Evaluate : `int_0^1x(1-x)^5dx`

Answer

To evaluate the integral

\[ \int_0^1 x(1-x)^5 \, dx, \]

we can use a substitution method. Let's go through the solution step by step.

### Step 1: Substitution
Let \( t = 1 - x \). Then, we have:

\[ dx = -dt. \]

### Step 2: Change the Limits
When \( x = 0 \), \( t = 1 - 0 = 1 \).

When \( x = 1 \), \( t = 1 - 1 = 0 \).

So, the limits of integration change from \( x: 0 \to 1 \) to \( t: 1 \to 0 \).

### Step 3: Rewrite the Integral
Now, substituting \( x = 1 - t \) into the integral, we get:

\[ \int_1^0 (1 - t) t^5 (-dt) = \int_0^1 (1 - t) t^5 \, dt. \]

### Step 4: Expand the Integral
Now, we can expand the integrand:

\[ \int_0^1 (1 - t) t^5 \, dt = \int_0^1 (t^5 - t^6) \, dt. \]

### Step 5: Integrate Term by Term
Now, we can integrate term by term:

\[ \int_0^1 t^5 \, dt - \int_0^1 t^6 \, dt. \]

Calculating these integrals:

\[ \int_0^1 t^5 \, dt = \left[ \frac{t^6}{6} \right]_0^1 = \frac{1}{6}, \]

\[ \int_0^1 t^6 \, dt = \left[ \frac{t^7}{7} \right]_0^1 = \frac{1}{7}. \]

### Step 6: Combine the Results
Now, substituting back into our expression:

\[ \int_0^1 (1 - t) t^5 \, dt = \frac{1}{6} - \frac{1}{7}. \]

### Step 7: Find a Common Denominator
To combine these fractions, we find a common denominator, which is 42:

\[ \frac{1}{6} = \frac{7}{42}, \quad \frac{1}{7} = \frac{6}{42}. \]

Thus,

\[ \frac{1}{6} - \frac{1}{7} = \frac{7}{42} - \frac{6}{42} = \frac{1}{42}. \]

### Final Answer
Therefore, the value of the integral is

\[ \int_0^1 x(1-x)^5 \, dx = \frac{1}{42}. \]

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