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I was doing a question that involved integration of a piecewise function.

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The question, as can be seen above, asks to evaluate the integral from 0 to pi. I worked it out by breaking the function into 2 integrals where the first had lower and upper limits of 0 and pi/2 respectively and adding it to the integral with lower and upper limits of pi/2 and pi respectively.

Now, my question is, how can we use FTC 2 on the first integral if it is not continuous over the domain [0, pi/2] as the pi/2 is not included in the sin (x).

I have tried my best to articulate my question but if there is any confusion, please do let me know.

Is there something I am missing? Can someone kindly clarify.

Thank you in advance.

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2 Answers 2

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You can still evaluate the integrals individually and then add them together. There is a jump discontinuity at $\frac{\pi}{2}$ due to the interval given in the piecewise, but this point will not impact the sum. That is because the thickness of a single point is intuitively $0$.

Consider the point $c$ (Note: c is finite):

$\int_c^{c} f(x) dx= F(c)-F(c)=0$ by the FTC.

Additionally, note that the function is integrable from that interval.

$\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 0-(-1) = 1$

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  • $\begingroup$ Hi, yes, I can see why. However, does this discontinuity affect our ability to use FTC 2 as it requires continuity over [a, b]? $\endgroup$
    – Oofy2000
    Aug 14 at 12:30
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    $\begingroup$ It does not because the sum of one point is, intuitively, $0$. In other words, a point has no thickness/width to sum. $\endgroup$ Aug 14 at 12:34
  • $\begingroup$ Oh ok. Thank you. So I would be correct in saying that in a case like this, there is some flexibility to the theorem? $\endgroup$
    – Oofy2000
    Aug 14 at 12:37
  • $\begingroup$ @Oofy2000 No because $\sin(x)$ is integrable/well defined/continuous from $0 \le x \lt \frac{\pi}{2}$ $\endgroup$ Aug 14 at 12:41
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    $\begingroup$ @Oofy2000 Added one. Here is a rigorous explanation, if interested: math.stackexchange.com/questions/511197/area-under-a-point $\endgroup$ Aug 14 at 13:13
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First prove this, from the definition of the integral:

Prop. Suppose $p\in[a,b]$, and $f:[a,b]\to\Bbb R$ satisfies $f(x)=0$ for all $x\in[a,b]\setminus\{p\}$. Then $f$ is integrable on $[a,b]$ and $\int_a^bf=0$.

Hence, even though $f(x)-\sin(x)$ does not quite vanish at every point, it follows that $$\int_0^{\pi/2}(f(x)-\sin(x))\,dx=0,$$hence $$\int_0^{\pi/2}f(x)\,dx=\int_0^{\pi/2}\sin(x)\,dx.$$

(Technicality: Since $\sin(x)$ and $f(x)-\sin(x)$ are both integrable on $[0,\pi/2]$, so is $f$.

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