Evaluate: `int_a^b sinx dx` as the limit of a sum.

Answer

To evaluate the integral \( \int_a^b \sin x \, dx \) as the limit of a sum, we can follow these steps:

### Step 1: Define the integral as a limit of a Riemann sum
We start by dividing the interval \([a, b]\) into \(n\) equal parts. The width of each subinterval is given by:
\[ h = \frac{b - a}{n} \]
The endpoints of these subintervals will be:
\[ x_0 = a, \quad x_1 = a + h, \quad x_2 = a + 2h, \ldots, \quad x_n = b \]

### Step 2: Write the Riemann sum
The Riemann sum for the function \(f(x) = \sin x\) over the interval \([a, b]\) can be expressed as:
\[ S_n = \sum_{i=0}^{n-1} f(x_i) \cdot h = \sum_{i=0}^{n-1} \sin(x_i) \cdot h \]
Substituting \(x_i = a + ih\), we get:
\[ S_n = \sum_{i=0}^{n-1} \sin(a + ih) \cdot h \]

### Step 3: Take the limit as \(n \to \infty\)
The integral can be represented as the limit of the Riemann sum:
\[ \int_a^b \sin x \, dx = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{i=0}^{n-1} \sin(a + ih) \cdot h \]

### Step 4: Substitute \(h\) in the limit
Substituting \(h = \frac{b - a}{n}\), we rewrite the limit:
\[ \int_a^b \sin x \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} \sin\left(a + i \frac{b - a}{n}\right) \cdot \frac{b - a}{n} \]

### Step 5: Recognize the form of the limit
As \(n \to \infty\), the sum approaches the integral:
\[ \int_a^b \sin x \, dx = \lim_{n \to \infty} \frac{b - a}{n} \sum_{i=0}^{n-1} \sin\left(a + i \frac{b - a}{n}\right) \]

### Step 6: Evaluate the integral
Using the Fundamental Theorem of Calculus, we can evaluate the integral:
\[ \int \sin x \, dx = -\cos x + C \]
Thus,
\[ \int_a^b \sin x \, dx = \left[-\cos x\right]_a^b = -\cos b + \cos a = \cos a - \cos b \]

### Final Answer
Therefore, we conclude that:
\[ \int_a^b \sin x \, dx = \cos a - \cos b \]

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