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Integrate sin(x) from 0 to pi.

Understand the Problem

The question is asking for the definite integral of the function sin(x) from 0 to π. This involves finding the area under the curve defined by sin(x) within the specified limits.

Answer

2
Answer for screen readers

The area under the curve of $sin(x)$ from 0 to $\pi$ is 2.

Steps to Solve

  1. Set up the definite integral

To find the area under the curve of $sin(x)$ from 0 to $\pi$, we set up the definite integral as follows:
$$ \int_0^{\pi} \sin(x) , dx $$

  1. Find the antiderivative of sin(x)

The next step is to calculate the antiderivative of $sin(x)$. The antiderivative of $sin(x)$ is $-cos(x)$.
Thus, we have:
$$ F(x) = -\cos(x) $$

  1. Evaluate the definite integral

Now we evaluate the definite integral using the antiderivative found. We substitute the upper limit $\pi$ and the lower limit 0 into $F(x)$:
$$ F(\pi) - F(0) = [-\cos(\pi)] - [-\cos(0)] $$

  1. Calculate cos(Ï€) and cos(0)

Next, we know that:
$$ \cos(\pi) = -1 $$
$$ \cos(0) = 1 $$

So we substitute these values back into our evaluation:
$$ F(\pi) - F(0) = [-(-1)] - [-1] $$

  1. Simplify the final result

Simplifying gives us:
$$ = 1 + 1 = 2 $$

The area under the curve of $sin(x)$ from 0 to $\pi$ is 2.

More Information

The definite integral of a function gives the area under the curve between two points. In this case, the area under the sine wave from 0 to $\pi$ results in a total area of 2.

Tips

  • Forgetting to use the correct limits when evaluating the definite integral.
  • Miscalculating the sine and cosine values.
  • Confusing the antiderivative of sine with other functions; always double-check the basic derivatives and integrals.
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