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I have always had a mental block towards this property and would be truly grateful if someone would please help me.

$$\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$$

Consider $$f(x) = x, for x\in [-3,1), f(x) = \dfrac{1}{x} for x \in [1,3]$$ Evaluate $$\int_{-3}^3 f(x)dx$$

Now, this is the standard method that I have come across: to break up the integral over intervals of integration from -3 to 1 and from 1 to 3. ie $$\int_{-3}^3 f(x)dx = \int_{-3}^1 xdx + \int_1^3 \dfrac{1}{x}dx$$

In my opinion, the first integral of RHS should be $$\lim_{\beta \to 1} \int_{-3}^\beta xdx$$

This is because f(x) = x ends at the point 'closest' to 1. At x = 1, the function changes to $$\dfrac{1}{x}$$ Thus, instead of 1 being the upper limit of the interval of integration, the limit has to be taken for the upper limit of the interval of integration.

Could someone please explain why we take 1 as the upper limit of the interval of integration, instead of taking the limit? Thanks a lot!

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Note that the given function $f$ is actually continuous, as the left- and right-hand limits of $f(x)$ at $x=1$ are equal to $1$. Thus, you can also say that $f(x) = x$ for $x$ in the closed interval $[-3, 1]$.

This wouldn't actually be a problem even in the case of a jump discontinuity in your piecewise function. If you dive down into working with the definition of the Riemann integral, it's not too hard to prove that changing the value of a function at finitely many points will not change the integral. Thus, you can freely alter the values at the endpoints without changing the value of the integral.

A different way of looking at it is that the function $$ F(\beta) = \int_{-3}^\beta x\,dx $$ is a differentiable and therefore continuous function due to one of the Fundamental Theorems of Calculus. This continuity means that $\displaystyle\lim_{\beta \to 1} F(\beta)$ will always converge to $F(1)$.

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  • $\begingroup$ Sir, firstly thanks very, very much for your help. I am indeed grateful to you.Sir, I tried to search for proofs of the fact that changing the value of a function at finitely many points will not change the integral. Unfortunately, I was unable to find any such proof. Since I have just begun learning Calculus, proving this seems beyond my scope. Sir, would it be possible for you to please prove this for me, or show how this could be proved? Please Sir? $\endgroup$ – Ishan Apr 28 '15 at 7:58
  • $\begingroup$ @Better World Informally, changing the value of a function at $k$ points can only change $2k$ of the terms in a Riemann sum (the point might be an endpoint). Since both functions are bounded, this alteration in the Riemann sum's value can be made arbitrarily small by choosing a fine enough partition. For a more formal argument, there is Proposition 1.35 in these course notes $\endgroup$ – Rolf Hoyer Apr 28 '15 at 8:30
  • $\begingroup$ Sir, if f(x) is continuous over [0,1), can we have the following integral?$$\int_0^1 f(x)dx$$ $\endgroup$ – Ishan Apr 28 '15 at 10:28
  • $\begingroup$ @BetterWorld You need to assume that $f(x)$ is bounded unless you're willing to deal with improper integrals (which may or may not converge in this case). With the bounded assumption the integral will exist, although I think the proof is at least a little bit involved. $\endgroup$ – Rolf Hoyer Apr 28 '15 at 10:38
  • $\begingroup$ Sorry Sir, I had forgotten to add that $f(x)$ was bounded. I'm sorry Sir, but I have another set of doubts. Sir let $g(x) = x$ for $x\in[0,50]$. This is bounded. However, $g(x)$ is discontinuous at $x = 3, 6, 9$ (at these points g(x) = 20). Then Sir, how will we evaluate $$\int_0^{10} g(x)dx $$ Will we break it up into $$\int_0^3 g(x)ds +\int_3^6 g(x)dx+\int_6^9g(x)dx+\int_9^{10}g(x)dx$$ Sir why can we not integrate this directly:$$\int_0^{10} g(x)dx$$ $\endgroup$ – Ishan Apr 28 '15 at 11:53

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