Let $\beta(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0$. If $\int_0^1\left(1-x^{10}\right)^{20} d x=a \times \beta(b, c)$, then $100(a+b+x)$ equals
Select the correct option:
A
1021
B
1120
C
2012
D
2120
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & I=\int_0^1 1 \cdot\left(1-x^{10}\right)^{20} d x \\ & x^{10}=t \\ & x=t^{1 / 10} \\ & d x=\frac{1}{10}(t)^{-9 / 10} d t \\ & I=\int_0^1(1-t)^{20} \frac{1}{10}(t)^{-9 / 10} d t \\ & I=\frac{1}{10} \int_0^1 t^{-9 / 10}(1-t)^{20} d t \\ & a=\frac{1}{10}\end{aligned}$
$b=\frac{1}{10} c=21$
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