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Question: If $I_1 = \int_0^1 (1-x^{50})^{100} dx$, $I_2 = \int_0^1 (1-x^{50})^{101} dx$ and $I_2 = \alpha I_1$...

If I1=∫01(1−x50)100dxI_1 = \int_0^1 (1-x^{50})^{100} dx, I2=∫01(1−x50)101dxI_2 = \int_0^1 (1-x^{50})^{101} dx and I2=αI1I_2 = \alpha I_1, then α\alpha equal

A

5049500\frac{5049}{500}

B

50515050\frac{5051}{5050}

C

50505051\frac{5050}{5051}

D

50505049\frac{5050}{5049}

Answer

c

Explanation

Solution

To find the value of α\alpha such that I2=αI1I_2 = \alpha I_1, we need to establish a relationship between I1I_1 and I2I_2. Let's define a general integral I(n)=∫01(1−x50)ndxI(n) = \int_0^1 (1-x^{50})^n dx. So, I1=I(100)I_1 = I(100) and I2=I(101)I_2 = I(101).

We will use integration by parts for I(n)I(n). Let u=(1−x50)nu = (1-x^{50})^n and dv=dxdv = dx. Then, du=n(1−x50)n−1(−50x49)dx=−50nx49(1−x50)n−1dxdu = n(1-x^{50})^{n-1} (-50x^{49}) dx = -50n x^{49} (1-x^{50})^{n-1} dx. And, v=xv = x.

Applying the integration by parts formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: I(n)=[x(1−x50)n]01−∫01x(−50nx49(1−x50)n−1)dxI(n) = \left[ x(1-x^{50})^n \right]_0^1 - \int_0^1 x \left( -50n x^{49} (1-x^{50})^{n-1} \right) dx

Evaluate the definite part: [x(1−x50)n]01=1⋅(1−150)n−0⋅(1−050)n=1⋅0n−0⋅1n=0−0=0\left[ x(1-x^{50})^n \right]_0^1 = 1 \cdot (1-1^{50})^n - 0 \cdot (1-0^{50})^n = 1 \cdot 0^n - 0 \cdot 1^n = 0 - 0 = 0 (since n≥1n \ge 1).

So, the integral becomes: I(n)=0−∫01(−50nx50(1−x50)n−1)dxI(n) = 0 - \int_0^1 \left( -50n x^{50} (1-x^{50})^{n-1} \right) dx I(n)=50n∫01x50(1−x50)n−1dxI(n) = 50n \int_0^1 x^{50} (1-x^{50})^{n-1} dx

Now, we need to express x50x^{50} in terms of (1−x50)(1-x^{50}) to relate it back to I(n)I(n) or I(n−1)I(n-1). We can write x50=1−(1−x50)x^{50} = 1 - (1-x^{50}). Substitute this into the integral: I(n)=50n∫01(1−(1−x50))(1−x50)n−1dxI(n) = 50n \int_0^1 \left( 1 - (1-x^{50}) \right) (1-x^{50})^{n-1} dx I(n)=50n∫01[(1−x50)n−1−(1−x50)n]dxI(n) = 50n \int_0^1 \left[ (1-x^{50})^{n-1} - (1-x^{50})^n \right] dx I(n)=50n[∫01(1−x50)n−1dx−∫01(1−x50)ndx]I(n) = 50n \left[ \int_0^1 (1-x^{50})^{n-1} dx - \int_0^1 (1-x^{50})^n dx \right]

Recognize the terms as I(n−1)I(n-1) and I(n)I(n): I(n)=50n[I(n−1)−I(n)]I(n) = 50n [I(n-1) - I(n)]

Now, we solve for I(n)I(n): I(n)=50nI(n−1)−50nI(n)I(n) = 50n I(n-1) - 50n I(n) I(n)+50nI(n)=50nI(n−1)I(n) + 50n I(n) = 50n I(n-1) I(n)(1+50n)=50nI(n−1)I(n) (1 + 50n) = 50n I(n-1) I(n)=50n50n+1I(n−1)I(n) = \frac{50n}{50n+1} I(n-1)

This is a reduction formula. We need to find α\alpha such that I2=αI1I_2 = \alpha I_1. Here, I2=I(101)I_2 = I(101) and I1=I(100)I_1 = I(100). Using the reduction formula with n=101n=101: I(101)=50×10150×101+1I(101−1)I(101) = \frac{50 \times 101}{50 \times 101 + 1} I(101-1) I2=50505050+1I(100)I_2 = \frac{5050}{5050 + 1} I(100) I2=50505051I1I_2 = \frac{5050}{5051} I_1

Comparing this with I2=αI1I_2 = \alpha I_1, we find: α=50505051\alpha = \frac{5050}{5051}