Interestingly this function is positive over the interval [ 1, sqrt 3 ) and negative over ( sqrt 3 , 2] so this integral calculates the "net" area between the function and the x-axis. If you want to determine the absolute area, you can split the integral into the two intervals above, BUT take the absolute value of the second half from (sqrt 3, 2). This results in an absolute area of approximately 0.5718.
Your voice is so relaxing and you solve the integral really calmly that is very soothing
Interestingly this function is positive over the interval [ 1, sqrt 3 ) and negative over ( sqrt 3 , 2] so this integral calculates the "net" area between the function and the x-axis. If you want to determine the absolute area, you can split the integral into the two intervals above, BUT take the absolute value of the second half from (sqrt 3, 2). This results in an absolute area of approximately 0.5718.
Your voice is amazing!
Nice Video