Today I learned about something known as the king's property which really helps in solving integrals and I wanted to know why does this property work. I dont know if this terminology is used elsewhere , so what im talking about is this property $$ \int_a^b f(x)dx = \int_a^b f(a+b-x)dx $$
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14$\begingroup$ It's called "change of variable" $x \mapsto a + b - x$ ... $\endgroup$– WhatsUpCommented Oct 8, 2020 at 15:45
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4$\begingroup$ Wouldn't this change of variables swap the limits of integration? NVM, when you take the differential, a minus sign appears so we can swap the limits back to make the minus sign positive. $\endgroup$– Nicholas RobertsCommented Oct 8, 2020 at 15:46
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$\begingroup$ This is a simple change of variable as others have indicated, but it is often useful to exploit the symmetries of a function to compute or transform an integral. $\endgroup$– Mark BennetCommented Oct 8, 2020 at 15:56
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1$\begingroup$ If you have a question about terminology, it is helpful to give more details of where the terminology was encountered. Your brief post only hints at what Readers might guess to be a calculus class of some kind. Better would be to identify a textbook or other course materials that use this terminology. If your interest is in understanding the idea, as opposed to the naming convention, focus on a particular problem would be useful. $\endgroup$– hardmathCommented Oct 9, 2020 at 18:59
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2$\begingroup$ Does anyone know who gave this rule $\endgroup$– Qui Gonn JinnCommented Nov 13, 2021 at 10:21
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4 Answers
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This property is essentially stating that it does not matter whether we integrate from left to right or from right to left.
One way of seeing why this must be the case is considering an interval partition $\mathcal{P}$ of $[a,b]$. For example, let's suppose that the partition contains the intervals separated by the points $a, c_1, ..., c_n, b$.
Then suppose that we apply to this partition the function $f(x) = a + b - x$. The result is another partition $\mathcal{P}'$ separated by the points $f(b), f(c_n), ..., f(c_1), f(a)$ - which corresponds to $a, a+b-c_n, ..., a+b-c_1, b$.
Hence, $\mathcal{P}'$ is another valid partition of $[a,b]$! Therefore the limits of the inferior and superior partition sums induced by both partitions must be equal - ie the equality you wrote.
answered Oct 8, 2020 at 15:59
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$\begingroup$ A very nice intuitive explanation! $\endgroup$ Commented Oct 8, 2020 at 16:02
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1$\begingroup$ Equivalently, if we horizontally flip the area under the curve, the area is unchanged. $\endgroup$– J.G.Commented Oct 8, 2020 at 16:16
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1$\begingroup$ Wow that was a nice explanation, thanks sir $\endgroup$ Commented Oct 8, 2020 at 16:31
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If $x=a+b-y$ then $$\int_{x=a}^{x=b}f(x)dx=\int_{y=b}^{y=a}f(a+b-y)d(a+b-y)=$$ $$=\int_{y=b}^{y=a}f(a+b-y)(-dy)=$$ $$=-\int_{y=b}^{y=a}f(a+b-y)dy=$$ $$=\int_{y=a}^{y=b}f(a+b-y)dy.$$ In the last integral we can replace each "$y$" with an "$x$" to get $\int_{x=a}^{x=b}f(a+b-x)dx.$ This is a different and separate change-of-variable, and is $not$ the earlier "$x=a+b-y$".
answered Oct 8, 2020 at 16:15
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Try the change of variables $u(x) = a+b-x$. Then $du=-dx$. Also, our limits of integration change accordingly: $u(a)=a+b-a = b$ and $u(b) = a+b-b = a$. So we get:
$$\int_{u(a)}^{u(b)}f(u)du = -\int_b^af(a+b-x)dx$$
However, we can interchange the limits of integration which creates an additional minus sign in front of the integral and thus getting rid of the minus sign that is there currently:
$$\int_a^bf(a+b-x)dx $$
answered Oct 8, 2020 at 15:55
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I'm late, other answers have already showed that this particular change of variable works. But, if you really want to see why, I think you could observe that you are integrating over the same interval two functions that are mirrored across the axis through the midpoint of that interval, $x_m=\frac{a+b}{2}$:
In fact, the new function, when evaluated at a certain distance $d$ from the mean of the interval's extremes, has the same value of the original function evaluated at distance $-d$:
$$\bigg[f(a+b-x)\bigg]_{x=\frac{a+b}{2}+d}=f\left(a+b-\frac{a+b}{2}-d\right)=f\left(\frac{a+b}{2}-d\right)=\bigg[f(x)\bigg]_{x=\frac{a+b}{2}-d}$$
