What is the evaluation of ∫ from 0 to π/2 of sin(x) dx?

To find the value of the integral of sin(x) from 0 to π/2, we start by recalling the fundamental rule of integration. The integral of sin(x) with respect to x is -cos(x). Therefore, we can set up the definite integral as follows:

∫ sin(x) dx = -cos(x) + C

For the definite integral from 0 to π/2, we evaluate the antiderivative at the upper limit and lower limit:

1. Calculate the antiderivative at π/2:

-cos(π/2) = -(0) = 0

2. Calculate the antiderivative at 0:

-cos(0) = -1

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

0 - (-1) = 0 + 1 = 1

Thus, the evaluation of the integral ∫ from 0 to π/2 of sin(x) dx is 1. This confirms that the correct choice is indeed 1.

This integral represents the area under the curve of sin(x) between these limits, which geometrically corresponds to the area of the quarter circle that lies above the x-axis, confirming that