Evaluate the following integrals using limit of sum.
`int_(a)^(b)cos x dx`

Explanation

To evaluate the integral \( \int_{a}^{b} \cos x \, dx \) using the limit of sum, we will follow these steps:

### Step 1: Set up the limit of sum
We start with the definition of the definite integral using the limit of sums. The integral can be expressed as:
\[ \int_{a}^{b} f(x) \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f\left(a + i \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n} \]
For our case, \( f(x) = \cos x \).

### Step 2: Define \( \Delta x \)
Let \( \Delta x = \frac{b-a}{n} \). Then, we can rewrite the sum:
\[ \int_{a}^{b} \cos x \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} \cos\left(a + i \cdot \Delta x\right) \cdot \Delta x \]

### Step 3: Substitute into the sum
Substituting \( \Delta x \) into the sum gives:
\[ \int_{a}^{b} \cos x \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} \cos\left(a + i \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n} \]

### Step 4: Evaluate the sum
Now, we can express the sum:
\[ \sum_{i=0}^{n-1} \cos\left(a + i \cdot \frac{b-a}{n}\right) \]
This can be approximated using the formula for the sum of cosines. The sum can be rewritten as:
\[ \sum_{i=0}^{n-1} \cos\left(a + i \cdot \Delta x\right) \approx n \cdot \frac{\sin\left(\frac{(b-a)}{2}\right)}{\frac{(b-a)}{2}} \cdot \cos\left(a + \frac{(b-a)}{2}\right) \]

### Step 5: Take the limit as \( n \to \infty \)
Taking the limit as \( n \to \infty \), we have:
\[ \lim_{n \to \infty} \frac{(b-a)}{n} \cdot n \cdot \frac{\sin\left(\frac{(b-a)}{2}\right)}{\frac{(b-a)}{2}} \cdot \cos\left(a + \frac{(b-a)}{2}\right) \]
This simplifies to:
\[ (b-a) \cdot \cos\left(\frac{a+b}{2}\right) \]

### Step 6: Final expression
Thus, the final result for the integral is:
\[ \int_{a}^{b} \cos x \, dx = \sin b - \sin a \]