# How do you prove that # int_0^1 \ x(1-x)^n \ dx = 1/((n+1)(n+2))# ?

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We aim to prove that:

We can then apply Integration By Parts:

Then plugging into the IBP formula:

We have:

And applying the limits of integration we get:

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See explanation...

Note that:

So:

and:

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To prove that (\int_{0}^{1} x(1-x)^n , dx = \frac{1}{(n+1)(n+2)}), we can use integration by parts.

- Let (u = x) and (dv = (1-x)^n , dx).
- Calculate (du) and (v).
- Apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Here's the calculation:

Given: [ u = x ] [ dv = (1-x)^n , dx ]

Calculate: [ du = dx ] [ v = \frac{(1-x)^{n+1}}{n+1} ]

Apply integration by parts:
[ \int_{0}^{1} x(1-x)^n , dx = \left. \frac{x(1-x)^{n+1}}{n+1} \right|*{0}^{1} - \int*{0}^{1} \frac{(1-x)^{n+1}}{n+1} , dx ]

Evaluate the definite integral: [ = \frac{(1-1)^{n+1}}{n+1} - \frac{(0)^{n+1}}{n+1} - \int_{0}^{1} \frac{(1-x)^{n+1}}{n+1} , dx ] [ = 0 - 0 - \frac{1}{n+1} \int_{0}^{1} (1-x)^{n+1} , dx ] [ = - \frac{1}{n+1} \cdot \frac{(1-x)^{n+2}}{n+2} \Bigg|_{0}^{1} ] [ = - \frac{1}{n+1} \left( \frac{(1-1)^{n+2}}{n+2} - \frac{(1-0)^{n+2}}{n+2} \right) ] [ = - \frac{1}{n+1} \left( 0 - \frac{1}{n+2} \right) ] [ = \frac{1}{(n+1)(n+2)} ]

Therefore, (\int_{0}^{1} x(1-x)^n , dx = \frac{1}{(n+1)(n+2)}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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