Respuesta :
Final Answer:
The integral of [tex]sin(3x) dx[/tex] from 0 to [tex]π[/tex]is equal to [tex]-cos(3π) - (-cos(0)) = -cos(3π) + cos(0).[/tex]
Explanation:
When finding the integral of [tex]sin(3x) dx[/tex] from [tex]0 to π[/tex], we employ the fundamental theorem of calculus, which states that the integral of a function over an interval can be found by evaluating the antiderivative of the function at the endpoints of the interval and taking the difference. In this case, the antiderivative of [tex]sin(3x) is -cos(3x)[/tex], so we evaluate this antiderivative at[tex]π and 0\\[/tex]. Thus, we have [tex]-cos(3π) - (-cos(0)).[/tex]
Now, let's simplify this expression. Since cosine has a period of [tex]2π, cos(3π) = cos(π), and cos(π) = -1. Also, cos(0) = 1.[/tex] Substituting these values into our expression, we get[tex]-(-1) - 1 = 1 + 1 = 2[/tex]. Therefore, the final answer is 2.
Therefore, the conclusion is that the integral of [tex]sin(3x) dx[/tex] from 0 to π is equal to 2. This result indicates the net area under the curve of [tex]sin(3x)[/tex] between the bounds of 0 and π.
