Evaluate the integral
`I = int_(0)^(5) sqrt(25 -x^(2))dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{5} \sqrt{25 - x^2} \, dx, \]

we can recognize that the integrand \(\sqrt{25 - x^2}\) resembles the equation of a semicircle. Specifically, it represents the upper half of a circle with radius 5 centered at the origin.

### Step 1: Identify the geometric interpretation
The equation \(x^2 + y^2 = 25\) describes a circle with radius 5. The integral we are evaluating corresponds to the area under the curve from \(x = 0\) to \(x = 5\).

### Step 2: Set up the integral using trigonometric substitution
To evaluate the integral, we can use the substitution \(x = 5 \sin \theta\). Then, \(dx = 5 \cos \theta \, d\theta\). The limits change as follows:
- When \(x = 0\), \(\theta = 0\).
- When \(x = 5\), \(\theta = \frac{\pi}{2}\).

### Step 3: Rewrite the integral
Substituting into the integral, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{25 - (5 \sin \theta)^2} \cdot 5 \cos \theta \, d\theta. \]

This simplifies to:

\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{25(1 - \sin^2 \theta)} \cdot 5 \cos \theta \, d\theta. \]

Since \(1 - \sin^2 \theta = \cos^2 \theta\), we can simplify further:

\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{25} \cdot \cos \theta \cdot 5 \cos \theta \, d\theta = 25 \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta. \]

### Step 4: Evaluate the integral
Now, we use the identity \(\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}\):

\[ I = 25 \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta = \frac{25}{2} \left[ \theta + \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{2}}. \]

Calculating the limits:

- At \(\theta = \frac{\pi}{2}\):

\[ \frac{\pi}{2} + \frac{1}{2} \sin(\pi) = \frac{\pi}{2} + 0 = \frac{\pi}{2}. \]

- At \(\theta = 0\):

\[ 0 + \frac{1}{2} \sin(0) = 0. \]

Thus,

\[ I = \frac{25}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{25\pi}{4}. \]

### Final Result
The value of the integral is

\[ \boxed{\frac{25\pi}{4}}. \]