`int_(0)^(pi) [2 sin x] dx`=

Options

A. `(2pi)/(3)`
B. `-(5pi)/(3)`
C. `-pi`
D. `-2pi`

Answer

To solve the integral \( \int_{0}^{\pi} 2 \sin x \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{\pi} 2 \sin x \, dx \]

### Step 2: Factor out the constant
Since 2 is a constant, we can factor it out of the integral:
\[ I = 2 \int_{0}^{\pi} \sin x \, dx \]

### Step 3: Evaluate the integral of \( \sin x \)
The integral of \( \sin x \) is:
\[ \int \sin x \, dx = -\cos x + C \]
Thus, we can evaluate:
\[ \int_{0}^{\pi} \sin x \, dx = \left[-\cos x\right]_{0}^{\pi} \]

### Step 4: Calculate the definite integral
Now we substitute the limits into the evaluated integral:
\[ \left[-\cos(\pi) - (-\cos(0))\right] = \left[-(-1) - (-1)\right] = 1 + 1 = 2 \]

### Step 5: Multiply by the constant
Now we substitute this result back into our equation for \( I \):
\[ I = 2 \cdot 2 = 4 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\pi} 2 \sin x \, dx = 4 \]