Integral of sin x from 0 to pi
Understand the Problem
The question is asking for the definite integral of the sine function from 0 to pi. We will evaluate the integral of sin(x) over the specified limits.
Answer
2
Answer for screen readers
The value of the definite integral is (2).
Steps to Solve
- Set up the definite integral
We need to evaluate the definite integral of the sine function from 0 to $\pi$. This is represented as follows:
$$ \int_0^{\pi} \sin(x) , dx $$
- Find the antiderivative of sine
The next step is to find the antiderivative of $\sin(x)$. The antiderivative is:
$$ -\cos(x) + C $$
Where $C$ is the constant of integration.
- Evaluate the definite integral
Now we will evaluate the antiderivative at the upper and lower limits:
$$ \left[-\cos(x)\right]_0^{\pi} = -\cos(\pi) - \left(-\cos(0)\right) $$
- Calculate the values
We know the values of cosine at these points:
- $\cos(\pi) = -1$
- $\cos(0) = 1$
Now replace these values:
$$ = -(-1) - (-1) = 1 + 1 = 2 $$
The value of the definite integral is (2).
More Information
The integral of the sine function represents the area under the curve from $0$ to $\pi$. This area is equal to 2, which is a fundamental result related to the sine function's behavior over this interval.
Tips
- Forgetting to change the signs while evaluating the definite integral.
- Confusing the antiderivative of sine with that of cosine. Remember that the antiderivative of $\sin(x)$ is $-\cos(x)$.
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