Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{1}x(1-x)^n\,dx$

Concept: Power Rule Integration — rewrite the integrand so that the power rule $\int x^m dx = \frac{x^{m+1}}{m+1}$ applies directly.

Let $I = \int_{0}^{1} x(1-x)^n \, dx$. Use the substitution $u = 1-x$, so $x = 1-u$ and $dx = -du$. When $x=0$, $u=1$; when $x=1$, $u=0$.

Now integrate term by term: …

Why This Approach Works

When you see a product like $x(1-x)^n$ integrated from $0$ to $1$, your first instinct should be symmetry. The interval $[0,1]$ and the factor $(1-x)^n$ practically beg for the substitution $u = 1-x$. This trick turns the integral into something you can handle with the Power Rule — no need for integration by parts or memorising Beta function formulas, though we'll note that connection.

The key insight: the integrand is a polynomial in $x$ (once you expand $(1-x)^n$), but the substitution keeps the limits simple and the algebra clean.

Step-by-Step Solution

1. Set up the substitution. Let $u = 1 - x$. Then $x = 1 - u$, and $dx = -du$. When $x = 0$, $u = 1$; when $x = 1$, $u = 0$. The integral becomes:

2. Simplify the limits. The negative sign in $dx = -du$ flips the limits back:

3. Rewrite the integrand. Expand $(1-u)u^n = u^n - u^{n+1}$. So:

4. Apply the Power Rule for integration. For any $m > -1$, $\int_{0}^{1} u^m \, du = \frac{1}{m+1}$. Here:

5. Combine the results.

6. Simplify the difference. …

Method: Reflection substitution $u=a-x$ to tame $x\,(a-x)^n$

For $\displaystyle\int_0^a x\,(a-x)^n\,dx$ over $[0,a]$, the substitution $u=a-x$ (equivalently the reflection property) turns the high power into a clean base and leaves an easy polynomial.

Steps

Step 1: Substitute $u=a-x$.

Then $x=a-u$, $dx=-du$; the limits swap, and the $-du$ is absorbed by reversing them.

Step 2: Rewrite the integrand.

$x\,(a-x)^n$ becomes $(a-u)\,u^{n}$, a sum of two power terms $a\,u^{n}-u^{n+1}$ (take $a=1$ as in the standard case). …

Common Mistakes

Mistake 1: Expanding $(1-x)^n$ with the binomial theorem for general $n$.

Why it's wrong: with $n$ symbolic, a full binomial expansion is unwieldy and error-prone; the substitution $u=1-x$ is far cleaner. Correct approach: set $u=1-x$ so the integrand becomes $u^n-u^{n+1}$.

Mistake 2: Sign/limit slip when substituting $dx=-du$. …