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As we know, if $f(x)$ is integrable on $[a,b]$, then the integral upper limit function $\int_{a}^{x} f(t)dt$ is continuous.

Given a integrable piecewise function, why is the upper limit function discontinuous? $$ f(x) = \begin{cases} x^2, & x \in[0,1) \\ x+1, & x \in[1,2] \end{cases} $$ Obviously, the upper limit function is not continuous.

gt6989b
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fin3574
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    But shouldn't this integral upper limit function be continuous? – fin3574 Mar 04 '20 at 19:39
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    Here is a related discussion on another question – gt6989b Mar 04 '20 at 19:43
  • Why do you think it's obvious the upper limit function is not continuous? The function $f(x)$ is, but if you check its integral, due to the possible need to break it up into separate parts, since it's basically just adding small amounts of "area" (including possibly negative area, but not in this case), you'll find it's actually continuous, even across the boundary at $x = 1$. – John Omielan Mar 04 '20 at 19:43
  • $\int_{0}^{x}t^2dt=\frac{1}{3}x^3,x\in[0,1)$ ,and$\int_{0}^{x}f(t)dt=1/2 x^2+x-\frac{7}{6},x\in[1,2]$ ，at x=1 the integral upper limit function is not continuous. – fin3574 Mar 04 '20 at 19:59
  • @fin3574 Thanks for explaining in your comment. Since you didn't use @ with my name after it, I didn't get notified you had responded. You have the values defined correctly. However, it is continuous since, although the expressions are different, their values at $x = 1$ match. I've explained this in more detail in my answer. – John Omielan Mar 04 '20 at 23:02

1 Answers1

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You have the piecewise function

$$f(x) = \begin{cases} x^2, & x \in[0,1) \\ x+1, & x \in[1,2] \end{cases} \tag{1}\label{eq1A}$$

Define for $x \in [0,2]$ what you call the "upper limit" function of

$$g(x) = \int_{0}^{x}f(t)dt \tag{2}\label{eq2A}$$

As you stated in your comment, for $x \in [0,1)$, you have

$$g(x) = \int_{0}^{x}t^2 dt = \frac{x^3}{3} \tag{3}\label{eq3A}$$

For $x \ge 1$, though, the integration involves using the $x^2$ for $f(x)$ given in \eqref{eq1A}, namely the first part for up to $1$, and then the second case of $x + 1$ for any part greater than $1$. Thus, for $x \in [1,2)$, you have

$$\begin{equation}\begin{aligned} g(x) & = \int_{0}^{1}t^2 dt + \int_{1}^{x}(t+1) dt \\ & = \left. \frac{t^3}{3} \right\rvert_{0}^{1} + \left. \left(\frac{t^2}{2} + t\right) \right\rvert_{1}^{x} \\ & = \frac{1}{3} + \left(\frac{x^2}{2} + x\right) - \left(\frac{1}{2} + 1\right) \\ & = \frac{x^2}{2} + x - \frac{7}{6} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

This matches the expression you gave in your comment. Note $g(x)$ is continuous on each sub-interval where it's defined. Also, from \eqref{eq3A}, you have

$$\lim_{x \to 1^{-1}}g(x) = \frac{1}{3} \tag{5}\label{eq5A}$$

In addition, from \eqref{eq4A}, you have

$$\lim_{x \to 1^{+}}g(x) = g(1) = \frac{1}{2} + 1 - \frac{7}{6} = \frac{1}{3} \tag{6}\label{eq6A}$$

This shows, by the definition of continuity, that $g(x)$ is continuous at $x = 1$ and, thus, for all $x \in [0,2]$.

In effect, $g(x)$ has become a piecewise function that is continuous at it's point of change, i.e., $x = 1$. If you were to graph it, you would see the $2$ parts would join at $x = 1$, but there would be a "kink" there. This is because for $x \lt 1$, you get $g'(x) = x^2$, so $\lim_{x \to 1^{-}}g'(x) = 1$, but for $x \gt 1$, you have $g'(x) = x + 1$, so $\lim_{x \to 1^{+}}g'(x) = 2$. Thus, the curve's slope makes a sudden change at $x = 1$. This shows that $g(x)$ is not differentiable at $x = 1$.

Another way to look at this is that, for $x \ge 1$, you are taking the cumulative "area" defined by the integral up to $x = 1$, and then adding to it the part which is greater than $1$, so there's a continuous change of the "area". In general, integration tends to have a "smoothing" effect on functions, such as this example shows where the function you are integrating is a discontinuous piecewise function but its integral is continuous.

John Omielan
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