Question

Derive a formula for \(\mathrm{I}_{\mathrm{n}}=\int_{0}^{1} \frac{(1-\mathrm{x})^{\mathrm{n}}}{\sqrt{\mathrm{x}}} \mathrm{d} \mathrm{x}\), (n is a positive integer).

Short answer

Answer: The general formula for the integral \(\mathrm{I}_n\) is given by \(\mathrm{I}_n = n - n(n-1)\int_0^1x^2(1-x)^{n-2}\mathrm{dx}\).

Step-by-step solution

Define the Functions

First, let's define the functions to integrate and differentiate: $$u = (1 - x)^n \quad \Rightarrow \quad \mathrm{du} = -n(1-x)^{n-1}\ \mathrm{dx}$$ $$\mathrm{dv} = \frac{1}{\sqrt{x}}\ \mathrm{dx} \quad \Rightarrow \quad v = 2\sqrt{x}$$

Integration by Parts

Let's use integration by parts on the given integral. Recall the formula for integration by parts: \(\int u \ dv = uv - \int v\ du\). Using the defined functions, we get: $$\mathrm{I}_{n} = \int_0^1 \frac{(1-x)^n}{\sqrt{x}}\ \mathrm{dx} = \left[(1-x)^n(2\sqrt{x})\right]_0^1 - \int_0^1 2\sqrt{x}\ (-n)(1-x)^{n-1}\ \mathrm{dx}$$

Simplify the Expression

Now, let's simplify the expression we got in the previous step: $$\mathrm{I}_{n} = 2n \int_0^1 x \cdot (1-x)^{n-1}\ \mathrm{dx}$$

Another Integration by Parts

We need to find the integral in the expression again. Let's do integration by parts again, but this time with new functions: $$u = (1-x)^{n-1} \quad \Rightarrow \quad \mathrm{du} = -(n-1)(1-x)^{n-2}\ \mathrm{dx}$$ $$\mathrm{dv} = x\ \mathrm{dx} \quad \Rightarrow \quad v = \frac{1}{2}x^2$$ Applying integration by parts again: $$\int_0^1 x \cdot (1-x)^{n-1}\ \mathrm{dx} = \left[\frac{1}{2}x^2(1-x)^{n-1}\right]_0^1 - \int_0^1 \frac{1}{2}x^2(-n+1)(1-x)^{n-2}\ \mathrm{dx}$$

Simplify the Expression and Combine with Previous Steps

Simplifying the expression again and combining with the expression from step 3: $$\mathrm{I}_{n} = 2n \left[\frac{1}{2}-\frac{n-1}{2}\int_0^1x^2(1-x)^{n-2}\mathrm{dx}\right]$$ Thus, the final formula for the integral \(\mathrm{I}_n\) becomes: $$\mathrm{I}_n = n - n(n-1)\int_0^1x^2(1-x)^{n-2}\mathrm{dx}$$