x = ?2 ----> vertical asymptote (non-detachable) x = 0 ------> bounce discontinuity (non-detachable) x = 2 ------> detachable x = 4 ------> non-detachable notice that there discontinuity at x = 0 is non-detachable even although f(x) is defined at x = 0 ----> f(0) = 4 (as stated in an answer above). A detachable discontinuity has no longer something to do with in spite of if function is defined at a particular element, yet has each and every thing to do with in spite of if shrink exists at that element. because of the fact shrink of f(x) as x procedures 0 from the left = a million/2 and shrink of f(x) as x procedures 0 from the main suitable = 4 then shrink of f(x) as x procedures 0 would not exist. consequently discontinuity at x = 0 is non-detachable

## Answers & Comments

## Verified answer

∫(x = 0 to 2) f(x) dx

= ∫(x = 0 to 1) f(x) dx + ∫(x = 1 to 2) f(x) dx

= ∫(x = 0 to 1) -1 dx + ∫(x = 1 to 2) 2 dx

= (-1)(1 - 0) + 2(2 - 1)

= 1.

I hope this helps!

x = ?2 ----> vertical asymptote (non-detachable) x = 0 ------> bounce discontinuity (non-detachable) x = 2 ------> detachable x = 4 ------> non-detachable notice that there discontinuity at x = 0 is non-detachable even although f(x) is defined at x = 0 ----> f(0) = 4 (as stated in an answer above). A detachable discontinuity has no longer something to do with in spite of if function is defined at a particular element, yet has each and every thing to do with in spite of if shrink exists at that element. because of the fact shrink of f(x) as x procedures 0 from the left = a million/2 and shrink of f(x) as x procedures 0 from the main suitable = 4 then shrink of f(x) as x procedures 0 would not exist. consequently discontinuity at x = 0 is non-detachable