Integrating a step function using antiderivativesLet f ( x ) = { 1 if 0 ≤ x ≤ 1 2 if 1 &lt; x ≤ 2 .Then ∫ 0 2 f ( x ) d x = 3 and an anti derivative of f(x) is F ( x ) = { x if 0 ≤ x ≤ 1 2 x if 1 &lt; x ≤ 2 But, F ( 2 ) − F ( 0 ) = 4 − 0 = 4 ≠ 3.. Why has this happened?Find an anti-derivative G(x) of f(x) such that G ( 2 ) − G ( 0 ) = 3, the correct answer.My attempt: F ( 2 ) − F ( 0 ) = 4 ≠ 3 because we need to break it up into two parts like this: ∫ 1 2 2 x d x + ∫ 0 1 x d x = x 2 | 0 2 + 1 2 x 2 | 0 1 = 4 − 1 + 1 2 − 0 = 3.5 but this is not right either.For the "Find an anti-derivative G(x)..."I'm completely lost but perhaps if we solve the first part then I'll understand what it's asking for there.

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Integrating a step function using antiderivativesLet   f  (  x  )  =      {                                                                      1                                                             if                                 0                ≤                x                ≤                1                                                                    2                                                             if                                 1                &amp;lt;                x                ≤                2                                                            .Then       ∫    0    2    f  (  x  )  d  x  =  3 and an anti derivative of f(x) is  F  (  x  )  =      {                                                                      x                                                             if                                 0                ≤                x                ≤                1                                                                    2                x                                                             if                                 1                &amp;lt;                x                ≤                2                                                            But,   F  (  2  )  −  F  (  0  )  =  4  −  0  =  4  ≠  3.. Why has this happened?Find an anti-derivative G(x) of f(x) such that   G  (  2  )  −  G  (  0  )  =  3, the correct answer.My attempt:   F  (  2  )  −  F  (  0  )  =  4  ≠  3 because we need to break it up into two parts like this:       ∫    1    2    2  x  d  x  +      ∫    0    1    x  d  x  =      x    2              |        0    2    +      1    2        x    2              |        0    1    =  4  −  1  +      1    2    −  0  =  3.5 but this is not right either.For the &quot;Find an anti-derivative G(x)...&quot;I&#039;m completely lost but perhaps if we solve the first part then I&#039;ll understand what it&#039;s asking for there.

candydulce168nlid · Accepted Answer

Step 1Your F is not really an antiderivative, because we don&#039;t have       F    ′    (  x  )  =  f  (  x  ) everywhere - in fact F′(1) doesn&#039;t exist at all!Even worse, F is not even an indefinite integral, because it has a jump discontinuity at 1.If you add an appropriate constant to one of the two cases in the definition of F, you can get rid of the jump discontinuity, and then it will actually be an indefinite integral (but still not an antiderivative) - if we define &quot;indefinite integral&quot; to mean a function that allows us to compute definite integrals by the       ∫    a    b    f  (  x  )    d  x  =  F  (  b  )  −  F  (  a  ) rule. (On the other hand, it seems to be more common to define &quot;indefinite integral&quot; simply as a synonym for &quot;antiderivative&quot;, and then getting rid of the jump doesn&#039;t produce one, of course).Step 2In fact f can&#039;t have any antiderivative because derivatives always satisfy the intermediate value property (by Darboux&#039;s theorem), but f doesn&#039;t do that.

Jayla Faulkner · Accepted Answer

Step 1Using the antiderivative to compute an integral is the (second) Fundamental Theorem of Calculus. Said theorem requires the function f which you want to integrate to be continuous on the whole interval (where you want to integrate). Your f is not continuous in [0,2] and hence the theorem doesn&#039;t work. In fact, f cannot have an antiderivative at point   x  =  1 because f has a simple discontinuity at point   x  =  1.We can find a function G such that       G    ′    =  f in [0,1] (considering left hand derivatives at   x  =  1) and       G    ′    =  f in [1,2] considering right hand derivatives at   x  =  1, and such that G computes the area below f in the interval [0,x]Step 2When considering the second half [1,2] , you must not forget the area below f in [0,1]. With this in mind put:  G  (  x  )  =      {                            x                                                    if                     x          ∈          [          0          ,          1          ]                                      1          +          2          (          x          −          1          )                                if                     x          ∈          [          1          ,          2          ]                        Where the   x  −  1 comes from considering the rectangle whose base is the segment [1,x]  for   x  &amp;gt;  1.Then   G  (  2  )  −  G  (  0  )  =  1  +  2  (  2  −  1  )  −  0  =  3  =      ∫          0              2        f  (  x  )  d  x, as desired.

Integrating a step function using antiderivatives Let f ( x ) = { <mt

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