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> I have the following equation:
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> $$2\int\_{0}^{1/n}(1-nx)^{2}dx=\frac{2}{n}\int\_{0}^{1}(1-x)^{2}dx.$$
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My question is: Is this a rule and where does it come from and if so when are you allowed to use it?

What I have done so far:

I calculated both the integrals by hand and from that I did see they are indeed equal but I am interested in the rule that is being used to see this immediately. I found:

$$2\int\_{0}^{1/n}(1-nx)^{2}dx=2\int\_{0}^{1/n}(1-2nx+n^{2}x^{2})dx=2\left[ x-nx^{2}+\frac{1}{3}n^{2}x^{3} \right]\_{0}^{1/n}= \\ 2\left[\left( \frac{1}{n}-n\frac{1}{n^{2}}+\frac{1}{3}n^{2}\frac{1}{n^{3}}\right)-0\right]=\frac{2}{3n},$$

and

$$\frac{2}{n}\int\_{0}^{1}(1-x)^{2}dx=\frac{2}{n}\int\_{0}^{1}(1-2x+x^{2})dx=\frac{2}{n}\left[ x-x^{2}+\frac{1}{3}x^{3} \right]\_{0}^{1}= \\ \frac{2}{n}\left[\left( 1-1+\frac{1}{3}\right)-0\right]=\frac{2}{3n}.$$

Maybe another example would also be nice to see how this rule works.