Key Concepts and Formulas

• Integration by Parts (IBP): The formula for integration by parts is $\int u \, dv = uv - \int v \, du$. For definite integrals, it is $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$. This technique is used to simplify integrals by transforming them into a form that is easier to solve.
• Reduction Formulas: Integrals of the form $\int f(x, n) dx$ where $n$ is an exponent can often be simplified using reduction formulas. These formulas express an integral with a higher exponent in terms of an integral with a lower exponent, or directly relate different integrals.
• Properties of Definite Integrals: Linearity of integration ($\int (af(x) + bg(x)) dx = a\int f(x) dx + b\int g(x) dx$) is used to split or combine integrals.

---

Step-by-Step Solution

We are given two definite integrals: $I_1 = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ $I_2 = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$ We need to find $\alpha$ such that $I_2 = \alpha I_1$.

Step 1: Rewrite $I_2$ to relate it to $I_1$

We can rewrite $I_2$ by separating one factor of $(1 - x^{50})$: $I_2 = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}} \cdot \left( {1 - {x^{50}}} \right)} dx$ Distributing the term $(1 - x^{50})$ inside the integrand: $I_2 = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}} - {x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ Using the linearity of integration, we split this into two integrals: $I_2 = \int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx - \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ The first integral is $I_1$. Let's call the second integral $J$: $I_2 = I_1 - J$ where $J = \int\limits_0^1 {{x^{50}}{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$

Step 2: Evaluate the integral $J$ using Integration by Parts

To evaluate $J$, we will use integration by parts. We need to choose $u$ and $dv$. A strategic choice here is to let $dv$ be the part that, when integrated, simplifies the expression, and $u$ be the part that, when differentiated, also leads to simplification. Let $u = x^{50}$ and $dv = (1 - x^{50})^{100} dx$. Then, $du = 50x^{49} dx$. To find $v$, we need to integrate $dv$. This seems complicated.

Let's try a different approach for integration by parts, which is often more effective for integrals of the form $\int x^m (a - bx^n)^p dx$. We can rewrite $J$ as: $J = \int_0^1 x^{50} (1 - x^{50})^{100} dx$ Let $u = (1 - x^{50})^{101}$ and $dv = x^{50} (1 - x^{50})^{-1} dx$. This doesn't look right.

Let's go back to $I_2 = I_1 - J$. We want to express $J$ in terms of $I_1$. Consider $I_2 = \int_0^1 (1 - x^{50})^{101} dx$. Let $u = (1 - x^{50})^{101}$ and $dv = dx$. Then $du = 101(1 - x^{50})^{100} (-50x^{49}) dx = -5050 x^{49} (1 - x^{50})^{100} dx$ and $v = x$. Applying integration by parts to $I_2$: $I_2 = [x(1 - x^{50})^{101}]_0^1 - \int_0^1 x (-5050 x^{49} (1 - x^{50})^{100}) dx$ $I_2 = (1(1 - 1^{50})^{101} - 0(1 - 0^{50})^{101}) + 5050 \int_0^1 x^{50} (1 - x^{50})^{100} dx$ The first term $[x(1 - x^{50})^{101}]_0^1$ evaluates to $0 - 0 = 0$. The integral term is exactly $J$: $I_2 = 0 + 5050 \int_0^1 x^{50} (1 - x^{50})^{100} dx$ $I_2 = 5050 J$

Step 3: Substitute $J$ back into the equation for $I_2$

From Step 1, we have $I_2 = I_1 - J$. From Step 2, we found $I_2 = 5050 J$. We can express $J$ in terms of $I_2$: $J = \frac{I_2}{5050}$. Substitute this into the equation from Step 1: $I_2 = I_1 - \frac{I_2}{5050}$

Step 4: Solve for $\alpha$

Rearrange the equation to solve for $I_2$ in terms of $I_1$: $I_2 + \frac{I_2}{5050} = I_1$ Combine the terms on the left side: $I_2 \left( 1 + \frac{1}{5050} \right) = I_1$ $I_2 \left( \frac{5050 + 1}{5050} \right) = I_1$ $I_2 \left( \frac{5051}{5050} \right) = I_1$ We are given that $I_2 = \alpha I_1$. Let's rearrange our equation to match this form: $I_1 = I_2 \left( \frac{5051}{5050} \right)$ This implies: $\frac{I_2}{I_1} = \frac{5051}{5050}$ Therefore, $\alpha = \frac{5051}{5050}$.

---

Common Mistakes & Tips

• Incorrect choice of $u$ and $dv$ in Integration by Parts: For integrals involving powers of $(a - bx^n)$, it's often beneficial to choose $u$ as the term with the higher exponent of the binomial and $dv$ as a simpler term (like $dx$ or $x^k dx$) that can be integrated easily.
• Algebraic Errors: Be careful with algebraic manipulations, especially when rearranging equations and combining fractions.
• Evaluating Boundary Terms: Ensure the boundary terms in integration by parts $[uv]_a^b$ are evaluated correctly, especially when the terms involve variables that become zero or one at the limits of integration.

---

Summary

The problem involves relating two definite integrals, $I_1$ and $I_2$, which differ by the exponent of the integrand. The approach used is to express $I_2$ in terms of $I_1$ and another integral, $J$, by splitting the integrand of $I_2$. Then, integration by parts is applied to $I_2$ itself, strategically choosing $u$ and $dv$ such that the resulting integral is $J$ and the boundary term evaluates to zero. This establishes a direct relationship between $I_2$ and $J$. Finally, substituting the relationship between $I_2$ and $J$ back into the initial expression for $I_2$ in terms of $I_1$ and $J$ allows us to solve for $\alpha$.

The final answer is $\boxed{\frac{5051}{5050}}$.