Using Riemann Integral for proof.

Question

Prove that f(x) = absolute value of x when x is not 0 and f(0)=9

on the closed interval [-1, 1] is integrable .

Answer 1

The graph looks just like the regular absolute value function, except there is a hole at (0,0) and it is replaced with a single dot at (0,9)

This is a removable discontinuity , or single point discontinuity, which by known theorems

are Riemann integrable. All of the legal mumbo jumbo states that the function can be integrated

via Riemann sum - limits if the function is bounded along each partition, provided the partitions

have only one point in common in their intersection.

So in this case, we can partition [-1,0] U [0,1].

The function is bounded by 0 and 9 in the left partition as well as the right partition.

Since these bounds (call supremum and infimum) agree, the function can be integrated

as the single point of discontinuity ( 0,9) in this case does not affect the area under the curve.

Feel free to google Reimann integrability proof for the gory details of the real analysis