`int_(0)^(pi) [2sin 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 can 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 \(-\cos x\). Therefore, we can evaluate:
\[ \int \sin x \, dx = -\cos x + C \]
Now, we apply the limits from \(0\) to \(\pi\):
\[ \int_{0}^{\pi} \sin x \, dx = \left[-\cos x\right]_{0}^{\pi} \]

### Step 4: Substitute the Limits
Now we substitute the upper and lower limits into the evaluated integral:
\[ = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 5: Multiply by the Constant
Now, we multiply the result by the constant we factored out earlier:
\[ I = 2 \cdot 2 = 4 \]

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

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