Question

S= интеграл сверху п снизу 0 sin x dx=????

Answer 1

$-\cos x|_0^\pi =-(-1-1)=2$

Answer 2

The expression you've written is a definite integral of the sine function from 0 to p (π). Let's evaluate it step by step:

∫[0 to π] sin(x) dx

The integral of sin(x) is -cos(x). So,

∫[0 to π] sin(x) dx = -cos(x) evaluated from x = 0 to x = π

Now, plug in the limits of integration:

= -cos(π) - (-cos(0))

Since cos(π) = -1 and cos(0) = 1, the expression simplifies to:

= -(-1) - 1

= 1 + 1

= 2

So, the value of the definite integral ∫[0 to π] sin(x) dx is 2.