How do you evaluate the definite integral ∫ sin 3x dx from [0, π]?

To evaluate the definite integral ∫ sin(3x) dx from 0 to π, we first need to find the antiderivative of sin(3x).

Finding the Antiderivative

The antiderivative of sin(3x) can be calculated using a simple substitution. We know that:

• The integral of sin(kx) is -1/k * cos(kx), where k is a constant.

In this case, k = 3. Thus, the antiderivative of sin(3x) is:

-1/3 * cos(3x)

Setting Up the Definite Integral

Now we can set up the definite integral:

∫ sin(3x) dx from 0 to π = [-1/3 * cos(3x)] from 0 to π

Evaluating the Limits

Next, we evaluate this expression at the upper and lower limits:

• At x = π: -1/3 * cos(3π) = -1/3 * (-1) = 1/3
• At x = 0: -1/3 * cos(0) = -1/3 * 1 = -1/3

Calculating the Final Result

Now, we subtract the value at the lower limit from the value at the upper limit:

(1/3) - (-1/3) = 1/3 + 1/3 = 2/3

Final Answer

The value of the definite integral ∫ sin(3x) dx from 0 to π is 2/3.