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Little earlier laid out an example of which can be found at: Explaining why $\Vert x_n\Vert _1=\frac{3}{4}$

I do not understand why: $$\int_{b-\frac{1}{n}}^{b-\frac{1}{2n}}\text{affin}=\text {area of triangle}$$

I hope someone will help me understand. Previously, thank you

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user145717
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  • What is the the geometric interpretation of the integral of a positive function on $[a,b]$? – Hamou Aug 19 '14 at 21:33
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    No idea what you mean by "affin". – Thomas Andrews Aug 19 '14 at 21:33
  • @Hamou: Very good to meet you, is it possible to tell the details of the task you have previously solved, for which thank you very much – user145717 Aug 19 '14 at 21:34
  • Sir @ Thomas Andrews: I've submitted a few example first, which you can find on but http://math.stackexchange.com/questions/903018/explaining-why-vert-x-n-vert-1-frac34 but do not understand why it is so choice – user145717 Aug 19 '14 at 21:36
  • I've used this formula just to simplify calculus. – Hamou Aug 19 '14 at 21:58

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Let $f$ affine positive in $[a,b]$ with $f(a)=0$ and $f(b)=k$. Then $f$ wrote $f(x)=\alpha x+\beta$, as $f(a)=0$ and $f(b)=k$ we obtain $\alpha=\dfrac{k}{b-a}$ and $\beta=\dfrac{-ak}{b-a}$, hence $f(x)=\frac{k}{b-a}x-\frac{ak}{b-a}$.
$\int_a^bf(x)dx=\dfrac{1}{2}\alpha(b^2-a^2)+\beta(b-a)=\frac{1}{2}k(b+a)-ka=\frac{1}{2}k(b-a)$.
Consider the three points in the plan $A(a,f(a))$, $B(b,0)$ and $C(b,f(b))$, the area of the triangle ABC: $\text{area}(ABC)=\dfrac{1}{2}AB\times BC=\frac{1}{2}k(b-a)=\int_a^bf(x)dx$.
You can see this on a drawing.

Hamou
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