Is f(x)=|x+2| integrable? Please give a reason behind your answer.
Yes, it is piecewise integrable.
http://www.mathwords.com/p/piecewise_continuous_function.htm
Could you explain that a bit more? I understand that is is continuous, but how does that relate to integration?
Sure! The concept of integrability is closely related to the continuity of a function. A function is said to be integrable if it can be integrated over a certain interval or region.
For piecewise continuous functions like f(x) = |x + 2|, the function may have different definitions or formulas for different intervals. In this case, we have two intervals to consider: x < -2 and x ≥ -2.
On the interval x < -2, the function can be written as f(x) = -(x + 2). This is because for x < -2, the expression inside the absolute value, x + 2, is negative, so the negative sign is needed to make the absolute value positive.
On the interval x ≥ -2, the function can still be written as f(x) = x + 2. This is because for x ≥ -2, the expression inside the absolute value, x + 2, is already non-negative, so there is no need for a negative sign.
Now, since both of these formulas are continuous functions (a linear function and an affine function, respectively), we can conclude that f(x) = |x + 2| is piecewise continuous.
The integrability of a piecewise continuous function can be determined by integrating each piece of the function over their respective intervals. In this case, integrating -(x + 2) on the interval x < -2 and integrating x + 2 on the interval x ≥ -2 will yield a finite result, meaning that f(x) = |x + 2| is integrable.