Key Concepts and Formulas

• Beta Function Definition: The Beta function is defined as \beta(\mathrm{m}, \mathrm{n})=\int_\limits0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, for $\mathrm{m}, \mathrm{n}>0$.
• Definite Integral Substitution: When performing a substitution in a definite integral, the limits of integration must be transformed according to the substitution rule.
• Differentiation of Power Functions: The derivative of $x^k$ is $kx^{k-1}$.

Step-by-Step Solution

The problem asks us to evaluate the definite integral \int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x and express it in the form $\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c})$, then find the value of $100(\mathrm{a}+\mathrm{b}+\mathrm{c})$.

Step 1: Analyze the Integral and Identify the Need for Substitution The given integral is I = \int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x. The Beta function definition is \beta(\mathrm{m}, \mathrm{n})=\int_\limits0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x. We observe that the term $(1-x^{10})^{20}$ in our integral does not directly match the $(1-x)^{\mathrm{n}-1}$ form of the Beta function. To make it resemble the Beta function, we need to transform the term $x^{10}$ into a simpler variable.

Step 2: Perform the Substitution Let $t = x^{10}$. This substitution is chosen because it directly addresses the non-linear term $x^{10}$ within the parenthesis, aiming to convert $(1-x^{10})$ into $(1-t)$.

Now, we need to find the differential $dx$ in terms of $dt$ and change the limits of integration.

1. Express $x$ in terms of $t$: From $t = x^{10}$, we get $x = t^{1/10}$.

2. Find $dx$ in terms of $dt$: Differentiating $x = t^{1/10}$ with respect to $t$: $\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{10} t^{\frac{1}{10}-1} = \frac{1}{10} t^{-\frac{9}{10}}$ Therefore, $\mathrm{d}x = \frac{1}{10} t^{-\frac{9}{10}} \mathrm{~d}t$.

3. Change the limits of integration:

   • When $x = 0$, $t = (0)^{10} = 0$.
   • When $x = 1$, $t = (1)^{10} = 1$. The limits of integration remain from 0 to 1, which is consistent with the Beta function definition.

Step 3: Rewrite the Integral in Terms of the New Variable $t$ Substitute $t = x^{10}$ and $dx = \frac{1}{10} t^{-\frac{9}{10}} \mathrm{~d}t$ into the integral $I$: I = \int_\limits0^1 \left(1-x^{10}\right)^{20} \mathrm{~d}x I = \int_\limits0^1 (1-t)^{20} \left(\frac{1}{10} t^{-\frac{9}{10}}\right) \mathrm{~d}t

Step 4: Rearrange and Match with the Beta Function Definition We can pull the constant factor $\frac{1}{10}$ out of the integral: I = \frac{1}{10} \int_\limits0^1 t^{-\frac{9}{10}} (1-t)^{20} \mathrm{~d}t Now, we compare this integral with the Beta function definition \int_\limits0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x. The integral is in the form \int_\limits0^1 t^{\mathrm{m}-1}(1-t)^{\mathrm{n}-1} \mathrm{~d}t, where:

• The term $t^{-\frac{9}{10}}$ corresponds to $t^{\mathrm{m}-1}$. So, $\mathrm{m}-1 = -\frac{9}{10}$, which implies $\mathrm{m} = 1 - \frac{9}{10} = \frac{1}{10}$.

• The term $(1-t)^{20}$ corresponds to $(1-t)^{\mathrm{n}-1}$. So, $\mathrm{n}-1 = 20$, which implies $\mathrm{n} = 20 + 1 = 21$.

The integral can thus be written as: $I = \frac{1}{10} \beta\left(\frac{1}{10}, 21\right)$

Step 5: Identify a, b, c and Calculate the Final Expression We are given that \int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x=\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c}). By comparing our result with this given form, we identify: $a = \frac{1}{10}$ $b = \frac{1}{10}$ $c = 21$

Now we need to calculate $100(\mathrm{a}+\mathrm{b}+\mathrm{c})$: $100(\mathrm{a}+\mathrm{b}+\mathrm{c}) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right)$ First, sum the values inside the parenthesis: $\frac{1}{10} + \frac{1}{10} + 21 = \frac{2}{10} + 21 = \frac{1}{5} + 21$ To add these, find a common denominator: $\frac{1}{5} + \frac{21 \times 5}{5} = \frac{1}{5} + \frac{105}{5} = \frac{1+105}{5} = \frac{106}{5}$ Finally, multiply by 100: $100 \left( \frac{106}{5} \right) = \frac{100}{5} \times 106 = 20 \times 106$ $20 \times 106 = 2120$

Common Mistakes & Tips

• Incorrectly identifying m and n: Always remember that the exponents in the Beta function definition are $m-1$ and $n-1$. Ensure you add 1 to the observed exponents to find $m$ and $n$. For instance, if you see $x^k$, then $m-1 = k$, so $m = k+1$.
• Forgetting to transform dx: When performing a substitution, it is crucial to express $dx$ in terms of the new differential ($dt$ in this case). Forgetting this will lead to an incorrect integral.
• Not changing integration limits: While in this specific problem the limits remained the same (0 to 1), this is not always the case. Always check and transform the limits of integration based on the substitution.

Summary

The problem requires transforming a given definite integral into the form of a Beta function. By making the substitution $t = x^{10}$, we successfully converted the integral into \frac{1}{10} \int_\limits0^1 t^{-\frac{9}{10}} (1-t)^{20} \mathrm{~d}t. Comparing this with the Beta function definition, we found $a = \frac{1}{10}$, $b = \frac{1}{10}$, and $c = 21$. The required expression $100(a+b+c)$ was then calculated to be 2120. This approach highlights the power of substitution in simplifying complex integrals and relating them to known special functions like the Beta function.

The final answer is $\boxed{\text{2120}}$.