0.545$, value $\approx (0.545)^{12}(0.455)^{10} \approx 1.2 \times 10^{-4}$, so area is about $10^{-4}$, so $1/182 \approx 5.5 \times 10^{-3}$ is too large.
So probably none. But given the options, the correct beta function value for $(12,10)$ is $\frac{1}{13 \cdot \binom{22}{11}}$? Actually $B(p,q) = \frac{(p-1)!(q-1)!}{(p+q-1)!}$.
For $p=13, q=11$, it's $\frac{12!10!}{23!} = \frac{1}{23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13} \cdot (12! \cdot 10!)$ This simplifies to $\frac{1}{23 \cdot 22 \cdots 13} = \frac{10!}{23 \cdot 22 \cdots 13} = \frac{3628800}{23 \cdot 22 \cdots 13}$. Compute denominator: $23 \cdot 22 = 506$, $\cdot 21 = 10626$, $\cdot 20 = 212520$, $\cdot 19 = 4037880$, $\cdot 18 = 72681840$, $\cdot 17 = 1235591280$, $\cdot 16 = 19769460480$, $\cdot 15 = 296541907200$, $\cdot 14 = 4151586700800$, $\cdot 13 = 53970627110400$. So $3628800 / 53970627110400 = 1/14873536 \approx 6.72 \times 10^{-8}$.
So (C) is $1/182 \approx 0.0055$,
not matching. So answer is (D) None of the above.
Step 3: Final Answer:
None of the above.