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- - **What can be said about an integral that is only continuous from one side**
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What can be said about an integral that is only continuous from one side1 Attachment(s) I need help figuring out what is happening when it is only continuous from one side or not continuous. |

From the description it looks like f(1/2) is the only value that matters. If it is continuous at that point, then the integral = f(1/2). If f is discontinuous then the integral is the average of the left and right values. |

Really? What examples did you have in mind? |

I would say that $$\int_0^1 f(x) \, \mathrm dx = \lim_{t \to \frac12^-} \int_0^t f(x) \, \mathrm dx + \lim_{t \to \frac12^+} \int_t^1 f(x) \, \mathrm dx$$ as long as both limits on the right hand side exist and are finite. In this case we get $$\int_0^1 f(x) \, \mathrm dx = \cancelto{0}{ \lim_{t \to \frac12^-} \int_0^t 0 \, \mathrm dx } + \lim_{t \to \frac12^+} \int_t^1 g(x) \, \mathrm dx = \lim_{t \to \frac12^+} \int_t^1 g(x) \, \mathrm dx$$ I don't know what you mean by the integral being continuous from one side. But if the integrand $g(x)$ is continuous on $\frac12 \le x \le 1$, the limit is just the obvious integral $$\int_0^1 f(x) \, \mathrm dx = \int_{\frac12}^1 g(x) \, \mathrm dx = G(1) - G(\tfrac12) $$ where $G(x)$ is a primitive (anti-derivative) of $g(x)$. If $g(x)$ is continuous everywhere (in the interval of integration) except at $x=\frac12$, then we get $$\int_0^1 f(x) \, \mathrm dx = \lim_{t \to \frac12^+} \int_{t}^1 g(x) \, \mathrm dx = \lim_{t \to \frac12^+} \big( G(1) - G(t) \big) = G(1) - \lim_{t \to \frac12^+} G(t)$$ as long as the limit exists and is finite. |

The integration is with respect to g (it's$\displaystyle \int_0^1\! fdg$), so the integration limits are values of g. |

Oh yes. I remember reading that a couple of days ago. |

Using g as defined it looks like you have a Stieltjes integral, where physicists would have an integral with a delta function at 1/2. |

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