Question

a function f is defined on [0,1] by f(x)=2rx, where 1/(r+1)<x<1/r then show that f€R[0,1] and it's integral is π^2/6​

Answer 1

Answer:

2/3

Step-by-step explanation:

Step 1: A mathematical phrase, rule, or law that establishes the link between an independent variable and a dependent variable (the dependent variable). In mathematics, functions exist everywhere, and they are crucial for constructing physical links in the sciences.

Step 2: A function is a type of rule that produces one output for a single input. Source of the image: Alex Federspiel. This is illustrated by the equation $y=x^2$. Any input for x results in a single output for y. Considering that x is the input value, we would state that y is a function of x.

Step 3:

$\begin{aligned}& \int_0^1 f(x) d x=\sum_{r=1}^{\infty} \int_{2^{-T}}^{2^{-(r-1)}} \frac{1}{2^{r-1}} d x \\& =\sum_{\mathrm{r}=1}^{\infty} \frac{1}{2^{\mathrm{r}-1}}\left[2^{-(\mathrm{r}-1)}-2^{-\mathrm{T}}\right] \\& =\sum_1^{\infty} 2^{-2(\mathrm{r}-1)}-\sum_1^{\infty} 2^{-2 \mathrm{r}+1} \\& =\left(2^2-2\right) \sum_1^{\infty} 2^{-2 \mathrm{r}}=2 \cdot \frac{1}{4} \cdot \frac{1}{1-1 / 4}=\frac{2}{3} .\end{aligned}$

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