my questions is how to solve the antiderivatove of 2abs(x) at interval -1
please help me!
i know that the derivative of abs(x) is x/abs(x)
thanks
please help me!
i know that the derivative of abs(x) is x/abs(x)
thanks
-
Let the absolute value of x be written |x|
A sketch of y = 2|x| is a V shape with vertex at origin and "wings" with slopes -2 and 2
It is therefore equivalent to y = -2x from -1 to zero and y = +2x from zero to +1
The total value assesed visually by area under curve is therefore 2 units.
Alternatively A = (-1 to 0)∫ -2x dx + (0 to 1)∫ +2x dx
A = (-x^2) from -1 to 0 plus (+x^2) from 0 to 1 = 2
Some, (but not all), discontinuous functions can be integrated.
In algebraic terms ∫ 2|x| dx = x|x| so inserting the limits required
(-1 to 1)∫ 2|x| dx = x|x| from -1 to 1= 1*|1| - (-1*|-1|) = 1 + 1
Regards - Ian
A sketch of y = 2|x| is a V shape with vertex at origin and "wings" with slopes -2 and 2
It is therefore equivalent to y = -2x from -1 to zero and y = +2x from zero to +1
The total value assesed visually by area under curve is therefore 2 units.
Alternatively A = (-1 to 0)∫ -2x dx + (0 to 1)∫ +2x dx
A = (-x^2) from -1 to 0 plus (+x^2) from 0 to 1 = 2
Some, (but not all), discontinuous functions can be integrated.
In algebraic terms ∫ 2|x| dx = x|x| so inserting the limits required
(-1 to 1)∫ 2|x| dx = x|x| from -1 to 1= 1*|1| - (-1*|-1|) = 1 + 1
Regards - Ian