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Definite Integral of Sin(x) --- is it 4 or 0? Watch

    • Thread Starter

    Very basic question.

    What's the definite integral of sin(x) with limits of 0 to 2pi.

    Is it 0 or 4?

    If the limits are 0 to pi it is 2; and if the limits are from pi to 2pi, then it is -2.

    Is the definite integral the sum of the area, in which case the absolute value of -2 is taken, so it adds up to 4, since area is always positive?

    Or is the definite integral not the area, but rather:

    ∫ sin(x)[0, 2π] dx = [cos(2π) - cos(0)] = 1-1 = 0

    Also, if I put it in the calculator, it also gives zero, so I'm leaning towards that idea right now.

    I'm making a Matlab program that numerically integrates a function, so I want to make sure if I need to keep or remove the ABS from this code:

    if mod(n,2) == 0; % if n is even
    S_s = (dx/3)*(abs(y_x(a)) + 4*sum (abs(y_x(a+dx:2*dx:b-dx))) + 2*sum(abs(y_x(a+2*dx:2*dx:b-2*dx))));
    else % if n is odd
    S_s = (dx/3)*(abs(y_x(a)) + 4*sum (abs(y_x(a+dx:2*dx:b-2*dx))) + 2*sum(abs(y_x(a+2*dx:2*dx:b-dx))));

    The integral of sin(x) from 0 to 2pi is 0. The area under the curve is 4.
    • Thread Starter

    (Original post by Bobifier)
    The integral of sin(x) from 0 to 2pi is 0. The area under the curve is 4.
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