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
- 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:
$$ \int_0^{\pi} \sin(x) , dx $$
- 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:
$$ F(x) = -\cos(x) $$
- 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 $F(x)$:
$$ F(\pi) - F(0) = [-\cos(\pi)] - [-\cos(0)] $$
- 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:
$$ F(\pi) - F(0) = [-(-1)] - [-1] $$
- 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.
