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

Answer

To solve the integral \( I = \int_0^1 x (1-x)^5 \, dx \), we can use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_0^1 x (1-x)^5 \, dx \]

### Step 2: Use substitution
Let’s use the substitution \( t = 1 - x \). Then, we differentiate:
\[ dt = -dx \quad \text{or} \quad dx = -dt \]
Now, we need to change the limits of integration. When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 0 \). Thus, the integral becomes:
\[ I = \int_1^0 (1-t) t^5 (-dt) = \int_0^1 (1-t) t^5 \, dt \]

### Step 3: Expand the integrand
Now, we can expand the integrand:
\[ I = \int_0^1 (t^5 - t^6) \, dt \]

### Step 4: Integrate term by term
We can now integrate each term separately:
\[ I = \int_0^1 t^5 \, dt - \int_0^1 t^6 \, dt \]
Using the power rule for integration, we have:
\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]
Thus:
\[ \int_0^1 t^5 \, dt = \left[ \frac{t^6}{6} \right]_0^1 = \frac{1^6}{6} - \frac{0^6}{6} = \frac{1}{6} \]
\[ \int_0^1 t^6 \, dt = \left[ \frac{t^7}{7} \right]_0^1 = \frac{1^7}{7} - \frac{0^7}{7} = \frac{1}{7} \]

### Step 5: Combine the results
Now substituting back into the equation for \( I \):
\[ I = \frac{1}{6} - \frac{1}{7} \]
To combine these fractions, we find a common denominator, which is 42:
\[ I = \frac{7}{42} - \frac{6}{42} = \frac{1}{42} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{1}{42} \]

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