`int_(2)^(4) x sqrt(6-x) dx`= ….

Answer

To solve the integral \( I = \int_{2}^{4} x \sqrt{6 - x} \, dx \), we can use a property of definite integrals. The property states that:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

In our case, \( a = 2 \) and \( b = 4 \), so \( a + b = 6 \). Thus, we can rewrite the integral as follows:

\[ I = \int_{2}^{4} x \sqrt{6 - x} \, dx = \int_{2}^{4} (6 - x) \sqrt{6 - (6 - x)} \, dx \]

This simplifies to:

\[ I = \int_{2}^{4} (6 - x) \sqrt{x} \, dx \]

Now, we can express \( I \) in terms of two integrals:

\[ I = \int_{2}^{4} x \sqrt{6 - x} \, dx + \int_{2}^{4} (6 - x) \sqrt{x} \, dx \]

Let’s denote the second integral as \( J \):

\[ J = \int_{2}^{4} (6 - x) \sqrt{x} \, dx \]

Thus, we have:

\[ I = I + J \]

To find \( J \), we can compute:

\[ J = \int_{2}^{4} (6\sqrt{x} - x\sqrt{x}) \, dx \]

This can be split into two separate integrals:

\[ J = 6 \int_{2}^{4} \sqrt{x} \, dx - \int_{2}^{4} x^{3/2} \, dx \]

Now, we compute each integral separately.

1. **Calculating \( \int_{2}^{4} \sqrt{x} \, dx \)**:

Using the formula for the integral of \( x^n \):

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} \]

For \( n = \frac{1}{2} \):

\[ \int \sqrt{x} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]

Thus,

\[ \int_{2}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{2}^{4} = \frac{2}{3} \left( 4^{3/2} - 2^{3/2} \right) \]

Calculating the values:

\[ 4^{3/2} = 8 \quad \text{and} \quad 2^{3/2} = 2\sqrt{2} \]

So,

\[ \int_{2}^{4} \sqrt{x} \, dx = \frac{2}{3} (8 - 2\sqrt{2}) = \frac{16}{3} - \frac{4\sqrt{2}}{3} \]

2. **Calculating \( \int_{2}^{4} x^{3/2} \, dx \)**:

Using the same formula:

\[ \int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2}{5} x^{5/2} \]

Thus,

\[ \int_{2}^{4} x^{3/2} \, dx = \left[ \frac{2}{5} x^{5/2} \right]_{2}^{4} = \frac{2}{5} (4^{5/2} - 2^{5/2}) \]

Calculating the values:

\[ 4^{5/2} = 32 \quad \text{and} \quad 2^{5/2} = 4\sqrt{2} \]

So,

\[ \int_{2}^{4} x^{3/2} \, dx = \frac{2}{5} (32 - 4\sqrt{2}) = \frac{64}{5} - \frac{8\sqrt{2}}{5} \]

Now substituting back into \( J \):

\[ J = 6 \left( \frac{16}{3} - \frac{4\sqrt{2}}{3} \right) - \left( \frac{64}{5} - \frac{8\sqrt{2}}{5} \right) \]

Calculating \( J \):

\[ J = \frac{96}{3} - \frac{24\sqrt{2}}{3} - \frac{64}{5} + \frac{8\sqrt{2}}{5} \]

Finding a common denominator (15):

\[ J = \frac{480}{15} - \frac{120\sqrt{2}}{15} - \frac{192}{15} + \frac{24\sqrt{2}}{15} \]

Combining terms:

\[ J = \frac{288}{15} - \frac{96\sqrt{2}}{15} = \frac{288 - 96\sqrt{2}}{15} \]

Now substituting back into \( I \):

\[ I = I + J \implies 2I = J \implies I = \frac{1}{2} J \]

Calculating \( I \):

\[ I = \frac{1}{2} \cdot \frac{288 - 96\sqrt{2}}{15} = \frac{288 - 96\sqrt{2}}{30} \]

Thus, the final answer is:

\[ I = \frac{32(3 - \sqrt{2})}{5} \]